WEBVTT mathematics/differential-equations/murray
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Hi and welcome back to the differential equations lectures here on www.educator.com, my name is Will Murray and we are going to be studying systems of linear differential equations.
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Before we jump into that, I think it would be good if we did a quick lecture on review of linear algebra because you really do have to remember your linear algebra concepts in order to solve these systems later on.
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In this lecture we are not really going to be talking about differential equations per se but we are just going to review some linear algebra concepts especially eigenvalues and eigenvectors.
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You will be using those to solve differential equations to subsystems of differential equations later on, let us jump right in here.
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Let me remind you first of all how you multiply matrices, I'm not really going to teach this to you from scratch, I assume you done a little bit of linear algebra before but this will be just a quick review of it is been a while since you done it.
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The idea here is we will start within by m x n matrix, what that means is m rows and n columns, this would be m and then the number of columns here is n and you can multiply that by an m x p matrix.
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Here the number of rows is n, n rows and p will be the number of columns and your answer will be in m x p matrix, you will get m rows and p columns, the way you want to think about that is you can multiply (m x n) x( n x p).
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The middle two numbers always have to match this n always has to match, and then in the answers those cancel each other away and you always get an m x p matrix.
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Maybe that is easier to think about if we actually put some numbers here, what I got with this example is this first matrix is a 2 x 4 matrix, the m is 2 rows and n is 4, 4columns. The second matrix is 4 x 3, n is 4 rows, p is 3 columns.
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When we multiply a 2 x 4 matrix with a 4 x 2 matrix, a 4 x 3 matrix, that is legit because the 4's matchup, you always have to have those inner two numbers match up.
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You multiply 2 x 4 x 4 x 3 matrix and you think of those 4's were cancelling each other out and the answer will be a 2 x 3 matrix, the m here is 2 rows and the p here is 3 columns, that is how you check out the sizes.
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The way you actually multiply them is you, to find an entry in the answer, you look at the corresponding row of the first matrix and the corresponding column of the second matrix.
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Here to find this entry, if this row were A, B, C, D and maybe then this row is E, F, G, H and this column were X, Y, Z, W, the way you get that entry first row times first column is you multiply A, B, C, D times X, Y, Z, W.
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This would be AX + PY + CZ +DW and you go through and you find each entry in the answer matrix the same way, for example the way you find this entry well that is the second row.
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I look at the second row and that is the second column, I look to the second column of the second matrix, second row of the of the first matrix, second column of the second matrix, maybe that is R, S, T, U.
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I'm multiplying E, F, G, H times R, S, T, U, to get that entry I would do ER + FS + GT + HU, I'm getting that from this row and this column and you go through and you find each entry in the answer matrix that way.
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We will do an example, if you are a little rusty on this, I think it will make a lot more sense when we do the example on actual numbers, let me go ahead and remind you some other things we are going to be studying.
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You can take determinants as follows, the determinant of a 2 x 2 matrix is really easy, you just cross multiply positive in that direction and cross multiply negative in that direction, you will get A x D down the main diagonal - B x C down the back diagonal.
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That is the determinant of a 2 x 2 matrix, determinant of a 3 x 3 matrix kind works the same way but it is a lot more complicated, again you try to go down the diagonals here, A, E, I and then you want to go B, F, we are running off the matrix.
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What you do is you copy the first two columns on the other side, think of it as wrapping around to the other side, that first column is A, D, G, I will copy that over here, ADG and that second column is B, E, H.
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Let me switch colors here so that you will be able to see what I'm doing, I got A, E, I positive, B, F, G positive and C, D, H positive, that is why I got the AEI, BFG, CDH positive from those three positive diagonals.
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Then I go down the negative diagonals CEG that is a negative, AFH that is a negative, and BDI that is a negative, if you look at whether were negative terms here that is AFH that comes from this one right here.
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The BDI comes from this one right here, that is a negative and CEG, it is a little hard to read now but it is coming from these terms right here, that is a negative.
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It is a little complicated to take a 3 x 3 determinant and there are other ways to do it, expansion all on row or column, you might have learned that in your linear algebra class and that is legitimate.
