Show that if A is invertible and diagonalizable,then A^−1 is

[charlies1902]

Show that if A is invertible and diagonalizable,
then A^−1 is diagonalizable. Find a 2 ×2 matrix
that is not a diagonal matrix, is not invertible, but
is diagonalizable.


Alright, I am having some trouble with the first part.
So far, I have this:
If A is diagnolizable then
A=PDP^-1 where P is the matrix who's columns are eigenvectors and D is the diagonal matrix of eigevenvalues of A.


(A)^-1=(PDP^-1)^-1
A^-1=PDP^-1

How's that?

 

[Dick]

Show that if A is invertible and diagonalizable,
then A^−1 is diagonalizable. Find a 2 ×2 matrix
that is not a diagonal matrix, is not invertible, but
is diagonalizable.Alright, I am having some trouble with the first part.
So far, I have this:
If A is diagnolizable then
A=PDP^-1 where P is the matrix who's columns are eigenvectors and D is the diagonal matrix of eigevenvalues of A.(A)^-1=(PDP^-1)^-1
A^-1=PDP^-1

How's that?

Just because D is diagonal doesn't mean D=D^(-1). And after you've fixed that, how do you know D is invertible?

 

[charlies1902]

Oh oops.
So
A^-1=P * D^-1 * P^-1

hmm, does D have to be invertible?
Can't you have eigen values of 0 and 2
so D looks like this:
0 0
0 2
which is not invertible?

 

[Dick]

Oh oops.
So
A^-1=P * D^-1 * P^-1

hmm, does D have to be invertible?
Can't you have eigen values of 0 and 2
so D looks like this:
0 0
0 2
which is not invertible?

They told you A is invertible. Doesn't that mean D has to be invertible? Can you prove that?

 

[charlies1902]

They told you A is invertible. Doesn't that mean D has to be invertible? Can you prove that?

I'm confused on why you would have to prove that D is invertible and if it is always invertible.
I'm calling D the diagonal matrix who's diagonal elements are the eigenvalues of A.

 

[Dick]

I'm confused on why you would have to prove that D is invertible and if it is always invertible.
I'm calling D the diagonal matrix who's diagonal elements are the eigenvalues of A.

D isn't invertible if it has a zero on the diagonal. But if it does then A has a zero eigenvalue and it's not invertible. I'm not sure whether you have to prove that or whether you can just say it. But it's not hard to prove.