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### 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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# Proof of formula for determining eigenvalues

Proof of formula for determining Eigenvalues. Created by Sal Khan.

## Want to join the conversation?

- Hmm - do more people see λI-A instead of A-λI? That's the way my professor is teaching it, and the way that David Lay's book lays it out.(17 votes)
- I see that way in college too, and that's the way it appears in Steinbruch's, Boldrini's and most of the books I checked.

But it makes no difference, it's just algebraic manipulation. You can do whatever you feel more confident to do.(29 votes)

- Heh, I know this might sound silly but what is the null space, and what does it mean by non-trivial? And how can we assume this information (although upon answering the first question, I may be able to figure it out myself)?

This may sound like a trolling question, but I'm serious. Where I've been going to school, the maths teachers seem to take more notice of how maths is done (i.e the formulas, ways of working out equations) more than what it means to be doing it (or how it can be applied). As for mathematical terms, we don't hear of them often as they are not seen as a big concern here. Working out maths is more important. If someone could help me out with this, it would be great(14 votes)- Null space of a matrix A is the set of all vectors v which satisfy the equation Av=0 (it is mentioned at about4:30), non-trivial in this context means the solution v=0 to the previous equation.(6 votes)

- So 2 questions.

1 - Do eigenvalues (and eigenvectors) only exist for an "n x n " matrix.

2 - Do eigenvalues (and eigenvecotors) only exist for a a matrix where the determinant is 0?(11 votes)- 1. Yes, eigenvalues only exist for square matrices. For matrices with other dimensions you can solve similar problems, but by using methods such as singular value decomposition (SVD).

2. No, you can find eigenvalues for any square matrix. The det != 0 does only apply for the A-λI matrix, if you want to find eigenvectors != the 0-vector.(7 votes)

- I don't understand why Det(λL-A) must equal 0 in order for (λL-a)V = 0 to be true.(11 votes)
- By definition of null space of matrix if vector v is not zero, column vectors of matrice are linearly dependent. Determinant for linearly dependent matrices are zero, singular matrice. Hope this helps.(8 votes)

- Why did sal only consider multiplying the identity matrix with v on the lamda side but not the v with A at3:04?(6 votes)
- Because he wants to put the vector v in matrix form using the Identity matrix and A is already in that form(9 votes)

- Could you add a video lesson on complex eigenvectors and complex eigenvalues?(5 votes)
- at4:10, just have a silly question: are addition and subtraction between a matrix and a scalar undefined?(3 votes)
- Yes, addition and subtraction between a scalar and a matrix (or even between matrices of different dimensions) is undefined. That is why previous to adding, the scalar is multiplied by the Identity Matrix, so that at the time of the addition, you are adding two matrices of the same size.(5 votes)

- I didn't quite understand why λI_n-A must have linearly dependent columns so we can resolve (λI_n-1)v=0... can somebody help me?(3 votes)
`(A-λI)v =`

For simplicity, let's simply replace**0**`A-λI`

with B.`Bv =`

Given this equation, we know that all possible values of v is the nullspace of B. If v is an eigenvector, we also know that it needs to be non-zero. A non-zero eigenvector therefore means a non-trivial nullspace since v would have to be**0****0**for a trivial nullspace. A non-trivial nullspace means linearly dependent column vectors.(2 votes)

- why can he assume that there is a nontrivial solution to the null-space, even if that is what we are looking for at5:50(1 vote)
- Because if there is a non-zero eigenvector, then there must be a nontrivial solution (what Sal explains at around1:40and5:20), on the other hand if there is no non-zero eigenvector, then the matrix λI-A has linearly independent columns, which (combined with the fact that it is square matrix) means that it is invertible and therefore det(λI-A) = 0 has no solution, so the assumption that there is a nontrivial solution can't be harmful.(5 votes)

- Are there exercise for this? I really like doing exercises on kahn academy instead of my math book, because i learn a lot quicker.(3 votes)

