eigenvalues and invertible matrix

Hello I'm having some issues with this current problem and I'm hoping that someone can help.

The problem states: Given A is an n x n invertible matrix such that lamda is equal to 1 is not an eigenvalue, I is the identity matrix, (A^1) is the inverse matrix of A. Explain why

a) (I-A^-1) is invertible

b) (A^-1)(I-A^-1) = (A-I)^-1

c) ((I-A^-1)^-1)(A^-1) = (A-I)^-1

Re: eigenvalues and invertible matrix

Quote:

Originally Posted by

**nivek0078** The problem states: Given A is an n x n invertible matrix such that lamda is equal to 1 is not an eigenvalue, I is the identity matrix, (A^1) is the inverse matrix of A. Explain why a) (I-A^-1) is invertible

Quote:

b) (A^-1)(I-A^-1) = (A-I)^-1

That is false. Choose for example

Re: eigenvalues and invertible matrix

A matrix is invertible if and only if 0 is not an eigenvalue. For any eigenvalue, , there exist some non-zero vector such that . In that case, and so . In other words, is an eigenvalue of A if and only if is an eigenvalue of , with the same eigenvectors. In particular, if 1 is not an eigenvalue of A, 1/1= 1 is not an eigenvalue of . (The fact that A is invertible means it does not have 0 as an eigenvalue so that reciprocal always exists.)

Then . The eigenvalues of are of the form . The fact that 1 is not an eigenvalue of A means 1 is not an eigenvalue of and so 1- 1= 0 is not an eigenvalue of which means that is invertible.

Unless I have made a silly computation error (always possible) (b) and (c) are NOT true! As a counter-example, look at .