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For me the easiest way is just to remember those three main diagonals and three back diagonals, main diagonals are all positive, back diagonals were all negative.
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The next concepts that we need to remind you about for linear algebra is eigenvalues and eigenvectors, the idea here is that A would be a square matrix, let me draw a little picture of this A would be a square matrix.
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V will be a column vector an n x 1 column vector, that is vector b and r is just a number, r is just a scaler and then V is still the same column vector.
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If you multiply an n x n matrix times an n x 1 vector, remember how those middle numbers cancel and you get an n x 1 vector, if A x V is the same as what you would have gotten just by multiplying V by a scaler.
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In that special situation that is the definition of an eigenvector, V is the eigenvector and R is the corresponding eigenvalue, that is what an eigenvector and eigenvalue mean.
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It is just means that A times that vector is equal to a scaler times that vector, this definition of eigenvectors does not really show you how to find them, let me go ahead and show you how to find an eigenvalue and eigenvector.
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The way do it is you look at the equation A - r(i)=0, remember (i) is the identity matrix, that means you look at the matrix A and you subtract an R down the main diagonal.
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You subtract off R's down the main diagonal and then you take the determinant of that matrix and set it equal to 0, what you will get is what is called the characteristic polynomial in R.
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This characteristic polynomial will just be a polynomial equation the you will solve and you will get one or more values for R, you will get a couple values for R and those values are the eigenvalues of the matrix.
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For each one of those eigenvalues you got a find the associated eigenvector, you take that R, you plug it back into A - R(i) and you try to solve the generic vector A - R(i) times V is equal to the 0 vector.
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Then you try to find those corresponding eigenvectors V, that is the way that works but I think it will make a lot more sense if we do some examples of eigenvalues and eigenvectors.
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One thing to notice, I will mention it here although we will see an example of it later is we might find more than one eigenvector, we might find multiple eigenvectors and we will see an example of this.
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You will see what that looks like, you might find multiple eigenvectors corresponding to a single eigenvalue, you will see what that looks like as we do some examples later on.
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I think it is time we jumped into some examples, first example we are just going to practice multiplying matrices, we got this matrix has 2 rows and 4 columns, that is a 2 x 4 matrix, two rows and four columns,
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This other matrix has 4 rows and 3 columns, that is a 4 x 3 matrix, that means it is legitimate to multiply these 2 matrices, a 2 x 4 times of 4 x 3, you have to think of the 4's as cancelling, that is going to give us a 2 x 3 matrix.
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I'm going to set up a 2 x 3 matrix as answer here, there is going to be two rows and three columns in my answer matrix and to get each entry, what I do is I multiply a row of the first matrix by column of the second matrix.
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For that first entry, the first row, first column, I have 1, 2, 0, I give myself a little more space here and then times 2, 2, 3, 0 and I'm going to add those up.
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1× 2 is 2, 2× 2 is 4, 2 + 4 is 6 and if I add those up I get a 6, that is why I put a 6 in that first entry there.
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In that next entry, I got still the first row, the second column, first row time second column again 1, 2, 0, -1 times the second column is 0, 7, 1 and -4 and then I add them up.
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I see I got 0, 4, 14, 0 and 4, that is 18 and means that second entry is in 18, for the last entry on the first row I look at the first row of the first matrix times the third column of the second matrix.
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It is always rows of the first matrix times columns of the second matrix, 1, 2, 0, -1 times 5, -3, × 2 and 1 and add them up and I get 5 - 6 -1, that is a negative, no -1-1 so that is -2, 6, 18, -2 in that entry .
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To find the second row, I take the second row of the first matrix and then I multiply that by each column of the second matrix, 3, -2, 0, 4 times 2, 2, 3, 0, add those up and I get the 6, -4, I get 2 for that first entry.
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Now bottom column, middle row, 3, -2, 0, 4, 0, 7, 1, -4, add those up I see I got 0, -14, 0, -16, -14 + -16=-30.
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Finally bottom row and the right hand column is 5, -3, 2, 1 add those up and I see I got 15, + 6, + 4, 15 + 6=21 + 4=25, I fill that one in there and that is my product matrix 6, 18, -2 and 2, -30, 25, that is my answer.
