For more information, please see full course syllabus of Linear Algebra

For more information, please see full course syllabus of Linear Algebra

### Change of Basis & Transition Matrices

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Change of Basis & Transition Matrices 0:56
- Change of Basis & Transition Matrices
- Example 1
- Example 2
- Theorem
- Example 3: Part A
- Example 3: Part B

### Linear Algebra Online Course

### Transcription: Change of Basis & Transition Matrices

*Welcome back to Educator.com and welcome back to linear algebra.*0000

*In the previous lesson, we talked about the coordinates of a particular vector and we realized that if we had two different bases that the coordinate vector with respect to each of those bases is going to be different.*0004

*So, as it turns out, it is not all together... it has to be this or that.*0018

*One basis is as good as another. We are going to continue that discussion today, deal with coordinates some more and we are going to talk about something called a transition matrix.*0022

*Where, if we are given the coordinates... if we are given both bases and if we are given the coordinates with respect to one basis, can we actually transform that and is there a matrix that actually does that.*0032

*The answer is yes, there is a matrix. It is called the transition matrix from one basis to another, and it ends up being a profoundly important matrix.*0045

*So, let us just dive right in.*0053

*The first thing I want to talk about it just two brief properties of the coordinates that we mentioned.*0058

*Their properties are exactly the same as that of vectors, so, it is going to be nothing new.*0066

*It is just the notation is, of course, slightly different because we have that little subscript s and t underneath the coordinate vector.*0070

*So, let us just write it out and start with that.*0076

*So, we have v + w, so if I add two vectors and take the coordinate with respect to a certain basis, well, I can treat that, I can just sort of separate them.*0080

*That is just going to be the coordinate vector with respect to s for v, + the coordinate vector with respect to s for w.*0098

*Again, it is something that you already know. The sum of two vectors is nothing new here.*0106

*If I have a vector v and I multiply by a constant, and if I have the coordinate vector with respect to a certain basis s, well, I can just go ahead and pull that constant out and multiply by the coordinate vector for that vector first, and then multiply by the constant.*0112

*So, just to have these properties to describe what it is that we are going to do in a minute.*0130

*Okay. I am actually going to go through something that I would not normally go through.*0137

*It is the derivation of where this thing called a transition matrix comes from, simply because I want you to see it.*0141

*It is not going to be particularly notationally intensive, but there are going to be indices, you know, there are some numbers and letters, things floating around.*0148

*So, it is really, really important to pay attention to where things are and what each number is doing.*0155

*Okay. Let us say we have two bases for a vector space that we call v, of course.*0161

*The first basis is going to be s, it is going to consist of vectors v1, v2, and so on all the way to vN.*0180

*And... we have t, which is another basis.*0195

*We will call these w... w1, w2, all the way to wN, and again they are bases for the same vector space, so they have the same number of vectors in them. That is the dimension of the vector space.*0200

*Now, choose some v in v, some random in the vector space.*0215

*Well, we can write this particular v with respect to this basis, let us choose this basis, t.*0228

*So, we can write... we can say that v = c1 × w1, c2 × w2, we have done this a thousand times... + cN × wN.*0235

*Okay, just a linear combination of the vectors in this basis t. Well, once I actually solve for these constants, c1, c2 through cN... what I end up with is the coordinate vector v with respect to the basis t.*0252

*That is what this t down here means, which is c1, c2, all the way to cN. Okay.*0268

*Now. Here is where it gets kind of interesting. So, let us watch very, very carefully. Let me put a little arrow right here to show what we are doing.*0279

*If I take v, and if I want to find the coordinate vector with respect to the basis s, I am just going to take this thing that I wrote, which is the ... so the left side, I put it in this notation... sub s.*0290

*Well, the left side is equal to the right side. I just happen to have written the right side with respect to this basis.*0310

*I am just going to write... I am basically just copying it.*0317

*c1w1, c2w2 + ... + cNwN, with respect to s. All of them is take this thing, and subjected it to this notation. Everything should be okay.*0324

