WEBVTT mathematics/linear-algebra/hovasapian
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Welcome back to Educator.com and welcome back to linear algebra.
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We have been investigating the structure of vector spaces and subspaces recently getting a little bit more deeper into it to understand what it is that is actually going on in the vector space.
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Today we are going to continue to talk about that of course, and we are going to talk about homogeneous systems.
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As you gleaned already, homogeneous systems are profoundly important, not only in linear algebra, but they end up being of central importance in the theory of differential equations.
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In fact, at the end of this lesson, I am going to take a little bit of a digression to talk about the nature of solutions in the field of differential equations as related to linear algebra.
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Linear algebra and differential equations, they are very, very closely tied together. So, it is very difficult to separate one from the other.
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Let us go ahead and get started.
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Okay, so, we know that the null space of... so what we have is this homogeneous system, for example, ax = 0, where a is some m by n matrix.
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This is just the matrix form of the normal linear system that we are used to, in this case a homogeneous system because everything on the right hand side is 0.
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x is, again, is a vector, so it is a n by m, n by 1 vector that is a solution to this particular equation.
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Now, we know that given this, we know that the null space, or the space of all solutions to this equation.
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In other words, all of the vectors that satisfy, all of the vectors x that satisfy this equation is a subspace of RN.
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Well, very important question, a very important problem is finding the basis for this subspace.
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If you recall what a basis is, a basis is a set of vectors that actually spans the entire space that we are dealing with and is also linearly independent.
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So, you remember, you might have a series of vectors that spans a space, but it might not be linearly independent.
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Or, you might have some vectors that are linearly independent, but they might not span the space.
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A basis is something that satisfies both of those properties.
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Again, it spans the space that we are talking about, and the vectors in it are linearly independent.
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In this case, the space that we are talking about is the null space. The space of solutions to the homogeneous system ax = 0.
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So, the procedure for finding a basis for the null space of ax = 0, where is a n by m, of course I will not go ahead and mention that.
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The first thing we do, well, we solve ax = 0 by Gauss Jordan elimination to reduced row echelon form, as always.
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Now, if there are no arbitrary constants, in other words, if there are no columns that have no leading entries, then, the null space equals the set 0 vector.
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What that means is that there is no basis, no null space. There is no null space, essentially -- well, there is, it is the 0 vector, but there is no basis for it.
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In other words, there is no collection of vectors. Okay.
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The dimension of the null space, which we called the nullity if you remember, equals 0.
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Okay. Our second possibility is if arbitrary constants do exist after we reduce it to reduced row echelon form.
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What that means is that if there are columns that do not have leading entries, those are... the x... the values corresponding to those columns, let us say it is the third and fifth column, so x3 and x5, I can give them any value I want. That is what the arbitrary constant means.
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So, if arbitrary constants exist, then, write the solution x = c1x1 + c2x2 + ... + ckxk, however many of these vectors and constants there are.
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Well, once you do that, the set s, which consists of this x1, this x2, all the way up to xk... x1, x2... xk... is a basis for the null space.
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Let me write space over here. So, again, what we do when we want to find the basis of a null space of a homogeneous system.
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We solve the homogeneous system with reduced row echelon form.
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We check to see if there are no columns that do not have a leading entry, meaning if all of the columns have a leading entry, there are no arbitrary constants.
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Our null space is the 0 vector. It has no basis, and the dimension of the null space, the nullity in other words equals 0.
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Let me go ahead and put nullity here. Nullity is the dimension of the null space. In other words it is the number of vectors that span that space, it is 0.
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If arbitrary constants do exist, meaning if there are columns that do not have a leading entry, then we can read off the solution for that homogeneous system.
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We can write it this way, and the vectors that we get end up being our basis.
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Let us do an example, and as always, it will make sense, hopefully.
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Let us see, so our example... find, not find the... find a basis.
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So, there is usually more than one basis. Find a basis. Basis is not necessarily unique.
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Find a basis for, and the nullity of the null space for the following system.
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(1,1,4,1,2), (0,1,2,1,1), this is a one, sorry about that, it looks like a 7, (0,0,0,1,2), (1,-1,0,0,2), and (2,1,6,0,1).
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x1, x2, x3, x4, x5 = (0,0,0,0,0).
