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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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# Eigenvalues of a 3x3 matrix

Determining the eigenvalues of a 3x3 matrix. Created by Sal Khan.

## Want to join the conversation?

- At9:30, wouldn't it have been way easier to factor the quadratic polynomial by grouping? We could have factored out a [lambda]^2 from the first two terms and a -9 from the last two, giving us ([lambda]^2 - 9) * ([lambda] - 3) = 0 which easily leads us to find that [lambda] can be 3 or - 3.(5 votes)
- Yes,

As Sal mentions there is (long and tedious) formula to factoring but in practice it is more of an art. The method you describe is elegant and when you can you should do it. The way Sal describes might be a little more algorithmic so if you can't see a clever way to do it is method is nice to know. But by all means use your way and continue down the path to becoming an algebra ninja.(4 votes)

- Can't you simply apply row operations to make it a triangular matrix, before subtracting? Seems like saving tons of work, since the eigenvalues of a triangular matrix are on the main diagonal(5 votes)
- Good question Danny,

The short answer is no, while it is true that row operations preserve the determinant of a matrix the determinant does not split over sums. We want to compute det(M-lambda I_n) which does not equal det(M)-det(lambda n).

The best way to see what problem comes up is to try it out both ways with a 2x2 matrix like ((1,2),(3,4)).(4 votes)

- Is there a way to calculate eigenvalues & eigenvectors with a regular graphing calculator? For instance, how would you calculate eigenvalues for a matrix larger than 3x3 in practice (without resorting to extremely tedious algebra by hand)?(2 votes)
- I have a Texas Instruments TI-85 (quite an old calculator and superseded now). It's the same as the calculator Sal uses a lot on the screen. On that there is a MATRX area where you can enter a matrix. Then you can choose the MATH submenu and choose the eigVl and eigVc menu items for eigenvalue and eigenvector respectively. Do you have a graphing calculator already? If so, what make?

Wolfram Alpha is great for doing these computations too. If you give it a 3x3 matrix, it'll tell you some properties (including characteristic polynomial, eigenvalues/vectors):

http://www.wolframalpha.com/input/?i=%7B%7B1%2C2%2C3%7D%2C%7B3%2C2%2C1%7D%2C%7B2%2C1%2C3%7D%7D&lk=4&num=1(5 votes)

- At10:25, shouldn't the possible divisors of 27 be +/-1, +/-3, +/-9, +/-27?(3 votes)
- Yes, Sal should've checked the negative divisors if none of the positive divisors had worked.(3 votes)

- Lucky for us, no need to calculate it by hand anymore, any mathematical software (matlab, mathematica, etc) or math lib of any programming language can do it for you, just type Eig(A)(4 votes)
- What name is the method that Sal use to factor out the characteristic polynomial in the video?(3 votes)
- Sal is applying the Rational Zeros Theorem. Here's a link: https://www.youtube.com/watch?v=YMyv9-9VXw4(2 votes)

- Does the fact that λ = 3 is a double root imply that the eigenspace E_3 will be of dimension 2? In other words, is the dimension of an eigenspace E_λ equal to the multiplicity of the λ root of the characteristic polynomial?(3 votes)
- the dimension is at least one and up to and including the multiplicity. in this case, either 1 or 2.(2 votes)

- in this video lesson the original matrix A row reduced to Isub3 identity matrix according to my trusty TI-84 buddy. So this makes me feel like sal sorta pulled a fast one on me. can someone explain and reassure me this isn't slight of math hand?(2 votes)
- You are correct, it does row reduce to identity: http://www.wolframalpha.com/input/?i=rowreduce%5B%7B-1%2C2%2C2%7D%2C%7B2%2C2%2C-1%7D%2C%7B2%2C-1%2C2%7D%5D.

