First of all, amazing video once again. They're helping me a lot. I study economics in Belgium, and we've been getting a lot of these problems (matrices, also integrals, geometry...) but I think our book (it's written by the uni itself) isn't sufficient. Is there anyone that could reccommend maybe a good, general book on mathematics on this level (college)?thank you!

try Linear Algebra and its applications by David C. Lays

I am unable to find vidios on diagonalization

It's not Khan, but here's a good one: https://youtu.be/13r9QY6cmjc

3 was a repeated root and the eigenspace resulted in a plane. Can the number of repeated roots be used to predict the dimension of the eigenspace?

If I recall, you can't use the number of repeated roots to find the dimension of the eigenspace, because it completely depends on the matrix A that you are finding eigenvalues for. Sometimes, after obtaining an eigenvalue of multiplicity >1, and then row reducing A-lambda(IdentityMatrix), the amount of free variables in that matrix matches the multiplicity. Other times, the amount of free variables is less than the multiplicity. All you can know, is that if an eigenvalue K has a multiplicity of n, then at most, the dimension of the eigenspace of the eigenvalue is n. If your dimensions of your eigenspaces match with your roots of the characteristic equation (including repeated ones), then you can diagonalize the matrix.

At 4:54, why v2 = a and v3 = b?

Sal is setting the free variables v2 and v3 to be arbitrary real numbers, so he can write v1, v2, and v3 as linear combinations of vectors and the arbitrary real numbers a, b.

In my book, the characteristic equation is A-lamda I, is that matters to switch the place of A and lamba I?

Good question, Both "A- lambda I=0" and "lambda I - A=0" work, in fact they are the same equation! To see why, look at the start of the video how Sal derives the formula. He starts with the definition of an eigenvalue, which is "Av=lambda v". He then subtracts one side of the equation from the other. It doesn't matter which side we subtract so we either end up with "Av- lambda I v=0" or "lambda I v - Av=0".

When is row reduce a matrix, it ends up being the identity matrix, hence the vector is [1,1,1] with no parameters.. Is this still considered a eigenvector?

Actually, if the row-reduced matrix is the identity matrix, then you have v1 = 0, v2 = 0, and v3 = 0. You get the zero vector. But eigenvectors can't be the zero vector, so this tells you that this matrix doesn't have any eigenvectors. To get an eigenvector you have to have (at least) one row of zeroes, giving (at least) one parameter. It's an important feature of eigenvectors that they have a parameter, so you can lengthen and shorten the vector as much as you like and it will still be an eigenvector.

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Linear algebra

Course: Linear algebra > Unit 3

Lesson 5: Eigen-everything

Eigenvectors and eigenspaces for a 3x3 matrix

Name: Eigenvectors and eigenspaces for a 3x3 matrix
Uploaded: 2011-02-20T16:47:24Z
Description: Eigenvectors and eigenspaces for a 3x3 matrix

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Eigenvectors and eigenspaces for a 3x3 matrix. Created by Sal Khan.