*Well, now I am going to use these properties. So, that is equal to c1 × w1, with respect to s + c2 × w2, with respect to s + so on + cN × wN, with respect to s.*0340

*Okay. Take a look at what we have done. A random vector v with respect to a basis t, and then, we want to find the coordinate vectors for... with respect to the basis s.*0368

*So, I have just taken this definition, and subjected it to the notation for the coordinate vector for s.*0384

*Then I use these properties up here, which you might call the linearity properties of these coordinate vectors, and just rewrite it.*0390

*Well, let us just see what this actually says... c1 × w1 with respect to s, w2 with respect to s... wN with respect to s.*0400

*Let us move forward, that is just this.*0409

*It says that coordinate vector v, with respect to s is equal to 1s, w2 respect to s, and so on and so forth.*0414

*I just set up, well let me actually finish up writing it and then I will tell you what it is we are doing here.*0434

*wN with respect to s, × c1, c2, all the way to cN.*0445

*So, the equation that I wrote on the previous slide is just this equation in matrix form.*0457

*What I am doing is I am taking the columns, I am taking each w1 in the basis t, I am expressing that vector with respect to the basis s, and whatever I get I am putting in as columns of my matrix.*0461

*Well, what this ends up being... this matrix that I get by doing that is precisely... well let me rewrite it.*0483

*So, we have the equation in front of us... p s... notice the arrow is going from right to left, not usually from left to right... × this thing, which is just coordinates of v with respect to t.*0496

*Okay, so what we have done, this is our ultimate goal. It is okay if you do not completely understand what it is that we did.*0520

*We will go through the procedure for how to find this matrix. This matrix right here, which is called the transition matrix from t to s. *0525

*There is a reason why I wrote it this way with the arrow going backwards from right to left. I will tell you what it is in a second.*0537

*It says that if I have a vector, and if I can find the coordinate vector with respect to a basis t, but I want to convert that to a coordinate vector with respect to the other basis that I have, s, I can multiply the coordinate vector with the one basis on the left by some matrix.*0543

*The transition matrix that takes it from t to s. This is why it is written this way. Notice this vector space -- I am sorry -- this coordinate vector with respect to s is on the left.*0566

*Here, the notation for the transition matrix has the s on the left, has the t on the right, because you are multiplying it by the coordinate vector for the basis t on the right.*0576

*It is just a way of... again, it is a notational device to remind us that we are going from the t basis to the s basis.*0588

*It is written this way simply because of how we wrote the equation. We wrote the coordinate vector with respect to s on the left side of the equality sign. That is why it is written this way.*0597

*Now, here is how we did it. We have the basis t which consists of vector w1, w2, and w3 and so on.*0609

*We take each of those vectors, we express them as coordinate vectors with respect to s, and we do that by solving this system, just like we did for the previous lesson.*0618

*Then, the coordinate vectors that we get, we just put them in columns and the final matrix that we get when we just put in all of the columns, that is our transition matrix.*0630

*Okay. Let us just do an example and I think it will all make sense.*0645

*Let us move forward here. This is going to be a bit of a long example notationally, but it should be reasonably straight forward.*0650

*Okay, now, let s = the set (2,0,1), (1,2,0), (1,1,1).*0659

*Okay. That is one basis for R3. Three vectors, three entries, it is R3.*0680

*T, let it be another set, let it be (6,3,3), (4,-1,3), (5,5,2).*0695

*So, we have two different bases. Okay, what we want to do is a, we want to compute the transition matrix.*0714

*What matrix will allow us to convert from t basis to s basis? The transition matrix from t to s, that is the first thing we want to do.*0725

*The second thing we want to do is we want to verify the equation that we just wrote. *0735

*That the coordinate with respect to basis s is equal to this transition matrix, multiplied by the coordinate for v with respect to t.*0740

*Okay. So, let us see what we have got here. Alright, so let us do the first thing first.*0755