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Let me remind you how we did that, we start out with a 2 x 4 matrix, 2 rows and 4 columns multiply it by a 4 x 3 matrix, in order to be able to multiply them, those numbers in the middle have to match up.
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Essentially that is saying that the length of these rows is equal to the length of these columns and they have to be equal so that you can match them up term by term.
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Those 4's match up and in the answer matrix you drop the 4's out and you make it 2 x 3, that is why I set this matrix two rows and three columns and to get each entry in the answer matrix.
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You look at the corresponding row of the first matrix multiply by the corresponding column of the second matrix.
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For example this 18, that is the first row second column, I go first row that is this one right here 1, 2, 0 -1 and second column 0, 7, 1, -4, I did 1, 2, 0 -1 x 0, 7, 1, -4 multiply them together and add them up that is where that 18 comes from.
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In each other entry comes the same way where you look at the corresponding row of the first matrix, column of the second matrix, multiply them together and get your answer.
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The second example we got a system of differential equations, we are not actually trying to solve it but we are trying to write it as a system of equations, we are trying to write it in matrix multiplication form.
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Let me show you how we can work that out, if you look at this middle column for example or middle row 3x + 2i + 12z + W, 3x + 2y + 12z + + w.
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Let me show you how you can write that in matrix multiplication form as the row 3, 2, 12, 1 multiplied by the column x, y, z, w because if we multiply that out term by term we get 3X +12 + 2y + 12z + w.
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What we are going to do is make all the coefficients of each of these equations rows of a matrix, we are going to multiply it by the column XYZW, let me write that out.
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I see in this first row here, I got an x, the coefficient is 1, I will think of that as 1x + 0y + 2z + 0w, 1, 0, 2, 0 and in the the second row I got 3, 2, 12, 1 reading the coefficients, third row is 1, 1, 5, 1, take that matrix and I multiply by the column XYZW.
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It might look like the sizes do not matchup there because the column is longer than the coefficient matrix but let me show you how they do, this coefficient matrix is 3 rows 4 columns that is a 3 x 4 matrix.
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This column vector is 4 rows, 1 column, that is a 4 x 1 matrix and the important thing to multiply matrices is that those internal numbers match up, that 4 matches up with a 4, that 4 does really does match up.
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We can multiply a 3 x 4 x 4 x 1 because those inside numbers match and our answer matrix is going to be the outside numbers 3 x 1 which means it is the three rows one column.
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A single column with three rows and I can see on the right here that I got my answers are 5, 16, 5, what I have done here is I have given you a matrix form for that system of equations.
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If you expanded out this matrix multiplication then it would turn into nothing more nothing less than these three equations here, let me remind you how that happened.
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For each of these equations I want to write it as a row times a column, what I was doing right here was illustrating with the second equation 3X + 2y + 12Z + w, we can write that as the row 3, 2, 12, 1 times the column XYZW.
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If we do that for every row, think of this first row as having a 0y and 0w, the coefficients are 1, 0, 2, 0, second row 3,2 ,12, 1, third row 1, 1, 5, 1 multiplied by the column X,Y, Z, W.
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We got a 3 x 4 matrix that can multiply by 4 x 1 matrix, think of the 4's cancelling and your answer will be a 3 x 1 matrix and those are the answers that we get over on the right hand side, the 5, 16, 5.
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In example 3, we are going to do something kind is similar to the previous example where we are going to write a system of differential equations in matrix multiplication form, here is the system we are given, x1′=2x1(t) - x2(t).
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A similar equation for x2′, the first thing I'm going to do is just not write the t's at least the parentheses t's, I'm going to write this as x1′=2x1 - x2 + e⁺t.
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I'm not changing anything there, I'm not cancelling the t's, we are going to remember that those t's are still in there, I'm just trying to write it a little more simply.
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Then x2′=2x1 + x2 + sin(t) and I'm going to focus on this part right here because I can write that as a matrix multiplication, the same way we did in the previous example, in example 2.
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You might want to go back and check on example 2 if you do not really remember how that one worked out, but I can write this as matrix multiplication if I put a column x1 x2, then my coefficients are 2, -1, 2 and 1.