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ilja.postel
Posted 12 years ago. Direct link to ilja.postel's post “First of all, amazing vid...”
First of all, amazing video once again. They're helping me a lot. I study economics in Belgium, and we've been getting a lot of these problems (matrices, also integrals, geometry...) but I think our book (it's written by the uni itself) isn't sufficient. Is there anyone that could reccommend maybe a good, general book on mathematics on this level (college)?thank you!
Button navigates to signup pageComment on ilja.postel's post “First of all, amazing vid...”
(21 votes)
Answer
- Shah Destructor
  Posted 12 years ago. Direct link to Shah Destructor's post “try Linear Algebra and it...”
  try Linear Algebra and its applications by David C. Lays
  Comment on Shah Destructor's post “try Linear Algebra and it...”
  (30 votes)
Zeeshan Ali
Posted 9 years ago. Direct link to Zeeshan Ali's post “I am unable to find vidio...”
I am unable to find vidios on diagonalization
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(17 votes)
Answer
- John Swift
  Posted 7 years ago. Direct link to John Swift's post “It's not Khan, but here's...”
  It's not Khan, but here's a good one:
  https://youtu.be/13r9QY6cmjc
  Button navigates to signup page
  (3 votes)
James LaFountain
Posted 8 years ago. Direct link to James LaFountain's post “3 was a repeated root and...”
3 was a repeated root and the eigenspace resulted in a plane. Can the number of repeated roots be used to predict the dimension of the eigenspace?
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(3 votes)
Answer
- Derek M.
  Posted 8 years ago. Direct link to Derek M.'s post “If I recall, you can't us...”
  If I recall, you can't use the number of repeated roots to find the dimension of the eigenspace, because it completely depends on the matrix A that you are finding eigenvalues for. Sometimes, after obtaining an eigenvalue of multiplicity >1, and then row reducing A-lambda(IdentityMatrix), the amount of free variables in that matrix matches the multiplicity. Other times, the amount of free variables is less than the multiplicity.
  All you can know, is that if an eigenvalue K has a multiplicity of n, then at most, the dimension of the eigenspace of the eigenvalue is n.
  If your dimensions of your eigenspaces match with your roots of the characteristic equation (including repeated ones), then you can diagonalize the matrix.
  Button navigates to signup page
  (8 votes)
Vasco
Posted 10 years ago. Direct link to Vasco's post “At 4:54, why v2 = a and v...”
At
4:54
, why v2 = a and v3 = b?
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(3 votes)
Answer
- Anthony Natoli
  Posted 9 years ago. Direct link to Anthony Natoli's post “Sal is setting the free v...”
  Sal is setting the free variables v2 and v3 to be arbitrary real numbers, so he can write v1, v2, and v3 as linear combinations of vectors and the arbitrary real numbers a, b.
  Comment on Anthony Natoli's post “Sal is setting the free v...”
  (4 votes)
Huiqin Hu
Posted 9 years ago. Direct link to Huiqin Hu's post “In my book, the character...”
In my book, the characteristic equation is A-lamda I, is that matters to switch the place of A and lamba I?
Button navigates to signup pageComment on Huiqin Hu's post “In my book, the character...”
(3 votes)
Answer
- Lucas Van Meter
  Posted 9 years ago. Direct link to Lucas Van Meter's post “Good question, Both "A- ...”
  Good question,
  
  Both "A- lambda I=0" and "lambda I - A=0" work, in fact they are the same equation! To see why, look at the start of the video how Sal derives the formula. He starts with the definition of an eigenvalue, which is "Av=lambda v". He then subtracts one side of the equation from the other. It doesn't matter which side we subtract so we either end up with "Av- lambda I v=0" or "lambda I v - Av=0".
