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In the last video we were able[br]to show that any lambda that

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satisfies this equation for some[br]non-zero vectors, V, then

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the determinant of lambda times[br]the identity matrix

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minus A, must be equal to 0.

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Or if we could rewrite this as[br]saying lambda is an eigenvalue

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of A if and only if-- I'll[br]write it as if-- the

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determinant of lambda times the[br]identity matrix minus A is

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equal to 0.

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Now, let's see if we can[br]actually use this in any kind

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of concrete way to figure[br]out eigenvalues.

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So let's do a simple 2[br]by 2, let's do an R2.

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Let's say that A is equal to[br]the matrix 1, 2, and 4, 3.

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And I want to find the[br]eigenvalues of A.

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So if lambda is an eigenvalue[br]of A, then this right here

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tells us that the determinant[br]of lambda times the identity

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matrix, so it's going to be[br]the identity matrix in R2.

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So lambda times 1, 0, 0, 1,[br]minus A, 1, 2, 4, 3, is going

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to be equal to 0.

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Well what does this equal to?

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This right here is[br]the determinant.

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Lambda times this is just lambda[br]times all of these

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terms. So it's lambda times 1[br]is lambda, lambda times 0 is

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0, lambda times 0 is 0, lambda[br]times 1 is lambda.

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And from that we'll[br]subtract A.

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So you get 1, 2, 4, 3, and[br]this has got to equal 0.

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And then this matrix, or this[br]difference of matrices, this

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is just to keep the[br]determinant.

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This is the determinant of.

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This first term's going[br]to be lambda minus 1.

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The second term is 0 minus[br]2, so it's just minus 2.

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The third term is 0 minus[br]4, so it's just minus 4.

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And then the fourth term[br]is lambda minus

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3, just like that.

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So kind of a shortcut to[br]see what happened.

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The terms along the diagonal,[br]well everything became a

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negative, right?

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We negated everything.

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And then the terms around[br]the diagonal, we've got

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a lambda out front.

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That was essentially the[br]byproduct of this expression

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right there.

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So what's the determinant[br]of this 2 by 2 matrix?

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Well the determinant of this is[br]just this times that, minus

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this times that.

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So it's lambda minus 1, times[br]lambda minus 3, minus these

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two guys multiplied[br]by each other.

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So minus 2 times minus 4[br]is plus eight, minus 8.

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This is the determinant of this[br]matrix right here or this

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matrix right here, which[br]simplified to that matrix.

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And this has got to[br]be equal to 0.

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And the whole reason why that's[br]got to be equal to 0 is

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because we saw earlier,[br]this matrix has a

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non-trivial null space.

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And because it has a non-trivial[br]null space, it

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can't be invertible and[br]its determinant has

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to be equal to 0.

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So now we have an interesting[br]polynomial

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equation right here.

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We can multiply it out.

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We get what?

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Let's multiply it out.

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We get lambda squared, right,[br]minus 3 lambda, minus lambda,

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plus 3, minus 8,[br]is equal to 0.

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Or lambda squared, minus[br]4 lambda, minus

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5, is equal to 0.

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And just in case you want to[br]know some terminology, this

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expression right here is known[br]as the characteristic

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polynomial.

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Just a little terminology,[br]polynomial.

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But if we want to find the[br]eigenvalues for A, we just

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have to solve this right here.

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This is just a basic[br]quadratic problem.

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And this is actually[br]factorable.

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Let's see, two numbers and you[br]take the product is minus 5,

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when you add them[br]you get minus 4.

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It's minus 5 and plus 1, so you[br]get lambda minus 5, times

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lambda plus 1, is equal[br]to 0, right?

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Minus 5 times 1 is minus 5, and[br]then minus 5 lambda plus 1

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lambda is equal to[br]minus 4 lambda.

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So the two solutions of our[br]characteristic equation being

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set to 0, our characteristic[br]polynomial, are lambda is

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equal to 5 or lambda is[br]equal to minus 1.

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So just like that, using the[br]information that we proved to

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ourselves in the last video,[br]we're able to figure out that

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the two eigenvalues of A are[br]lambda equals 5 and lambda

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equals negative 1.

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Now that only just solves part[br]of the problem, right?

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We know we're looking[br]for eigenvalues and

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eigenvectors, right?

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We know that this equation can[br]be satisfied with the lambdas

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equaling 5 or minus 1.

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So we know the eigenvalues, but[br]we've yet to determine the

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actual eigenvectors.

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So that's what we're going[br]to do in the next video.

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