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This is our homogeneous system. We are looking for this right here.
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We are looking for all vectors, x1, x2, x... all vectors whose components x1, x2, x3, x4, x5... that is what we are looking for.
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It is a solution space. So, we are going to try to find the solution, and when we do, we are going to try to write it as a linear combination of certain number of vectors.
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Those certain number of vectors are going to be our basis for that solution space.
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Okay. So, let us go ahead and just resubmit it to reduced row echelon form.
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What we end up getting is the following... (1,0,2,1,0), (0,1,2,0 -- oops, I forgot a number here.
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(1,0,2... let me go back... what I am doing here is... I am going to take the augmented matrix.
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I am going to take this matrix and I am going to augment it with this one, so let me actually rewrite the whole thing.
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It is (1,1,4,1,2,0), (0,1,2,1,1,0), (0,0,0,1,2,0), (1,-1,0,0,2,0), (2,1,6,0,1,0). Okay.
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So, this is the matrix that we want to submit to reduced row echelon form.
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I apologize, I ended up doing just the matrix a.
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When we submit it to reduced row echelon form using our software, we get the following.
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(1,0,2,0,1,0), (0,1,2,0,-1,0), (0,0,0,1,2,0), (0,0,0,0,0,0), and of course the last... not of course... but... there is no way of knowing ahead of time.
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What we end up with is something like this. So, now let us check to see which columns actually do not have leading entries.
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This first column does, this second column does, this third column does not... so, third.
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The fourth column has a leading entry, it is right there... the fifth column... okay.
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So the third and the fifth columns, they are arbitrary constants.
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In other words, these of course correspond to x1, x2, x3, x4, and x5.
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So, x3 and x5, I can let them be anything that I want.
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I am going to give them arbitrary parameters. So, let me go ahead and write x3... I am going to call x3 R.
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It just stands for any number.
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And... x5 is going to be S, any parameter. Columns with non-leading entries.
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Now, when I solve for, like for example if I take x1, it is going to end up being the following.
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x1 = -2R - S, and here is why... that row... well this is x1 + 2x3 + x5.
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So, x1 = -2x3 - x5.
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Well, -2x3 -x5 is -R - S, that is where this x1 comes from.
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I am just solving now for the x1, x2 and x4. That is where I am going to get the following equations.
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Then, I am going to rewrite these in such a way that I can turn them into a vector that I can read right off.
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So, x2 here is equal to -2R + S, because here, this is 2R over here, this is -S, it becomes +S when I move it over.
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x5, our last one, is equal to -2S, and here is the reason why: because it is -- I am sorry, this is x4 not x5, x5 is S.
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x4 + 2s = 0, therefore x = -2s.
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So, this is our solution. Notice our infinite number of solutions, now we are going to rewrite this in order... x1, 2, 3, 4, 5 as opposed to how I wrote it here, which was just reading it off, make it as simple as possible.
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So, let me move forward. What I end up with is x1 = -2R - S, x2 = -2R + S, x3 = R, x4 = -2S, and x5 = S.
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So, now x1, 2, 3, 4, 5... this is the vector that I wanted. This is the arrangement.
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Well, take a look. R and S are my parameters. This is equivalent to my following.
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I can write this as this vector, which is just x1 x2 in vector form... is equal to... I pull out an R and this becomes (-2, -2, 1, 0, 0).
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That is what this column right here is... -2R, -2R, R, 0, R, 0, R.
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Then this one right here, I pull out an S... +S × (-1,1,0,-2,1). There you go.
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This particular homogeneous system has the following solution, as it is some arbitrary number × this vector + some arbitrary number × this vector.
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Well, these are just constants. Any sort of number that I can imagine from the real numbers... any number at all, so this is a linear combination.
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Therefore, that vector and that vector form a basis for our solution space.
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Therefore, we have our basis is the set (-2,-2,1,0,0), and (-1,1,0,2,1).
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This is the basis of the null space. Well, the dimension is, well how many are there... there are 2.
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So, the null space has a dimension 2. The nullity is 2.
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So, notice what it is that we actually had. We had the system of 1, 2, 3, 4, 5... 1, 2, 3, 4, 5, we had this 5 by 5 system.
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So, R5 all the way around. Well, within that 5 dimensional space, 2 of those dimensions are occupied by solutions to the homogeneous system.