  Button navigates to signup page
  (4 votes)
pranay0604
Posted 10 years ago. Direct link to pranay0604's post “When is row reduce a matr...”
When is row reduce a matrix, it ends up being the identity matrix, hence the vector is [1,1,1] with no parameters.. Is this still considered a eigenvector?
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(3 votes)
Answer
- S W
  Posted 10 years ago. Direct link to S W's post “Actually, if the row-redu...”
  Actually, if the row-reduced matrix is the identity matrix, then you have v1 = 0, v2 = 0, and v3 = 0. You get the zero vector. But eigenvectors can't be the zero vector, so this tells you that this matrix doesn't have any eigenvectors.
  To get an eigenvector you have to have (at least) one row of zeroes, giving (at least) one parameter. It's an important feature of eigenvectors that they have a parameter, so you can lengthen and shorten the vector as much as you like and it will still be an eigenvector.
  Button navigates to signup page
  (3 votes)
Rachael Crozier
Posted 10 years ago. Direct link to Rachael Crozier's post “Is the intersection of an...”
Is the intersection of any two eigenspaces of the same matrix equal to the zero vector?
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(4 votes)
Answer
- InnocentRealist
  Posted 7 years ago. Direct link to InnocentRealist's post “Yes. It can be proven tha...”
  Yes. It can be proven that the eigenbasis vectors for a matrix are a linearly independent set (See "https://math.stackexchange.com/questions/29371/how-to-prove-that-eigenvectors-from-different-eigenvalues-are-linearly-independe"), therefore no eigenbasis vectors are in the span of any others, and so they have only the 0 vector in common, like all vector spaces.
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  (1 vote)
Anthony Natoli
Posted 9 years ago. Direct link to Anthony Natoli's post “Lambda = 3 is a repeated ...”
Lambda = 3 is a repeated root of the characteristic polynomial which Sal solved in the previous video, but lambda = -3 is not a repeated root. Does a repeated eigenvalue such as 3 in this case cause something funky to happen or something to degenerate with the eigenspaces? I am making an analogy with differential equations which, when they have a repeated root of the characteristic equation, change to having (Ax+B)sin( ...) instead of just coefficients C1, C2 in front of sin(...) and cos( ...), respectively. With eigenvalues, does the repeated root cause one eigenspace to degenerate to a line and the other stay a plane in this 3-dimensional example? Or does the repeated room cause the eigenspaces to be orthogonal as Sal noted in this example at the end of this video?
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(4 votes)
Answer
Hemen Taleb
Posted 8 years ago. Direct link to Hemen Taleb's post “Are the eigenspaces of 2 ...”
Are the eigenspaces of 2 lambdas always orthogonal if the case is such that one of the lambdas is just the negative of the other lambda (3 and -3 in this case)? or was this just a coincidence?
Thanks
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(3 votes)
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Rohan Kotwani
Posted 6 years ago. Direct link to Rohan Kotwani's post “Are the vectors of the nu...”
Are the vectors of the null space also eigen vectors?
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(3 votes)
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- sh3nster
  Posted a year ago. Direct link to sh3nster's post “Yes, the eigenvalue can b...”
  Yes, the eigenvalue can be 0. In this example, however, the null space is only the zero vector, which, by definition, cannot be an eigenvector.
  Button navigates to signup page
  (1 vote)

