00:01
In this problem, we have the matrix 4, 2, 2, 1.
00:05
Now, the first we find the eigenvalues, those are found by solving the determinant a minus lambda i equals to 0.
00:16
So, this is 4 minus lambda, this is 2, 2, 1 minus lambda, this must be 0.
00:23
So, from here we get 4 minus lambda into 1 minus lambda minus 4 equals to 0.
00:29
This is 4 minus 5 lambda plus lambda square minus 4 equals to 0, or we get lambda is 0 or 5.
00:40
So, these are the two eigenvalues.
00:42
Next, we find the eigenvectors.
00:47
So, first we find the eigenvector corresponding to lambda equals to 0.
00:52
So, this is nothing but first we find a minus lambda i times the eigenvector must be a 0 vector.
01:03
So, a minus lambda 1 i is 2, 2, 1 times v, let that be x, y, this is 0, 0...