This means A is invertible. The matrix that has to be singular however is (λI - A).(2 votes)

- in my book it says the det( A- lambda*I)=0

im guessing both are correct, but why is this?

book is from Pearson so pretty sure that it is correct, also my proffesor taught it that way as well(2 votes)- Yes, they mean the same thing. If det(λ
**I**-**A**) = 0, then det(**A**- λ**I**) = 0. This is because, if you multiply a matrix, like λ**I**-**A**, by a scalar, like -1, so that you get**A**- λ**I**, the determinant of the new matrix is just the determinant of the old matrix times that scalar raised to the power of the number of dimensions of the matrix. Since in this case, it is a 3 by 3 matrix, the new determinant is the old determinant, 0, multiplied by the scalar, -1, raised to the 3rd power, which just leaves 0 again.(2 votes)

- Hi,

1) is there a way to verify the eigenvalues somehow? That way I could see if I didn't make a silly mistake.

2) When the multiplicity are all one for the eigenvalues (not this example), is that proof enough to say that the matrix is diagonalizable? for example (l - 2) (l - 3) (l + 3)

3) if instead of doing Saurus I just calculate three determinants and at a certain point I have

(lambda = L)

-1.(L-4)(-L-6) this would be (-L+4)(-L-6). Would it be ok to write (L-4)(L+6)?

Help with any of the 3 questions is highly appreciated!

Thanks(1 vote)- For the first question, I'm going to assume you're taking det(A-lambda*I) correctly. If you solve the characteristic equation set = 0, the roots you find you should be able to double check by plugging them back into the characteristic equation itself and getting 0. Besides that, just make sure you're taking the determinant correctly. Also, if you take that eigenvalue and find an associated eigenvector, you should be able to use the original matrix (lets say A) and multiple A by the eigenvector found and get out the SAME eigenvector (this is the definition of an eigenvector).

For the second question: Yes. If you have 3 distinct eigenvalues for a 3x3 matrix, it is diagonalizable because we will have 3 distinct associated eigenvectors. To generalize, an nxn matrix will be diagonalizable if you have n distinct eigenvectors *Note:* we could have less than n*distinct*eigenvalues and end up with n eigenvectors through multiplicities, meaning A would still be diagonalizable in that scenario.(3 votes)