Video transcript

In the last video we set out to find the eigenvalues values of this 3 by 3 matrix, A. And we said, look an eigenvalue is any value, lambda, that satisfies this equation if v is a non-zero vector. And that says, any value, lambda, that satisfies this equation for v is a non-zero vector. Then we just did a little bit of I guess we could call it vector algebra up here to come up with that. You can review that video if you like. And then we determined, look the only way that this is going to have a non-zero solution is if this matrix has a non-trivial null space. And only non-invertible matrices have a non-trivial null space. Or, only matrices that have a determinant of 0 have non-trivial null spaces. So you do that, you got your characteristic polynomial, and we were able to solve it. And we got our eigenvalues where lambda is equal to 3 and lambda is equal to minus 3. So now, let's do-- what I consider the more interesting part-- is actually find out the eigenvectors or the eigenspaces. So we can go back to this equation, for any eigenvalue this must be true. This must be true but this is easier to work with. And so, this matrix right here times your eigenvector must be equal 0 for any given eigenvalue. This matrix right here-- I've just copied and pasted from above. I marked it up with the Rule of Sarrus so you can ignore those lines-- is just this matrix right here for any lambda. Lambda times the identity matrix minus A ends up being this. So let's take this matrix for each of our lambdas and then solve for our eigenvectors or our eigenspaces. So let me take the case of lambda is equal to 3 first. So if lambda is equal to 3, this matrix becomes lambda plus 1 is 4, lambda minus 2 is 1, lambda minus 2 is 1. And then all of the other terms stay the same, minus 2, minus 2, minus 2, 1, minus 2 and 1. And then this times that vector, v, or our eigenvector v is equal to 0. Or we could say that the eigenspace for the eigenvalue 3 is the null space of this matrix. Which is not this matrix. It's lambda times the identity minus A. So the null space of this matrix is the eigenspace. So all of the values that satisfy this make up the eigenvectors of the eigenspace of lambda is equal to 3. So let's just solve for this. So the null space of this guy-- we could just put in reduced row echelon form-- the null space of this guy is the same thing as the null space of this guy in reduced row echelon form. So let's put this in reduced row echelon form. So the first thing I want to do-- let me just do it down here. So let me-- I'll keep my first row the same for now. 4 minus 2, minus 2. And let me replace my second row with my second row times 2 plus my first row. So minus 2 times 2 plus 1 is 0. 1 times 2 plus minus 2 is 0. 1 times 2 plus minus 2 is 0. This row is the same as this row. So I'm going to do the same thing. Minus 2 times 2 plus 4 is 0. 1 times 2 plus 2 is 0. And then 1 times 2 plus minus 2 is 0. So the solutions to this equation are the same as the solutions to this equation. Let me write it like this. Instead of just writing the vector, v, let me write it out. So v1, v2, v3 are going to be equal to the 0 vector. 0, 0. Just rewriting it slightly different. And so these two rows, or these two equations, give us no information. The only one is this row up here, which tells us that 4 times v1 minus 2 times v2-- actually this wasn't complete reduced row echelon form but close enough. It's easy for us to work with-- 4 times v1 minus 2 times v2 minus 2 times v3 is equal to 0. Let's just divide by 4. I could've just divided by 4 here, which might have made it skipped a step. But if you divide by 4 you get v1 minus 1/2 v2 minus 1/2 v3 is equal to 0. Or, v1 is equal to 1/2 v2 plus 1/2 v3. Just added these guys to both sides of the equation. Or we could say, let's say that v2 is equal to-- yeah I don't know, I'm going to just put some random number-- a, and v3 is equal to b, then we can say-- and then v1 would be equal to 1/2 a plus 1/2 b. We can say that the eigenspace for lambda is equal to 3, is the set of all of vectors, v1, v2, v3, that are equal to a times times-- v2 is a, right? So v2 is equal to a times 1. v3 has no a in it. So it's a times 0. Plus b times-- v2 is just a. v2 has no b in it. So it's 0. v3 is 1 times-- so 0 times a plus 1 times b. And then v1 is 1/2 a plus 1/2 b. For any a and b, such that a and b are members of the reals. Just to be a little bit formal about it. So that's our-- any vector that satisfies this is an eigenvector. And they're the eigenvectors that correspond to eigenvalue lambda is equal to 3. So if you apply the matrix transformation to any of these vectors, you're just going to scale them up by 3. Let me write this way. The eigenspace for lambda is equal to 3, is equal to the span, all of the potential linear combinations of this guy and that guy. So 1/2, 1, 0. And 1/2, 0, 1. So that's only one of the eigenspaces. That's the one that corresponds to lambda is equal to 3. Let's do the one that corresponds to lambda is equal to minus 3. So if lambda is equal to minus 3-- I'll do it up here, I think I have enough space-- lambda is equal to minus 3. This matrix becomes-- I'll do the diagonals-- minus 3 plus 1 is minus 2. Minus 3 minus 2 is minus 5. Minus 3 minus 2 is minus 5. And all the other things don't change. Minus 2, minus 2, 1. Minus 2, minus 2 and 1. And then that times vectors in the eigenspace that corresponds to lambda is equal to minus 3, is going to be equal to 0. I'm just applying this equation right here which we just derived from that one over there. So, the eigenspace that corresponds to lambda is equal to minus 3, is the null space, this matrix right here, are all the vectors that satisfy this equation. So what is-- the null space of this is the same thing as the null space of this in reduced row echelon form So