00:01
In this question, we are going to diagonalize a if possible.
00:14
So we first need to find eigenvalues of a.
00:18
So we compute the characteristic polynomial of a, which is determinant a minus lambda i.
00:28
So it is determinant 2 minus lambda 2, negative 1, 1, 3 minus lambda, 2, 1, 2 ,000, 2, 1, 2, 1, 2, 2, 1, 2, 2, 2, 2 minus lambda.
00:42
If we expand it, this will be negative lambda cube plus 7 lambda square minus 11 lambda plus 5.
01:01
And this can be factored as negative lambda minus 1 square, lambda minus 5.
01:12
So the eigenvalues of a are 1 ,1, 5.
01:25
Note that the eigenvalue 1 repeat twice.
01:33
So next we are going to find eigenvectors.
01:42
So we look at a minus lambda i, and we first consider 1.
01:48
So a minus i.
01:50
This is 1, 2, negative 1.
01:56
1, 1, 2, 1, 2, 1, 2, 1, 1...