## Video transcript

I've got a transformation, m
that's a mapping from Rn to Rn, and it can be represented
by the matrix A. So the transformation of x
is equal to A times x. We saw in the last video it's
interesting to find the vectors that only get scaled
up or down by the transformation. So we're interested in the
vectors where I take the transformation of some
special vector v. It equals of course,
A times v. And we say it only gets scaled
up by some factor, lambda times v. And these are interesting
because they make for interesting basis vectors. You know, the transformation
matrix in the alternate basis-- this is one of
the basis vectors. It might be easier to compute. Might make for good coordinate
systems. But they're in general, interesting. And we call vectors v that
satisfy this, we call them eigenvectors. And we call their scaling
factors the eigenvalues associated with this
transformation and that eigenvector. You know, hopefully from that
last video, we have a little bit of appreciation of
why they're useful. But now in this video let's at
least try to determine what some of them are. You know, based on what we know
so far, if you show me an eigenvector I can verify that it
definitely is the case, or an eigenvalue. I could verify the case. But I don't know a systematic
way of solving for either of them. So let's see if we can come
up with something. So in general, we're looking for
solutions to the equation A times v is equal
to lambda v. It's equal to lambda
times the vector. Now one solution might
immediately pop out at you, and that's just v is equal
to the 0 vector. And that definitely is a
solution, although it's not normally considered to be an
eigenvector just because one, it's not a useful
basis vector. It doesn't add anything
to a basis. It doesn't add really the amount
of vectors that you can span when you throw the
basis vector in there. And also, it's not clear what
is your eigenvalue that's associated with it. Because if v is equal
to 0, any eigenvalue will work for that. So normally when we're looking
for eigenvectors, we start with the assumption that we're
looking for non-zero vectors. So we're looking for vectors
that are not equal to the 0 vector. So given that, let's see if we
can play around with this equation a little bit and see
if we can at least come up with eigenvalues maybe
in this video. So we subtract Av from both
sides, we get the 0 vector is equal to lambda v
minus A times v. Now, we can rewrite v as-- v is
just the same thing as the identity matrix times v, right?
v is a member of Rn. The identity matrix n by n. You just multiply and we're
just going to get v again. So if I rewrite v this way, at
least on this part of the expression-- and let me swap
sides-- so then I'll get lambda times-- instead of v
I'll write the identity matrix, the n by n identity
matrix times v minus A times v is equal to the 0 vector. Now I have one matrix times v
minus another matrix times v. Matrix vector products,
they have the distributive property. So this is equivalent to the
matrix lambda times the identity matrix minus A
times the vector v. And that's going to be
equal to 0, right? This is just some matrix
right here. And the whole reason why I made
this substitution is so I can write this as a matrix
vector product instead of just a scalar vector product. And that way I was able to
essentially factor out the v and just write this whole
equation as essentially, some matrix vector product
is equal to 0. Now, in order-- if we assume
that this is the case, and we're assuming-- remember,
we're assuming that v does not equal 0. So what does this mean? So we know that v is a member
of the null space of this matrix right here. Let me write this down. v is a
member of the null space of lambda I sub n minus A. I know that might look a little
convoluted to you right now, but just imagine this
is just some matrix B. It might make it simpler. This is just some matrix
here, right? That's B. Let's make that substitution. Then this equation just becomes
Bv is equal to 0. Now, if we want to look at the
null space of this, the null space of B is all of the vectors
x that are a member of Rn such that B times
x is equal to 0. Well, v is clearly one
of those guys, right? Because B times v
is equal to 0. We're assuming B solves this
equation and that gets all the way to the assumption that B
must solve this equation. And v is not equal to 0. So v is a member of the null
space and this is a nontrivial member of the null space. We already said the 0 vector is
always going to be a member of the null space, and it
would make this true. But we're assuming
v is non-zero. We're only interested in
non-zero eigenvectors. And that means that this guy's
null space has to be nontrivial. So this means that the null
space of lambda In minus A is nontrivial. The 0 vector is not
the only member. And you might remember before,
that the only time-- let me write this in general. If I have some matrix--
I don't know. I've used A and B. Let's say I have
some matrix D. D's columns are linearly
independent if and only if the null space of D only contains
the 0 vector. Right? So if we have some matrix here
whose null space does not only contain the 0 vector,
then it has linearly dependent columns. And I just wrote that there to
kind of show you what we do know and the fact that this
one doesn't have a trivial null space tells us that we're
dealing with linearly dependent columns. So lambda In minus A-- it looks
all fancy, but this is just a matrix-- must have
linearly dependent columns. Or another way to say that is,
if you have linearly dependent columns, you're not invertible,
which also means that your determinate
must be equal to 0. All of these are true. If your determinate is equal to
0, you're not going to be invertible. You're going to have linearly
dependent columns. If your determinate is equal
to 0, then that also means that you have nontrivial members
in your null space. And so, if your determinate is
equal to 0 that means there's some lambdas for which this is
true, for non-zero vectors v. So, if there are some solutions,
if there are some non-zero vector v's that satisfy
this equation, then this matrix right here must
have a determinate of 0. And it goes the other way. If this guy has a determinate of
0, then there must be-- or if there's some the lambdas
that make this guy have a determinate of 0, then those
lambdas are going to satisfy this equation. And you could go
the other way. If there's some lambdas that
satisfy this, then those lambdas are going to make this
matrix have a 0 determinate. Let me write this. Av is equal to lambda v for
non-zero v's if and only if the determinate of lambda
In minus A is equal to the 0 vector. No, not the 0 vector. Sorry, it's just equal to 0. The determinate is just
a scalar factor. And so that's our
big takeaway. And I know what you're saying
now, how is that useful for me, Sal? You know, we did all of
this manipulation. I talked a little bit about
the null spaces. And my big takeaway is, is that
in order for this to be true for some non-zero vectors
v, then lambda has to be some value. So if I take the determinate of
lambda times the identity matrix minus A, it has
got to be equal to 0. And the reason why this is
useful is that you can actually set this equation up
for your matrices, and then solve for your lambdas. And we're going to do that
in the next video.