*Let us go ahead and compute this transition matrix. So, we said that in order to compute the transition matrix, we have to take... so we are going from t to s. *0768

*That means we take the vectors in the basis t, (6,3,3) (4,-1,3), (5,5,2), and we express each of these vectors with respect to the basis s.*0779

*Again, these are just vectors in R3. They are random vectors, but they do form a basis and that forms a basis.*0791

*So, we want to change, we want to be able to write these vectors as a linear combination, each of these as a linear combination of these 3. That is what we are doing.*0796

*So, let us write that down. So, we have... we will take this one first, right?*0808

*Let us actually label these. No, that is okay, we do not need to label them.*0815

*So, we want (6,3,3) to equal some constant a1 × (2,0,1) + a2 × (1,2,0).*0824

*Actually, let us not, let us choose a different letter here. Let us choose b1, and we will make this c1 × (1,1,1).*0845

*So, this is one of the things that we want. We can solve this system. We can just solve this column, this column, this... let me write it on the other side.*0857

*We are accustomed to seeing it on the right, let us go ahead and be consistent... (6,3,3). That is one thing.*0865

*Okay. The other thing we want to do, is we want to express this one. The second vector in the basis t.*0876

*Again, t to s, so we want to take the vectors in t, the second vector expressed as a linear combination of these two.*0882

*So, this time we have a different set of constants, we will call them a2 × (2,0,1) + b2 × (1,2,0) + c2 × (1,1,1) = (4,-1,3).*0890

*Now we want to express the third vector in t as a linear combination of these.*0908

*So, we will take a3 ×, well, (2,0,1) + b3 × (1,2,0) + c3 × (1,1,1).*0914

*That is going to equal (5,5,2). So, we solve this system, we solve this system, we solve this system.*0932

*Well, this system, for each of these, the left hand side, these columns are the same, (2,0,1), (1,2,0), (1,1,1).*0941

*So, we can take all three of these and do them simultaneously. Here is what it looks like.*0948

*All we are doing is taking (2,0,1), (1,2,0), and (1,1,1) and then we are augmenting the (6,3,3).*0959

*(2,0,1), (1,2,0), (1,1,1), augmenting it with (4,-1,3).*0965

*And this one... (2,0,1), (1,2,0), (1,1,1)... augmenting it with (5,5,2).*0970

*Well, we can do all of the augmentations simultaneously. We can just add three columns and then do our matrix in reduced row echelon form. Here is what it looks like.*0974

*So, we get (2,0,1), (1,2,0), (1,1,1), and then we have our augmented... we have (6,3,3), (4,-1,3), and we have (5,5,2). Okay.*0985

*When we subject this to reduced row echelon form, let me go horizontally actually, we end up with the following.*1007

*We end up with (1,0,0), (0,1,0), (0,0,1), and we end up with (2,1,1), (2,-1,1), (1,2,1).*1016

*So, this, right here, in the red -- oh, I did not get red, oops -- right here, that is our transition matrix.*1035

*It is the columns of the vectors in t, expressed as... these are the coordinates of those vectors with respect to the s basis.*1046

*That is what we did, just like the previous lesson, so our transition matrix from t to s = (2,1,1), (2,-1,1), (1,2,1).*1058

*There you go. That is the first part. Okay.*1074

*Now, we want to confirm that that equation is actually true. In other words, we want to confirm this equation.*1077

*That the coordinate vector of some random vector v with respect to s is equal to this transition matrix that we just found × the coordinate vector with respect to the basis t.*1095

*Okay. Well, let us let v... well, let us choose a random vector. We will let v - (4,-9,5).*1105

*Okay, so now the first thing that we want to do is... again we are verifying so we are doing a left hand side, we are going to do a right-hand side.*1120

*We are verifying this. We need to check to see if this is actually equal. So, we need to do this side, and we need to do this side.*1128

*Okay. First of all, let us find the coordinate of this vector with respect to t, that is this right here. Okay.*1136

