00:01
We are given a matrix a.
00:02
We are asked to show that this matrix is non -defective, and then to find e to the a -t.
00:13
So our matrix a is the 2x2 matrix with entries 1 ,203, and therefore the characteristic polynomial of a, which i'll call p of lambda, is equal to 1 minus lambda, 3 minus lambda minus 2 times 0, just simply 0.
00:50
So when our characteristic polynomial is equal to 0, we have our characteristic equation.
00:57
And we see that the solutions to this equation are the eigenvalues lambda 1 equals 1, and lambda 2 equals 3.
01:12
If v1 is an eigenvector at a with eigenvalue the lambda 1.
01:18
So first of all, because we have eigenvalues, it follows that a is non -defective.
01:39
And if we have an identector v1 with eigenvalue lambda 1, then it follows that a minus lambda 1 i v1 is equal to the 0 vector.
01:58
So this implies that a minus identity matrix, which is simply 0 to 0 ,0, 2, times our eigenvector with the.
02:23
V1 and 1 2 is equal to the 0 matrix.
02:31
And this then implies that v1 is equal to 0.
02:41
This is what both equations tell us.
02:44
And therefore we can have that v11 can be any real number.
02:49
Let's just pick v1 to v1.
02:53
Then we obtain that our function, that our eigenvector v1 is the vector 1 -0.
03:05
If v2 is an eigenvector for eigenvalue lambda 2, then we have that a minus lambda 2 ib2 is equal to the 0 vector.
03:20
And since lambda 2 is equal to 3 follows that this matrix is going to be 1 minus 3 or negative 2, 0, and 3 minus 3, which is another 0, times our vector v2, which is components v21, b2, is equal to the 0, is equal to the 0 matrix.
04:01
The zero vector.
04:07
So we get that negative 2 v21 plus 2 v2 is equal to 0.
04:20
Therefore we have that v22 is equal to v21...