# [SOLVED] eigenvalue help!

• March 24th 2009, 07:45 PM
lord12
[SOLVED] eigenvalue help!
If a is an eigenvalue of A. show that a^(-1) is an eigenvalue of A^-1. I don't know where to start.
• March 24th 2009, 08:34 PM
Prove It
Quote:

Originally Posted by lord12
If x is an eigenvalue of A. show that a^(-1) is an eigenvalue of A^-1. I don't know where to start.

Do you mean that if $a$ is an eigenvalue of $A$, show that $a^{-1}$ is an eigenvalue of $A^{-1}$?
• March 24th 2009, 09:04 PM
lord12
yes!
• March 24th 2009, 09:41 PM
Prove It
Quote:

Originally Posted by lord12
yes!

If $a$ is an eigenvalue of $A$, then

$A\mathbf{x} = a\mathbf{x}$ for some eigenvector $\mathbf{x}$.

Notice that $a\mathbf{x} = aI\mathbf{x}$, where $I$ is the Identity matrix.

Therefore

$A\mathbf{x} - aI\mathbf{x} = \mathbf{0}$

$(A - aI)\mathbf{x} = \mathbf{0}$.

For a nonzero $\mathbf{x}, A - aI = \mathbf{0}$.

Therefore $A = aI$.

If $A$ is nonsingular, then

$A^{-1} = (aI)^{-1}$.

By definition of inverses

$AA^{-1} = aI(aI)^{-1} = I$

$aI(aI)^{-1} = I$

$I(aI)^{-1} = a^{-1}I$

$(aI)^{-1} = a^{-1}I$

$A^{-1} = a^{-1}I$.

Therefore $a^{-1}$ is an eigenvalue of $A^{-1}$.