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I'm looking over here at the coefficients 2, -1, 2, 1 and if I multiply out that matrix multiplication, I would see I had 2, -1 times x1 x2, that would give me 2x1 - x2.
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I also want to add on a column for the e⁺t and sin(t), on the left I see I have x1′, x2′, now we are going to introduce a little notation here, we use bold x, x with a bar over it to mean the column vector x1 x2.
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If we do that then we will call X′ would just be the derivatives of each of those, x1′ x2′ and that lets me rewrite this equation on the left I have X vector X′ is equal to 2, -1, 2, 1 times vector X.
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Because we have x1 x2 + e⁺t and sin(t) and from there I have a nice matrix multiplication form of my differential equations, let me remind you how we arrived at that.
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First we suppressed each of these t's, it is not that they are not there anymore, we are just not going to write them anymore, we are not cancelling them, we are not going to write them anymore.
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We got x1′ is equal to 2x1 minus x2 + e⁺t, x2′ is 2x1 + x2 + sin(t) and then we are going to write this in vector form where we write these coefficients as a matrix.
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This is just like the previous example, example 2, if this step looks a little strange to you, sorting this out into matrix form, you might want go back and look at example 2 again, it might make this make a little more sense.
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We wrote x1 x2 as a vector, wrote the coefficients 2, -1 and 2, 1, separate out the e⁺t and sin(t) as its own vector and then we have X1′ and x2′, we are going to call X vector x a shorthand for x1 x2.
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Vector X a shorthand for x1′ x2′, that means on the left we have vector X′, still that same coefficient matrix 2, -1 and 2,1 and then vector x e⁺t sin(t) .
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That is the vector form for this system of differential equations, what we are going to do in the next set of lectures here on www.educator.com is we are going to start solving these things.
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We are going to figure out how to solve this system of differential equations using eigenvalues and eigenvectors.
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We are not quite there yet, if you are excited to jump forward and figure out how to solve these, you might want to skip to the next lecture which is on how to solve systems of differential equations.
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What we are going to do now is do a little more practice finding eigenvalues and eigenvectors and just make sure that we are up to speed on that.
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In example 4, we are going to find the eigenvalues and eigenvectors of the matrix 2, 3, -1 and 6, and remember the way do that is you look at the determinant of A - R(i) and you set it equal to 0.
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Of course R(i), (i) is just the matrix 1, 0, 0, 1 that is the identity matrix and R(i) is just R 0, 0, R, effectively what you are doing here is you are subtracting R's off the main diagonals.
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A- R(i) is equal to 2-R(3) - 1 6 - R, we are just subtracting R's off the main diagonals and the determinant of that, you cross multiply, let me do that in another color so it does not get too confusing.
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We cross multiply positive in the forward direction, negative in the backwards direction and we get 2- R x 6 - R - 3, it is +3, and we set that equal to 0.
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I'm going to expand that out, I see I'm going to have a +R² -6r -2r, that is -8r and then +12 + 6 × 2+ there is another 3 there,+15 is equal to zero.
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If I factor that I can see the that factors into r -3 and r -5, my eigenvalues there are 3 and 5, those are both eigenvalues of this matrix, this is common by the way, you expect a matrix to have more than one eigenvalues in general.
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Let us see how we find the corresponding eigenvectors, what you do is you plug A- R(i) back into the matrix for each value of R and then you solve for A- R(i)V is equal to zero.
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When r=3, A- R(i) we will do 2-3, 3 -1, 6-3 and that gives us -1, 3, -1, 6 - 3 is 3 and we are going to try to solve A- R(i) x V is equal to 0.
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If you remember from your linear algebra class, what you have to do now is to row reduce the matrix, I can see right away that this bottom rows can go to zero because it is the exact same as the top row.
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I have zeros in the bottom row, I can subtract top row from the bottom one, maybe I will multiply the top row by -1 to make a little more cleaner, 1 -3 and since I'm trying to solve A- R(i) x V=0.
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I'm going to write that as x x y =0 and if I expand that out I see I got x -3y is equal to 0, the way solve this in your linear algebra class is through the use of free parameters, I'm going to say y=t.
