Okay in this video. I'm going to look at an example of finding eigenvalues and eigenvectors for a [two-By-two] Matrix
so this certainly generalizes [to] any n by n matrix but
the the arithmetic can get very tedious especially if you're [doing] it by hand, but this is the basic process, so
To solve for the eigenvalues Lambda sub I and the corresponding eigenvectors x sub I have an n by [N]
Matrix a will do the following so the first thing we're going to do is we're going to multiply
the n by n Identity Matrix by the scalar Lambda
Step two Subtract the Identity Matrix
multiple that you found in Step [one] from the Matrix a
Step three we'll find the determinant of the Matrix and the difference
So you'll basically just find the determinant of the matrix that you find in step two
Then we'll solve for the for the values of Lambda that satisfy the equation. [we] look at the determinant of a minus lambda
times the Identity Matrix
equals the zero vector
So again, you're just basically once you find the determinant. It says set it equal to zero and
solve and then we're going to solve for the corresponding vector to each Lambda, so
let's let's do that here, so
a few steps but
should be able to hopefully go through it relatively quickly, so
[just] to write everything down [okay], so we're going to find the eigenvalues
and eigenvectors of the Matrix A with entries
[seven] [three] [three] negative [one]
Okay, [so] the first step again. You know we'll just take
Lambda and multiply that by the Identity Matrix
So that's just going to be lambda again the identity Matrix for a two-By-two Matrix will have entries 1
0 0 1
So that just gives us lambda 0 0 and lambda
So again, you'll just have lambdas along the diagonal of your identity Matrix well
It will no [longer] be an identity Matrix specials that have lambdas along the diagonal and zeros everywhere else after you multiply the identity made
So step two. We'll take our Matrix a and subtract Lambda Times I
so in this case our Matrix egg and has entries 7 3 3
Negative [1] and it says just subtract
Lambda 0 0 and Lambda
And again remember you to subtract respective entries so the first the top left Entry will be 7 minus lambda
Then we would have 3 minus 0 which is just 3
We'll have 3 minus 0 which again will just give us a 3 and then we'll have negative 1 minus lambda
Ok so those are our [4] entries now
[step] 3 the way, we had it labeled a second ago. We're going to find the determinant of this Matrix
So we're going to find the determinant of the Matrix with entries 7 minus lambda
3 3 and negative 1 minus Lambda and
Recall to find the determinant [of] a 2x2 Matrix
We take the top left entry multiply it by the bottom right entry so we'll have 7 minus lambda
multiplied by Negative 1 minus Lambda
Then we use a minus sign
We take the bottom left entry multiply that by the top right Entry so we'll have 3 [multiplied] [by] 3
Okay, so let's see if we can't simplify this we'll have negative 7
[looks] like Minus 7 times Lambda
Will have a positive lambda and a negative lambda and negative Lambda will be positive Lambda Squared
Then we'll have minus 9
So I'm going to write that as lAmbda squared. It looks like we're going to have a
Negative 6 Lambda and then negative 7 minus 9 that's going to give us just negative 16
Okay, so step [four]. I think I can
Put it in here we take that equation
[2] Lambda Squared Minus 6 Lambda
minus 16 we set that equal to 0 and we solve for
[4] Lambda
So no need to waste paper so lambda squared minus six lambda minus sixteen that's going to factor that
[just] factors as I guess Lambda minus 8
multiplied by Lambda Plus, [two]
so we'll set that equal to zero [well] if you set each factor equal to zero we'll get the solutions of Lambda equals eight and
By setting the second factor equal to zero we'll get lambda equals negative two
Okay, so now we've got our our values for for Lambda
Those are going to be our
Eigenvalues okay, so these are the eigenvalues
And now we're we need to find the corresponding eigenvectors all right, so what we do now
Step 5
We look at that that that Matrix we form the a minus lambda
I so that was again the matrix that we found here, so [I'm] going to rewrite that so we've got 7 minus lambda
3 3 and negative 1 minus Lambda
And now we've got to do a few more computations. Okay, so the first thing
I'm going to do is I'm going to start with my lambda value of positive 8 we substitute that in
So we'll get 7 minus 8 we still have a 3 a 3 then, we'll have negative 1 minus 8 and
[that's] going to give us
negative 1 3 3 and Negative 9
So now what we do
I'm going to call I think in my original notes
Maybe I called it a that our original
Matrix was a so let me maybe call this I don't know I'll call this the Matrix B. Now what we do is
[We] solve the
Equation will take Matrix B
Multiplied by the [vector] x equals zero we have to solve this so what we want to do is we want to solve
We've got negative 1 times 3 times 3 times negative 9
the Matrix
the with with the Corresponds to the vector x here, we'll [write] little x sub 1 and x sub 2
And then we've got our [0] vector here, so to solve this I'm just going to use a little bit of Row reduction
so we can write this as negative 1 3 3
Negative 9
There's my zeros. I'm just going to do some row reduction
So to solve this it looks like to me we could take 3 times row one
Add that to row [two] to give us our new row two
So if we do that the first row nothing's going to change. We'll have negative 1 3 [and] 0 so
Negative 1 Times 3 Plus 3 will give us 0
3 times 3 will be 9 plus negative 9 will give us 0 and we still have a 0 at the bottom
So again, this is now going to correspond to x 1 and x 2 so that gives us our relationship
it says that it says that negative x 1
Plus 3 times x 2
equals 0 and we can simply add the x sub 1 [over] and write that as 3 times x sub 2 equals x sub 1
And now all you have to do is just pick some values [ok] so the easiest ones to me. I'm going to let
x sub 2 Equal 1
and if we let x sub 2
Equal 1 well, then We'll Have 3 [times] 1
Which will give us our value for x sub 1 so?