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That is what is going on here. If you think about it, what you have is any time you have 2 vectors, you have essentially a plane, what we call a hyper-plane.
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Because, you know, we are talking about, you know, a 5-dimensional space.
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But, that is all that is going on here. So, we solve a homogeneous system, we reduce the solution to this, and because it is a linear combination of vectors, the 2 vectors that we get actually form a basis for our solution space.
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We could how many vectors we get to and that is the dimension of our null space, which is a subspace of all of it.
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Let us do another example. Let us take the system 2, let us go (1,0,2), (2,1,3), (3,1,2) × x1, x2, x3 = (0,0,0).
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We set up the augmented matrix for the homogeneous system which is (1,0,2,0, (2,1,3,0), (3,1,2,0), just to let you know, that is what the augment is.
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We subject it to reduced row echelon form, and we get the following... We get (1,0,0,0), (0,1,0,0), (0,0,1,0).
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There are no columns here that do not have leading entries. Basically what this is saying is that x1 is 0, x2 is 0, x3 is 0. That is our solution.
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We have x1 = 0, x2 = 0, x3 = 0, all of these are equivalent the vector = 0 vector.
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When we have the 0 vector as the solution space, only the trivial solution, we have no basis for this system.
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The nullity is 0. The dimension of the solution space is 0. The only solution to this is the 0 vector.
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Okay. Now, let us go ahead and talk... take a little bit of a digression and talk about the relation between the homogeneous and non-homogeneous systems.
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So, we have been dealing with homogeneous systems, but as we know there is also the associated non-homogeneous system. Some vector v.
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So, that is the non-homogeneous and this is the homogeneous version of it.
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0 on the right, or some b on the right. There is a relationship that exists between the two. This relationship becomes profoundly important not only for linear algebra, but especially for differential equations.
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Because, often times, we will have a particular solution to a differential equation, but we want to know some other solution.
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But, maybe we cannot actually solve the equation. Believe it or not, the situation comes up all the time.
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As it turns out, we do not have to solve the equation. We can solve a simpler equation. The homogeneous version, which is actually quite easy to solve for differential equations, and here is the relationship.
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Okay. If, some vector x < font size="-6" > b < /font > is a solution, some solution that I just happen to have, to the system ax = b, and x₀ is a solution to the associated homogeneous system, okay?
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Then, if I add these two, x < font size="-6" > b < /font > and x₀, then the some of those 2 as a solution is also the solution. I should say also.
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Is also a solution to ax = b. So, if I happen to have the non-homogeneous system and if I happen to know some solution for it, and if I happen to know some solution to the homogeneous system, which is usually pretty easy to find, then I can add those 2 solutions.
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That is a third solution, or that is a second solution to ax = b.
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That is actually kind of extraordinary. There is no reason for it to actually be that way, and yet there it is.
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What is more, every solution to the non-homogeneous system, ax = b, every single solution can be written as a particular solution, which we called x < font size="-6" > b < /font > + something from the null space.
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Symbolically, x < font size="-6" > b < /font > + x₀, what this says is that when I have the system ax = b, and I happen to know a solution to it and I also happen to know the solution to the homogeneous system, all the solutions.
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In other words, I happen to know the null space every single solution to the non-homogeneous system consists of some particular solution to the non-homogeneous system plus something from the null space.
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So, if I have, I mean obviously I am going to have maybe an infinite number of solutions to the null space, all I have to do is take any one of those solutions from the null space, add it to the particular solution that I have for the non-homogeneous, and I have my collection of every single solution to the non-homogeneous system.
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That is amazing. What makes this most amazing is that usually when we are dealing with a non-homogeneous system, more often than not, you can usually just guess a solution.
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We can check it, and if it works, that is a good particular solution.
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Then, what we do is instead of solving that equation, we actually end up solving the homogeneous system, which is pretty easy to solve.
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Then we just take any solution we want from the homogeneous system, add it to the solution that we guessed, and we have any solution that we want for the particular problem at hand.
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That is extraordinary. It is a technique that you as engineers and physicists will use all the time.
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Very, very important for differential equations. Okay.
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We will go ahead and stop it there today. Thank you for joining us at educator.com, we will see you next time.