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## Linear algebra

### Course: Linear algebra > Unit 3

Lesson 5: Eigen-everything- Introduction to eigenvalues and eigenvectors
- Proof of formula for determining eigenvalues
- Example solving for the eigenvalues of a 2x2 matrix
- Finding eigenvectors and eigenspaces example
- Eigenvalues of a 3x3 matrix
- Eigenvectors and eigenspaces for a 3x3 matrix
- Showing that an eigenbasis makes for good coordinate systems

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# Eigenvectors and eigenspaces for a 3x3 matrix

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

## Want to join the conversation?

- 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!(21 votes)
- try Linear Algebra and its applications by David C. Lays(30 votes)

- I am unable to find vidios on diagonalization(17 votes)
- It's not Khan, but here's a good one:

https://youtu.be/13r9QY6cmjc(3 votes)

- 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?(3 votes)
- 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.(8 votes)

- At4:54, why v2 = a and v3 = b?(3 votes)
- 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.(4 votes)

- In my book, the characteristic equation is A-lamda I, is that matters to switch the place of A and lamba I?(3 votes)
- 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".(4 votes)

- 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?(3 votes)
- 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.(3 votes)

- Is the intersection of any two eigenspaces of the same matrix equal to the zero vector?(4 votes)
- 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.(1 vote)

- 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?(4 votes)
- 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(3 votes) - Are the vectors of the null space also eigen vectors?(3 votes)
- 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.(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.