# Thread: eigenvalues and invertible matrix

1. ## 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

2. ## Re: eigenvalues and invertible matrix

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
\displaystyle \displaystyle\begin{aligned}I-A^{-1}\mbox{ is not invertible }&\Rightarrow \det (I-A^{-1})=0\\&\Rightarrow\det (A^{-1}-1I)=0\\&\Rightarrow 1\mbox{ is eigenvalue of }A^{-1}\\&\Rightarrow 1^{-1}=1\mbox{ is eigenvalue of }(A^{-1})^{-1}=A\mbox{ (Absurd).} \end{aligned}

b) (A^-1)(I-A^-1) = (A-I)^-1
That is false. Choose for example $\displaystyle A=\mbox{diag }(2,3).$

3. ## Re: eigenvalues and invertible matrix

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

Then $\displaystyle (I- A^{-1})v= v- A^{-1}v= v- \frac{1}{\lambda}v= (1- \frac{1}{\lambda})v$. The eigenvalues of $\displaystyle I- A^{-1}$ are of the form $\displaystyle 1- \frac{1}{\lambda}$. The fact that 1 is not an eigenvalue of A means 1 is not an eigenvalue of $\displaystyle A^{-1}$ and so 1- 1= 0 is not an eigenvalue of $\displaystyle I- A^{-1}$ which means that $\displaystyle I- A^{-1}$ is invertible.

Unless I have made a silly computation error (always possible) (b) and (c) are NOT true! As a counter-example, look at $\displaystyle A= \begin{pmatrix}2 & 1 \\ 0 & 2\end{pmatrix}$.