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I got x - 3t=0, t is a free parameter, X is equal to 3t and xy as a vector is equal to t and 3t, I mixed those up because the x is 3t and y was t and we could factor out t and we just get a 3(1) there.
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This will work for any value of t, any value of t will give us an eigenvector, but to make it as easy as possible, let us just take T is equal to one, my eigenvector for r=3 is 3, 1.
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I'm going to do the same thing for R=5, again I'm going to plug in A- R(i), I'm going to subtract 5 off the main diagonal from the original matrix 2 -5 is -3, 3, -1, 6 - 5 is 1.
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I'm going to do row reduction on this matrix, think the easiest things to switch the rows and multiply the bottom row by -1 as I switch, that will give me 1 and -1, you notice that the top row was just a multiple of the bottom row.
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I'm going to drop that right out, I'm doing several steps of row reduction in my head here, if you are little rusty on row reduction, you might want go back and check the www.educator.com lectures on linear algebra.
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You will see lots of practice on row reduction, I'm doing this a little fast here and I'm also going to multiply this times XY and get zero and that tells me the x - y is equal to zero.
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If I use a free parameter y=t, X would have to equal y, X would be t as well and XY would be t times 1, 1 and again we can just take T to be 1, our xy would be, our eigenvector would be 1,.
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Let me summarize here, the eigenvalues are 3 and 5, each one has a corresponding eigenvector, the vector for 3 was 3, 1 and the vector for 5 was 1, 1.
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Let me summarize this and teach you a little trick that sometimes can be helpful here, what we did here was we started out by working out the determinant of A- R(i) which means you subtract an R off the main diagonal.
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That is what we did here, that is what this - R and that -R came from, take the determinant means cross multiply, you will get 2-r x 6-r -3, that negative and that negative cancel each other out and gave us a +3 there.
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Expanded that out, we got a polynomial which factored pretty easily into 3 and 5, and then for each one of those we plug those back in as the -R and we got matrix and then we reduce that matrix using our row operations for linear algebra.
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You might want to review your row operations for linear algebra if you are having trouble reducing a matrix and then we used free parameters to set up a system of equations y=t, x would have to be 3t and then we get xy is t x 3(1).
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We can just take t=1 to be 3 x 1, now let me show you a little bit of a shortcut you could have done there, what we are looking for is a vector XY, when we multiply 1 A-3 times it will get zero.
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Sometimes what you can do is a shortcut here is you just put 1 in for the bottom component y, and then you see what the top component would have to be in order to make it come out to be zero when you multiply it by this vector.
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Here I'm looking at one times something -3 8 1=0 and I see that it is something would have to be a 3 and my eigenvector is 3,1.
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That is a little trick the you can sometimes employ, it means you put one in for the bottom component and then you just figure out what the top component would have to be in order for it to come out to be zero when you multiply it by this row here.
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Let us try that out with r=5, if we look at A- R(i) subtracting R off the main diagonal we get -3 here and 1 here, the other entries are the same and that reduces our row operations to 1, -1.
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If we just take that y entry to be 1 and I think what would that X entry have to be in order to get 0 when we multiply this row by it, I see that if that x were 1, then 1× 1 - 1× 1 would give us 0.
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That would be a little shortcut to getting the eigenvector as 1-1, I might use that shortcut in some of our future examples but the summary of all this is that we have to eigenvalues, 3 and 5.
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Each one corresponds one eigenvector, the 3 corresponds the 3-1 and the 5 corresponds to 1-1, on our next example we got a 3 x 3 matrix, -1, -2, -2, 1, 2, 1 and -1, -1, 0.
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We want to find the eigenvalues and eigenvectors of this matrix, this one is going to be quite a bit messier algebraically than the previous example but the principles are still exactly the same.
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Let me walk you through that, remember we want to figure out A- R(i), now R(i) means you would you have R's down the main diagonal, what we are going to do is we are going to subtract R's from the main diagonal.
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I'm going to go ahead and write those in, A- R(i) on the original matrix will be -1 - R 2 - R and 0 - R and now we got take the determinant of that matrix, I'm going to highlight things in greens so it does not get too messy here.
