00:01
In this exercise, we have to say if the matrix a is diagonalizable.
00:07
For that, we're going to take a look at its eigenvalues.
00:13
So in this case, the determinant of a minus lambda i, we can compute it by using minor zanco factors, but it is easy to do that by expanding along the last row.
00:30
So we have 1 minus lambda times minus 1 minus lambda times minus 1 minus lambda times minus 1 minus lambda squared.
00:50
From here we can see that there are two different solutions for this equation.
01:00
Lambda 1 equals 1 and lambda 2 equals minus 1.
01:05
And notice that lambda 2 has algebraic multiplicity of 3.
01:16
So this matrix is going to be diagonalizable if we can show that the, if we can prove, we can show that the eigen space corresponding to lambda 2 has dimension 3.
01:37
So define the eigenvectors corresponding to lambda 2, we have to think about the following system.
01:48
A minus lambda i times v equals 0, where lambda equals minus 1.
01:56
And this corresponds to the augmented matrix, 0 -1 -0 -0 ,000, 0...