eigenvalues and invertible matrix

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September 20th 2012, 04:55 AM
nivek0078

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
September 20th 2012, 06:47 AM
FernandoRevilla

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

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

Quote:

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

That is false. Choose for example $A=\mbox{diag }(2,3).$
September 20th 2012, 07:07 AM
HallsofIvy

Re: eigenvalues and invertible matrix

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

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

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