It says if we let x sub 2 equal 1 it says x sub 1 is going to equal 3
That's going to be the eigenvector that
Corresponds to the Eigen value of 8 so the Eigenvector
The Eigenvector in this case will be the vector with entries three and one
Okay, so that eigenvector again goes with the eigenvalue of eight
And you can check you know if you want to you can you can take the the very?
The Matrix that we started with seven three three negative one
To break this down you can check that if you take that Sturdy Matrix
seven three three
Negative one if we multiply that by our eigenvector again which has entries three and one
That'll be the same thing as when we take our
Eigenvalue, which is 8 and?
Multiply that by the eigenvector of three and one you can check that on the left side if you do this, we'll get twenty four
And eight as the entries and on the right we'll get well twenty four and eight
Just like we should so you can always check in
And again right seven times three is going to be twenty one three times one is three
So twenty one plus three will give us 24
Three times three is nine plus a negative one well that's 9 minus one which gives us eight on the left
So hey in fact that does does work out?
ok so
Halfway there now all we have to do. We're just going to do the same process except now
I'm going to use our lAmbda value of what was it I guess negative two and
Do the same thing we're just going to do
Just like the exact same thing so now if we let lAmbda equal negative two
well in this case, we're going to get
Will have 7 minus negative two we've got an entry of three
Three we've got negative one minus negative two
and
That's going to give us nine three three
And I guess we've got negative one plus two which will give us an entry of one
And just like before what we do is
We have to solve that that matrix equation, so I'm just going to skip writing down this step
I'm just going to go [ahead] and jump to the row reduction right so once we found our Matrix
We just basically did row reduction to find the relationship, so I'm going to do the same thing. I'm going to do the row reduction
So we've got nine
three three
and one
So different ways to do it to me. It looks like that
The most simple way is [to] take negative [three] times row two and add that to row one to get our new Row one
So if we do that row two is not going to change
And again you can check negative three times three plus nine. I'll give us zero negative three times one plus
Three will give us zero also have a zero
zeros at the top, right
again this corresponds to X1 and X2
So now we can write our equation. We've got three x sub [one] plus x sub [two] equals
0 or [equivalently] that x sub [two] equals negative three times x sub 1
so the same thing you can just pick values [if] we let x sub 1
Equal [one] well in that case
we'll get next sub 2 equals negative 3 times 1 or negative 3 so it says the
Eigenvector the other eigenvector
Will be the eigenvector
1 comma or 1 and negative 3 and again that corresponds to the Eigen value of
Negative 2 so
That's all there is to it. So you know obviously quite a few steps when I say all there is to it
I don't mean to make it sound like there's nothing going on
There's definitely quite a few steps, but you know no no no one of them was terribly terribly bad again
You know this is something you definitely don't want to be doing by [hand] [you]
[3x3] [one] even finding the determinant of 3x3 Matrix can already be a little obnoxious
so once you get to anything beyond that you certainly and
you're not going to be expected to do these by hand you know a 4 by 4 [Matrix] or a 5 by 5 [Matrix] in
class
on a quiz or a test unless the instructor is just a
real Masochist, so
so again, then you just turn it you know you'll use some sort of computer software to software package and
It'll grind these out in a split second, but again. It's always good [just] to kind of know what's going on
what's happening and
so
So maybe I'll do a three by three Matrix example. If people really want to see it again
It's just going to be the same process just long and drawn-out
Again the only thing that's going to be much different is just going to be finding that determinant
It's going to take more effort to do some row reduction there'll be a little bit more work on these equations
But other than that again, it'll be the same process so [alright]. I hope this makes and sense on how to find
eigenvalues and eigenvectors