*Well, we need to set up the following. c1w1 + c2w2 + c3w3 = our (4,-9,5).*1148

*Well, let us take our columns, which are our basis t, so we get the following. We get (6,3,3), (4,-1,3), (5,5,2).*1171

*It is going to be (4,-9,5).*1187

*Convert to reduced row echelon form, we get (1,0,0), (0,1,0), (0,0,1), and we get (1,2,-2). Okay.*1194

*So, the coordinates of v with respect to the basis t is equal to (1,2,-2). That is part of the right hand side.*1209

*Well, we have the transition matrix, that is this, so let us circle what we have. We have that. that is our coordinate with respect to t.*1223

*We have our transition matrix, so we have the right-hand side. Now, we need to find the left hand side, do a multiplication, and see if they are actually equal to confirm that equation.*1232

*Okay. Now, let us move to the next page, so we want to find with respect to s. Well, with respect to s we set up the columns from the vectors in the basis s.*1241

*So, we get (2,0,1), (1,2,0), (1,1,1), and we are solving for (4,-9,5).*1261

*Reduced row echelon, when you do that, you end up with... I will actually write that out... and I put (4,-5,1)... so now we have the left hand side.*1276

*Now, what we want to do is we want to check, is (4,-5,1)... does it equal that transition matrix × the coordinate vector with respect to... yes, as it turns out, when I do the multiplication on the right hand side, I end up with (4,-5,1).*1293

*So, yes. It is verified.*1329

*So again, our equation is this. If I have some coordinate vector with respect to a basis t, and I want to find the coordinates with respect to another basis s, I multiply on the left with something called the transition matrix.*1333

*That will give me the coordinates with respect to s, and the columns of that transition matrix are the individual basis vectors for the basis t expressed as coordinate vectors with respect to the basis, s.*1353

*That is what this notation tells me. Okay.*1372

*Now, that is exactly what we did. We were given two bases, t and s, we took the basis, the vectors in the basis from t, we expressed them as coordinate vectors with respect to the basis s, and that which we got, we set up as columns in a matrix.*1379

*That matrix that we get is the transition matrix. That allows us to go from 1 basis to another, given one coordinate, or another.*1405

*Okay. Let us see. Let us continue with a theorem here.*1417

*s = v1, v2... vN. And, t = w1, w2... wN. Okay.*1431

*Let s and t be two bases for an n-dimensional vector space. Okay.*1456

*If p, from t to s, is the transition matrix... transition matrix from t to s... then, the inverse of that transition matrix from t to s is the transition matrix from s to t.*1471

*So, if I have 2 bases, and if I calculate a transition matrix from t to s, I can take that matrix, I can take the inverse of that matrix, and that is going to be the transition matrix from s to t.*1526

*So, I do not have to calculate it separately. I can if I want to, but really all I have to do is take the inverse of the matrix that I found.*1539

*That is the relationship between these two. Okay.*1546

*Also, the transition matrix which we found is non-singular.*1552

*Of course, invertible. Non-singular means invertible. Okay.*1566

*Okay. Let us see what we have got here.*1576

*So, let us continue with our previous example. Let us recall the conditions... we said that s is equal to (2,0,1), (1,2,0), (1,1,1).*1580

*And... t is equal to (6,3,3), (4,-1,3), and (5,5,2).*1605

*Okay. What we want to do is we want to compute the transition matrix from s to t, directly.*1618

*So, we can do it directly, and the other way we can do it is to take the inverse of the transition matrix from t to s that we already found. That is going to be the second part of this.*1632

*So, the first part a will be computed directly, and the second part, we want to show that this thing from s to t is actually equal to the inverse of the matrix from t to s.*1647

*Make sure you look at these very, very carefully to make sure you actually know which direction we are going in.*1661

*Well, in order to calculate it directly, we take the... so this, we are going from s to t, alright?*1666

*So, let us go with vectors in s... so we are going to write (6,3,3), so in other words, we are going to express... so this is from s to t.*1680