## Video transcript

We figured out the eigenvalues
for a 2 by 2 matrix, so let's see if we can figure
out the eigenvalues for a 3 by 3 matrix. And I think we'll appreciate
that it's a good bit more difficult just because the math
becomes a little hairier. So lambda is an eigenvalue
of A. By definition, if and only if--
I'll write it like this. If and only if A times some
non-zero vector v is equal to lambda times that non-zero
vector v. Let we write that for
some non-zero. I could call it eigenvector v,
but I'll just call it for some non-zero vector v or
some non-zero v. Now this is true if and only if,
this leads to-- I'll write it like this. This is true if and only if--
and this is a bit of review, but I like to review it just
because when you do this 10 years from now, I don't want you
to remember the formula. I want you to just remember the
logic of how we got to it. So this is true if and only if--
let's just subtract Av from both sides-- the 0 vector
is equal to lambda- instead of writing lambda times v, I'm
going to write lambda times the identity matrix times v. This is the same thing. The identity matrix
times v is just v. Minus Av. I just subtracted Av from both
sides, rewrote v as the identity matrix times v. Well this is only true if and
only if the 0 vector is equal to lambda times the identity
matrix minus A times v. I just factored the vector v out
from the right-hand side of both of these guys, and
I'm just left with some matrix times v. Well this is only true-- let
me rewrite this over here, this equation just in a form
you might recognize it. Lambda times the identity
matrix times A. This is just some matrix. This matrix times v has got
to be equal to 0 for some non-zero vector v. That means that the null space
of this matrix has got to be nontrivial. Or another way to think about it
is that its columns are not linearly independent. Or another way to think about it
is it's not invertible, or it has a determinant of 0. So lambda is the eigenvalue of
A, if and only if, each of these steps are true. And this is true if and only
if-- for some at non-zero vector, if and only if, the
determinant of lambda times the identity matrix minus
A is equal to 0. And that was our takeaway. I think it was two videos
ago or three videos ago. But let's apply it now to
this 3 by 3 matrix A. We're going to use the 3
by 3 identity matrix. So we want to concern ourselves
with-- lambda times the identity matrix is just
going to be-- times the 3 by 3 identity matrix is just
going to be-- this is, let me write this. This is lambda times the
identity matrix in R3. So it's just going to be
lambda, lambda, lambda. And everything else is
going to be 0's. The identity matrix had 1's
across here, so that's the only thing that becomes
non-zero when you multiply it by lambda. Everything else was a 0. So that's the identity
matrix times lambda. So lambda times the identity
matrix minus A is going to be equal to-- it's actually pretty straightforward to find. Everything along the diagonal is
going to be lambda minus-- let's just do it. Lambda minus minus 1-- I'll
do the diagonals here. Lambda minus minus 1
is lambda plus 1. And then 0 minus 2-- I'll do
that in a different color. 0 minus 2 is minus 2. 0 minus 2 is minus 2. 0 minus 2 is minus 2. Let's do this one. 0 minus 2 is minus 2. 0 plus or minus minus 1 is
0 plus 1, which is 1. And then let's just
do this one. 0 minus minus 1. That's one. Let me finish up the diagonal. And then you have
lambda minus 2. And then you have
lambda minus 2. So lambda is an eigenvalue
of A if and only if the determinant of this matrix
right here is equal to 0. Let's figure out its
determinate. And the easiest way, at least
in my head to do this, is to use the rule of Sarrus. So let's use the rule of
Sarrus to find this determinant. So I just rewrite these
rows right there. I could just copy and
paste them really. I just take those two rows. And then let me paste them,
put them right there. It's a little bit too close
to this guy, but I think you get the idea. And now the rule of Sarrus I
just take this product plus this product plus this product
and then I subtract out this product times this product
times this product. We'll do that next. So this product is lambda plus
1 times lambda minus 2 times lambda minus 2. That's that one there. And then plus, let's see,
minus 2 times minus 2. That's plus 4. And then we have minus 2 times
minus 2 plus 4 times 1. So that is plus 4 again. And then we do minus this column
times this column. Minus this column minus this
column and then-- or I shouldn't say column,
but diagonal really. So we say minus 2
times minus 2. Let me write this. Minus 2 times minus
2, which is 4. Times lambda minus 2. That was this diagonal. And then we have minus-- what
is this going to be? Going to be minus 1 times
lambda plus 1. So minus lambda plus 1. And then you go down
this diagonal. Minus 2 times minus 2 is 4. So it's going to be 4 times
lambda minus 2 and we're subtracting. So minus 4 times
lambda minus 2. And let's see if we
can simplify this. So this blue stuff over here--
let's see, these guys right here become an 8 and then
this becomes-- this becomes lambda plus 1. Times-- if I multiply these two
guys out, lambda squared minus 4 lambda. Minus 2 lambda and then