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But it is going to be a little bit complicated because remember for determinant of a 3 x 3 matrix, you got 3-4 diagonals and then 3 back diagonals, it does get a little messy we have plus on each of the four diagonals, minus on each of the back diagonals.
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Let us go ahead and figure each one of those out, the first main diagonal is -1 - r x 2 - r x 0 - r which is just -r and now the next diagonal is -2 x 1 x -1 that gives us -2 x 1 x -1.
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I'm wrapping around the other side, maybe I will go ahead and write those columns to make that a little more obvious, on that first column I got -1 - r, 1 and -1.
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On the second column I got -2, 2 - r and -1 and if I look that second column, that second diagonal here that is + 2, that gives me a 2 there and I will put a + 2 here.
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That third column or that third diagonal gives me a +2, I'm going to work back diagonals, that back diagonal right there is negative and I had two other negative, the whole thing is negative.
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-2× 2 - r, that is that first back diagonal, that second back diagonal again remember that each one of these is negative, or at least we are going to multiply by -1 and then see what it comes out to be.
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That second back diagonal is -1 - r x -1 and then there is another negative it is plus -1 -r and that third back diagonal is -2x 1 x -r , there are 2 negatives in there and then there is another negative because it is a back diagonal.
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The whole thing is negative, -2 x r because there is a negative, there is a negative and there is a negative, you put them all together you get three negatives and it simplifies down to a single -2r.
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We get a fairly complicated expression here, let us expand things out as best we can, this -R I'm just going to write it as -r here and I'm going to multiply these two terms together.
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I see I got -r, ir is r² -2r + r is -r and -1 x 2 is -2, let me collect some terms here, I see I have a +4 that is +2 + 2 - 2 x 2 so -4 + 2r and then -1 -r -2r and again if I multiply this -r through.
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I see I have -r³ + r² + 2r and let me collect some terms, the +4 -4 cancels, I see I got a -1 and 2r - 2r cancel, but I still have a -r here and if I simplify that all down, I got -r³ + r² + 2r - r is +r -1.
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Remember you want to set that equal to 0, I'm going to find the determinant of A -R(i) =0, I'm going to set this equal to zero and then I'm going to multiply both sides by -1, I do not really like having a negative sign on the leading term there.
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I'm going to go r³ - r² - r + 1, I got that by multiplying both sides by -1 there and that is equal to 0, now I'm going to solve this by grouping, I have to somehow factor cubic polynomial and it is usually useful to do that by grouping.
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I'm going to write this as r³ - r² and then there is this r-1 is equal to zero, in fact that is r² x r - 1, -r -1=0 and I get (r² - 1) x r -1 factoring out r -1 is equal to zero.
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But r² - 1, I know I can factor that into r -1 x r +1 and when I finally factor this all the way down, I see I have two roots of r=1 and one root of r= -1, those are my eigenvalues for my matrix 1 and -1.
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Now I still need to go through and find the corresponding eigenvectors of the matrix and I need to find eigenvectors for each one of those, for r=1 and for r= -1.
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I'm going to start with r=1 actually I think I'm going to start with r= -1 and I'm going to plug that back in to the matrix, I'm going to plug that back into the -r for each one of these.
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Since r is -1 that means I'm adding one to each one of those entries, let me write that out by adding 1 to each one of those entries -1+ 1 that will give you 0 and I have -2, -2 1 now I have a main diagonal 2 + 1 will give you 3 and 1, -1, 0 + 1 gives me 1.
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I want to find the eigenvectors corresponding to this matrix which means I have to do row operations on this matrix, I think what I'm going to do is I'm going to bring this third row up to the top and then I'm going to use that to clear out the other rows.
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I will put 1, 3, 1 on the top row, switch that down to 0, -2, -2 and I'm going to use that third row that 1, 3, 1 row to clear out this -1, I'm going to add one copy of the middle row to the bottom row.
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These are elementary row operations that we learned how to do in linear algebra, if you are a little rusty on this, I got some lectures on linear algebra here on www.educator.com.