let's put it in reduced row echelon form. So the first thing I want to do, I'm going to keep my first row the same. I'm going to write a little bit smaller than I normally do because I think I'm going to run out of space. So minus 2, minus 2, minus 2. Actually let me just do it this way. I will skip some steps. Let's just divide the first row by minus 2. So we get 1, 1, 1. And then let's replace this second row with the second row plus this version of the first row. So this guy plus that guy is 0 minus 5 plus minus-- or let me say this way. Let me replace it with the first row minus the second row. So minus 2 minus minus 2 is 0. Minus 2 minus minus 5 is plus 3. And then minus 2 minus 1 is minus 3. And then let me do the last row in a different color for fun. And I'll do the same thing. I'll do this row minus this row. So minus 2 minus minus 2 is a 0. Minus 2 plus 2. Minus 2 minus 1 is minus 3. And then we have minus 2 minus minus 5. So it's minus 2 plus 5. So that is 3. Now let me replace-- and I'll do it in two steps. So this is 1, 1, 1. I'll just keep it like that. And actually, well let me just keep it like that. And then let me replace my third row with my third row plus my second row. It'll just zero out. If you add these terms, these all just become 0. That guy got zeroed out. And let me take my second row and divide it by 3. So this becomes 0, 1, minus 1. And I'm almost there. I'll do it in orange. So let me replace my first row with my first row minus my second row. So this becomes 1, 0, and then 1 minus minus 1 is 2. 1 minus minus 1 is 2. And then in the second row is 0, 1, minus 1. And then the last row is 0, 0, 0. So any v that satisfies this equation will also satisfy this guy. This guy's null space is going to be the null space of that guy in reduced row echelon form. So v1, v2, v3 is equal to 0, 0, 0. Let me move this. Because I've officially run out of space. So let me move this lower down where I have some free real estate. Let me move it down here. This corresponds to lambda is equal to minus 3. This was lambda is equal to minus 3, just to make us-- it's not related to this stuff right here. So what are all of the v1s, v2s and v3s that satisfy this? So if we say that v3 is equal to t. If v3 is equal to t, then what do we have here? We have-- this tells us that v2 minus v3 is equal to 0. So that tells us that v2 minus v3-- 0 times v1 plus v2 minus v3 is equal to 0. Or that v2 is equal to v3, which is equal to t. That's what that second equation tells us. And then the third equation tells us, or the top equation tells us, v1 times 1-- so v1 plus 0 times v2 plus 2 times v3 is equal to 0. Or v1 is equal to minus 2v3 is equal to minus 2 times t. So the eigenspace that corresponds to lambda is equal to minus 3 is equal to the set of all the vectors, v1, v2 and v3, where-- well, it's equal to t times-- v3 is just t. v3 was just t. v2 also just ends up being t. So 1 times t. And v1 is minus 2 times t. For t is any real number. Or another way to say it is that the eigenspace for lambda is equal to minus 3 is equal to the span-- I wrote this really messy-- where lambda is equal to minus 3 is equal to the span of the vector minus 2, 1, and 1. Just like that. It looks interesting. Because if you take this guy and dot it with either of these guys, I think you get 0. Is that definitely the case? Take minus 2 times 1/2, you get a minus 1 there. Then you have a plus 1. That's 0. And then minus 2 times 1/2. Yeah. You dot it with either of these guys you get 0. So this line is orthogonal to that plane. Very interesting. So let's just graph it just so we have a good visualization of what we're doing. So we had that 3 by 3 matrix, A. It represents some transformation in R3. And it has two eigenvalues. And each of those have a corresponding eigenspace. So the eigenspace that corresponds to the eigenvalue 3 is a plane in R3. So this is the eigenspace for lambda is equal to 3. And it's the span of these two vectors right there. So if I draw them, maybe they're like that. Just like that. And then the eigenspace for lambda is equal to minus 3 is a line. It's a line that's perpendicular to this plane. It's a line like that. It's the span of this guy. Maybe if I draw that vector, that vector might look something like this. And it's the span of that guy. So what this tells us, this is the eigenspace for lambda is equal to minus 3. So what that tells us-- just to make sure we are interpreting our eigenvalues and eigenspaces correctly-- is look, you give me any eigenvector, you give me any vector in this, you give me any vector right here, let's say that is vector x. If I apply the transformation, if I multiply it it by a, I'm going to have 3 times that. Because it's in the eigenspace where lambda is equal to 3. So if I were to apply a times x, a times x would be just 3 times that. So that would be a times x. That's what it tells me. This would be true for any of these guys. If this was x, and you took a times x, it's going to be 3 times as long. Now these guys over here, if you have some vector in this eigenspace that corresponds to lambda is equal to 3, and you apply the transformation. Let's say that this is x right there. If you took the transformation of x, it's going to make it 3 times longer in the opposite direction. It's still going to be on this line. So it's going to go down like this. And that would be a times x. It would be the same, it'd be 3 times this length, but in the opposite direction. Because it corresponds to lambda is equal to minus 3. So anyway, we've, I think, made a great achievement. We've not only figured out the eigenvalues for a 3 by 3 matrix, we now have figured out all of the eigenvectors. Which are-- there's an infinite number-- but they represent 2 eigenspaces that correspond to those two eigenvalues, or minus 3 and 3. See you in the next video.