*So, we want to take the vectors in s and express them as a linear combination of these vectors.*1700

*These vectors are the ones that actually form the matrix over here... (6,3,3), (4,-1,3), (5,5,2), and we augment with the (2,0,1), (1,2,0), (1,1,1).*1706

*Again, we are going from s to t. We want to express the vectors in s as linear combinations of these. That is why s is on the augmented side, and t is over on this side. Okay?*1725

*These are three linear equations. This, this augment, that, that augment, this, this augment. Okay?*1738

*When we subject to reduce row echelon form, we end up with (1,0,0), (0,1,0), (0,0,1).*1749

*We end up with -- nope, we do not end up with stray lines -- we have (3/2,-1/2,-1,1/2,-1/2,0,-5/2,3/2,2) make these as clear as possible.*1760

*Therefore, q transition matrix from s to t is equal to (3/2,-1/2,-1,1/1,-1/2,0,-5/2,3/2,2).*1793

*Okay. So, that is the direct computation of u(s(t).*1819

*Now, let us go back to blue. Now, let us calculate the inverse to show that that equals the inverse of that. Yes.*1827

*Okay. So, now, let us see. We want to take, in order to find... so we have... let us recall what the transition matrix from t to s was.*1844

*We had (2,1,1), (2,-1,1), (1,2,1), okay? That was our transition matrix.*1859

*Now, you recall, when you actually, in order to find the inverse of the matrix, you set up this system... (2,1,1), (2,-1,-1), (1,2,1).*1881

*Then, you put the identity matrix here, (0,1,0), (1,0,0)... yes.*1898

*Then, of course, if you do reduced row echelon form, this right side ends up being the inverse of that.*1910

*In this particular case, we do not need to do that. If we say that one thing is the inverse of another, all that I have to really do is multiply them and see if I end up with the identity matrix.*1915

*So, part b, in order to confirm that that is the case, all I have to do is I have to take the transition matrix of t from s to t, and I multiply by the matrix from t to s, to see if I get the identity matrix.*1926

*While I do that, I have of course my (3/2,-1/2,-1,1/2,-1/2,0,-5/2,3/2,2).*1949

*Multiply that by our transition matrix (2,1,1), (2,-1,-1), (1,2,1), and I can only hope I have not messed up my minus signs or anything like that.*1970

*As it turns out, when I do this, I get (1,0,0), (0,1,0), (0,0,1), which is the identity matrix - n-dimensional, the 3 by 3 identity matrix, so this is not n, this is 3.*1982

*That confirms that q, s to t = inverse of t to s.*2000

*When I find a transition matrix from t to s, if I want the transition matrix from s to t, all I do is take the inverse. That is what we have done here.*2014

*Thank you for joining us Educator.com to discuss transition matrices, we will see you next time.*2023

1 answer

Last reply by: Professor Hovasapian

Wed Jan 27, 2016 4:15 PM

Post by Hen McGibbons on January 16 at 02:04:48 PM

where did you learn how to teach? i am a new tutor and I am trying to help my students as much as possible. do you follow the principles of any books or mentors? or did you develop your own teaching style?

1 answer

Last reply by: Professor Hovasapian

Sun Dec 7, 2014 6:45 PM

Post by Nkosi Melville on December 4, 2014

for example if i am trying to find a transition matrix corresponding to a change of basis from the standard

basis {e1, e2} to the ordered basis {u1, u2}

Would the vectors u1 and u2 be the matrix i turn into reduced row echelon form?

1 answer

Last reply by: Professor Hovasapian

Mon Apr 15, 2013 11:36 PM

Post by Matt C on April 14, 2013

At the end when you confirm problem the identity matrix is correct, but I think you wrote the identity matrix wrong at the 31:50 mark.

At the 31:50 mark, doesn't the right side have to be equal to the identity matrix ([1,0,0], [0,1,0], [0,0,1]) you have ([1,0,0], [0,1,0], [1, 0,0]). I wrote them as columns.