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You want to check those out and just remind yourself how to do elementary row operation.
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If I add this 1, 3, 1 to -1, -1, 1 what I'm going to get is 0, 2, 2, that is already looking a little better, I created 0's in my first column, take that one more step, I'm going to get 1, 3, 1 in the first row.
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In that second row I can divide by -2, I will get 0, 1, 1 and then I can use that 1-1 to clear that bottom row, I could subtract it completely and just get 0, 0.
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I think I'm going to take it one more step, I'm going to use this 1 to clear away this 3, I'm going to subtract 3 copies of the second row from the first row.
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That will give me 1, 0 and I subtracted 3 copies so that is 1, 0, -2, 0, 1, 1 and 0, 0, 0, I want to see what x, y, z would have to be in order to give me 0, 0, 0 there.
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If I look at that, I see that I have several equations here, from the first one I got x + 0y - 2z is = 0, from the second equation I got 0x + y + z=0, I'm going to build my eigenvector and remember how I said in the previous example.
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You can sometimes start by just taking the last entry to be one, and that works this time, my z is equal to 1 then y + z=0, y would have to be -1 and x - 2z =0, my x would have to be 2.
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I'm going to fill that in, 2 and then my y is -1, and my z is 1, if you are uncomfortable with that, if you are not quite sure how I did that, what you can do is you can solve this by free parameters.
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You just set z to be an arbitrary free parameter t and let me go back wards and say since y + z =0, you will get y=-t, since x - 2z=0 you will get x = 2t and you will get x, y, z = 2t/t and you can factor out the t.
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You will just get 2, -1, 1 x t, if you take t=1, you will end up with the same eigenvector that we found by my little trick of just taking the last entry to be 1.
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What we got so far is we found the eigenvalues -1 and 1, we found an eigenvector corresponding to one of the eigenvalues for -1, we still need to find an eigenvector corresponding to the other eigenvalue r=1.
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Let me recap what we did here on this side before we move on and find the eigenvectors corresponding to the other eigenvalue.
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I will go over this in red here because it looks like I already used up the blue and the green pastel pen. What we first did was we looked at A-R(i), that means we subtracted and r on the main diagonal here.
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And we found the determinant which meant this cross multiplying 3 main diagonals and 3 back diagonals, the main diagonals each one of them is going to be plus on the main diagonals, the 4 diagonals.
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When you do the back diagonals, each one is going to be negative, you are multiplying positively all these 4 diagonals, that is what we are doing here and then multiplying with an extra negative factor all these back diagonals.
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That gets to be messy and you have to be very careful not to make a mistake as you work through all those but it simplifies down, we worked out all the algebra here and simplify it all the way down to r³ + r² + r -1.
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We multiply everything here by -1, both sides by -1 in order to get a nice, positive coefficient on the r³, then we could group the terms and factor out r - 1 on each term.
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We factor out r - 1 by the whole thing, r² - 1 that factored down into r + 1 r- 1 and when we looked at the factors we got a double root at r=1 and a single root at r=-1, those are my 2 eigenvalues 1 and -1.
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I'm going to find eigenvectors for each one, r=-1 we plugged that back into the matrix which means we are subtracting -1 along the main diagonal which effectively means we are adding 1 on the main diagonals.
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That is where we got the 0, 3, 1 on the main diagonal, all the other entries are just the original matrix, we are row reducing this according to our rules for linear algebra and we row reduced it to reduced row echelon form.
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From there we are trying to solve for x, y, and z, you can use free parameters the same way you learned in linear algebra or you can use my little trick of starting with z=1 and then working backwards to figure out what x and y should be.
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Either way you end up with this as the eigenvector and remember that is just for the eigenvalue r=1, we still need to go back and find the eigenvector corresponding to the eigenvalue r=1, that is what we are going to do on the next slide.
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That should have been an r, our eigenvalues are r 1 and -1, we already found the eigenvectors corresponding to -1, that one is already done on the previous side, you can go back and check that out if you are just jumping to this part.
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We still need to find the eigenvectors corresponding to r=1, the way we do that is we plug in A-R(i), which means you subtract r on the main diagonal, I will do -1 there, -1 there, and -1 there.