minus 2 lambda. So minus 4 lambda. Plus 4. And then I have this
plus 8 here. And then I have-- let's see. I have minus 4 times lambda. Let me just multiply
everything out. So I have minus 4 lambda plus 8
minus lambda minus 1 minus 4 lambda plus 8. And then let me simplify
this up a little bit. So this guy over here--
let's see. The constant terms, I have an 8,
I have a minus 1, I have an 8 and I have an 8. So that's 24 minus 1. So that is a 23. And then the lambda terms
I have a minus 4 lambda. I have a minus lambda and
I have a minus 4 lambda. So it's minus 8, minus 1. So I have minus 9 lambda. Plus 23. And now I have to simplify
this out. So first I can take lambda and
multiply it times this whole guy right there. So it's going to be lambda cubed
minus 4 lambda squared plus 4 lambda. And then I can take this
one and multiply it times that guy. So plus lambda squared. Minus 4 lambda plus 4. And now of course, we have
these terms over here. So we're going to have
to simplify it again. So what are all of our
constant terms? We have a 23 and we
have a plus 4. So we have a 27. Plus 27. And then, what are all
of our lambda terms? We have a minus 9 lambda and
then we have a-- let's see. We have a minus 9 lambda, we
have a plus 4 lambda, and then we have a minus 4 lambda. So these two cancel out. So I just have a
minus 9 lambda. And then, what are my lambda
squared terms? I have a plus lambda squared
and I have a minus 4 lambda squared. So if you add those two
that's going to be minus 3 lambda squared. And then finally, I have only
one lambda cubed term, that right there. So this is the characteristic
polynomial for our matrix. So this is the characteristic
polynomial and this represents the determinant for
any lambda. The determinant of this
matrix for any lambda. And we said that this has to be
equal to 0 if any only if lambda is truly an eigenvalue. So we're going to set
this equal to 0. And unlucky or lucky for us,
there is no real trivial-- there is no quadratic. Well there is, actually, but
it's very complicated. And so it's usually
a waste of time. So we're going to have to do
kind of the art of factoring a quadratic polynomial. I got this problem out of a book
and I think it's fair to say that if you ever do run into
this in an actual linear algebra class or really, in an
algebra class generally-- it doesn't even have to be in the
context of eigenvalues, you probably will be dealing
with integer solutions. And if you are dealing with
integer solutions, then your roots are going to be factors
of this term right here. Especially if you have a
1 coefficient out here. So your potential roots-- in
this case, what are the factors of 27? So 1, 3, 9 and 27. So all these are potential
roots. So we can just try them out. 1 cubed is 1 minus 3. So let me try 1. So if we try a 1, it's 1 minus
3 minus 9 plus 27. That does not equal 0. It's minus 2 minus
9 is minus 11. Plus 16. That does not equal 0. So 1 is not a root. If we try 3 we get 3
cubed, which is 27. Minus 3 times 3 squared
is minus 3 times 3, which is minus 27. Minus 9 times 3, which
is minus 27. Plus 27. That does equal 0. So lucky for us, on our second
try we were able to find one 0 for this. So if 3 is a 0, that means that
x minus 3 is one of the factors of this. So that means that this is going
to be x minus 3 times something else. Or I should say,
lambda minus 3. So let's see what the
other root is. So if I take lambda minus 3 and
I divide it into this guy up here, into lambda cubed minus
3 lambda squared minus 9 lambda plus 27, what do I get? Lambda goes into lambda cubed
lambda squared times. Lambda squared times that. Lambda squared times lambda
is lambda cubed. Lambda squared times minus 3
is minus 3 lambda squared. You subtract these guys,
you get a 0. You get 0. And then we can put here--
well, we could do it either way. We could put it down
the minus 9. We could bring down
everything really. So now you have minus
9 lambda plus 27. You can almost imagine we just
subtracted this from this whole thing up here. And we're just left with
these terms right here. And so lambda minus
3 goes into this. Well lambda minus 3 goes
into 9 lambda. It goes into 9 lambda
minus 9 times. So I'll just write
minus 9 here. Minus 9 times lambda minus 3
is minus 9 lambda plus 27. So it went in very nicely. So you get to 0. Our characteristic polynomial
has simplified to lambda minus 3 times lambda squared
minus 9. And of course, we're going to
have to set this equal to 0 if lambda is truly an eigenvalue
of our matrix. And this is very
easy to factor. So this becomes lambda minus 3
times-- lambda squared minus 9 is just lambda plus 3 times
lambda minus 3. And all of that equals 0. And these roots, we already
know one of them. We know that 3 is a root and
actually, this tells us 3 is a root as well. So the possible eigenvalues of
our matrix A, our 3 by 3 matrix A that we had way up
there-- this matrix A right there-- the possible eigenvalues
are: lambda is equal to 3 or lambda is
equal to minus 3. Those are the two values that
would make our characteristic polynomial or the determinant
for this matrix equal to 0, which is a condition that we
need to have in order for lambda to be an eigenvalue of a
for some non-zero vector v. In the next video, we'll
actually solve for the eigenvectors, now that we know
what the eigenvalues are.