Find the eigenvalues of the matrix A = eginbmatrix 5 0 ...

Find the eigenvalues of the matrix A = egin{bmatrix} 5 & 0 & 3 \ -2 & 2 & 0 \ 0 & 0 & 1 end{bmatrix} Characteristic polynomial (type in your answer as a*lambda^2+b*lambda+c, if your polynomial is alambda^2 + blambda + c) is The eigenvalues are (type your answer as [lambda=a, lambda=b] if answer is lambda = a, lambda = b)

Transcript

00:01 We are going to find the eigenvalues of the matrix a equals 5, 0, 3, negative 2, 2, 0, 0, 0, 1.

00:11 So we find first the characteristic polynomial of a, which we must give as an answer indeed.

00:23 Did and we call that characteristic polynomial of a p a of lambda that is the independent variable for the polynomial will be lambda and that is equal to the determinant of a minus lambda times the identity of order 3 let's call it i3 and that will be p a of lambda that is the characteristic polynomial of a.

00:56 And then we got to calculate the determinant of the matrix 5 minus lambda 0, 3, negative 2, 2 minus lambda, 0, and 0, 0, 1 minus lambda.

01:20 So we got to calculate this determinant.

01:25 We can't develop that determinant using the last row because we have two zeros there.

01:32 So we get zero times some determinant that is zero minus zero times another determinant that is zero again.

01:40 So we only get one minus lambda, which preserves its sign because the location of that coefficient one minus lambda inside the matrix is three, 3, 3, that is third row, third column, and the indices 3 and 3 add up to 6, which is an even number, so in that case the coefficient preserves its sign.

02:03 And that is multiplied by the determinant of the submatrix we obtained by removing the third row, third column, which is the location of the coefficient 1 minus lambda, and then the sub matrix we get is 5 minus lambda 0 negative 2 2 minus lambda so the characteristic polynomial of a in lambda is 1 minus lambda times the the determinant of this matrix here is 5 minus lambda times 2 minus lambda, sorry, minus 0, because the product of negative 2 and 0 is 0.

02:59 So we get, finally, 1 minus lambda times 5 minus lambda times 2 minus lambda.

03:08 And this is a third degree polynomial, but the good thing here is that we obtained the characteristic polynomial in a factorized way already.

03:18 It's completely factorized out.

03:20 So looking at or inspecting the polynomial, we can say which are the eigenvalues of x...

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