My method is:Suppose that A is a square matrix and is an eigenvalue of A.
(i) Show that is an eigenvalue of for all positive integers n.
(ii) Suppose A is invertible. Show is non-zero and that is an eigenvalue of
I need to show that
which is what I needed to show.
Is this right?
For part 2:
If A is invertible then exists.
Like before .
This is where my method goes awry. cannot equal zero or the matrix A is not invertible.
What am I doing wrong??
This implies that . I thought that if the matrix A was multiplied by an eigenvector then only a multiple of this eigenvector was produced. How do you know that the scalar in front of the eigenvector will be it's eigenvalue? Since you know that it's going to be a multiple of the eigenvector, did you just decide to make this the eigenvalue since the eigenvector can just be scaled down?
Then by definition:
You may have a different convention about this, the other one has:
but the argument will still work with a little adaptation