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I will see how that works out, A-R(i) is equal to, now -1 minus -1 is -2, -2, -2, 1, 2-1 is 1, 1 there, -1, -1, 0-1 is -1.
00:51:09.000 --> 00:51:31.000
What I see is basically 3 copies of the same row, they are all multiples of 1, 1, 1, if I pull that 1, 1, 1 up to the top I can get 1,1, 1 in the top row and then I can use that and multiples of that row to kill all the other rows, send all the other rows to 0.
00:51:31.000 --> 00:51:48.000
I'm trying to find vectors x, y, z, that when I multiply that x, y, z I get 0, all that really tells me is that x + y + z=0, I really do not have very much information about x, y, and z.
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I'm going to use my free parameters to solve that, I'm going to set up z=t, I still do not have enough information to solve for y, I'm going to set up another free parameter s there.
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Now I can solve for x because I have x + y + z=0, that is telling me that x + s + t=0, my x would be -s -t.
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My x, y, z vector is equal to, my x was -s -t, my y was s and my z was t, what I like to do is factor that out and get an s part and a t part.
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I'm going to get an s vector and I'm going to get a t vector, I see the coefficients of s are -1, 1, and 0, the coefficients of t here are -1, and there is no t in the middle so 0, there is a 1t in the bottom.
00:53:06.000 --> 00:53:29.000
The vectors that I get are s x -1, 1, 0, and t x -1, 0, 1, my eigenvectors, looks like I have 2 independent eigenvectors, if I just take different combination of s and t.
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These are corresponding to the eigenvalue r=1, my eigenvectors are -1, 1, and 0, and -1, 0, and 1.
00:54:04.000 --> 00:54:15.000
I'm finally finished with finding eigenvalues and eigenvectors there, but let me recap how we worked all that out.
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The first thing we did in this problem was on the previous side, but we started out with A-R(i) and we took its determinant, that was a bit messy and we ended up subtracting r's on the main diagonal.
00:54:27.000 --> 00:54:42.000
And expanding out a 3 x 3 determinant which meant multiplying 3 things down the positive diagonal directions and then 3 negative diagonal directions.
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It is quite messy and we got a cubic polynomial and then we solve that out, we got r=1 or -1 and on the previous side we plugged r=1 back in and we found an eigenvector on the previous side.
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That one worked out ok, what we are doing here is we are finding our eigenvectors for r=1, we are plugging in r=1 into A-R(i) here, we subtract 1 down the main diagonal, that gives us this matrix.
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When you look at this matrix, we really have 3 copies of the same row because they are all multiples of 1, 1, 1, that means we can use the 1, 1, 1 row to wipe out the other 2 rows.
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We just end up with the equation x + y + z=0, that does not give us very much information about x, y, and z, we have to use free parameters to solve it.
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To start out, we always work backwards from z, I will start out with z = t, then I still cannot figure out what y is, I set y = s and then finally if I know what s and t are, I can solve for x in terms of t and s.
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x= -s -t, if I plug those back in, the x was -s -t, y was s, the z was , that is what I figured out over here, plugging those back in.
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Then I can factor out in to an s vector and t vector, I factor out the coefficients of s here and the coefficients of t here.
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Those are 2 independent eigenvectors corresponding to r=1, those are the 2 eigenvectors that I associate with the eigenvalue r=1.
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I found all the eigenvalues and eigenvectors of this matrix, that is the end of our review of linear algebra lecture here on the differential equation series on www.educator.com.
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If that was not enough linear algebra to get you completely comfortable with finding eigenvalues and eigenvectors, what you might want to do is go and look at the linear algebra lectures here on www.educator.com.
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You will find a lot more examples and a lot of this stuff work out in a lot more detail, what we are going to do in the differential equation lectures is we are going to start using some of this ideas to solve systems of differential equations.
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On the next lecture we will look at systems of differential equations, we will write them in matrix form and then we will use eigenvalues and eigenvectors to solve them.
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That is what we have to look forward to next on www.educator.com.
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My name is Will Murray and you have been watching the differential equations lectures. Thanks a lot for joining us, bye.