# [SOLVED] eigenvalue help!

• Mar 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.
• Mar 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 $\displaystyle a$ is an eigenvalue of $\displaystyle A$, show that $\displaystyle a^{-1}$ is an eigenvalue of $\displaystyle A^{-1}$?
• Mar 24th 2009, 09:04 PM
lord12
yes!
• Mar 24th 2009, 09:41 PM
Prove It
Quote:

Originally Posted by lord12
yes!

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

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

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

Therefore

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

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

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

Therefore $\displaystyle A = aI$.

If $\displaystyle A$ is nonsingular, then

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

By definition of inverses

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

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

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

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

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

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