Eigenvalues

• Apr 7th 2012, 08:22 AM
renolovexoxo
Eigenvalues
Suppose A is nonsingular. Prove that the eigenvalues of A^-1 are the reciprocals of the eigenvalues of A.

I was able to come up with a proof for eigenvalues of A being eigenvalues of A^-1, but I'm unsure about how to prove the opposite way to prove the equality.
• Apr 7th 2012, 08:31 AM
Prove It
Re: Eigenvalues
You should know \displaystyle \begin{align*} \mathbf{A}\mathbf{x} = \lambda \mathbf{x} \end{align*}. Now, if \displaystyle \begin{align*} \mathbf{A} \end{align*} is invertible, then \displaystyle \begin{align*} \mathbf{A}^{-1} \end{align*} exists and \displaystyle \begin{align*} \mathbf{A}^{-1}\mathbf{A} = \mathbf{I} \end{align*}. Anyway...

\displaystyle \begin{align*} \mathbf{A}\mathbf{x} &= \lambda \mathbf{x} \\ \mathbf{A}^{-1}\mathbf{A}\mathbf{x} &= \mathbf{A}^{-1}\lambda \mathbf{x} \\ \mathbf{I}\mathbf{x} &= \lambda \mathbf{A}^{-1}\mathbf{x} \\ \mathbf{x} &= \lambda \mathbf{A}^{-1}\mathbf{x} \\ \frac{1}{\lambda}\mathbf{x} &= \mathbf{A}^{-1}\mathbf{x} \end{align*}

Since we have the product of two matrices (in this case \displaystyle \begin{align*} \mathbf{A}^{-1}\mathbf{x} \end{align*}) equal to a scalar multiple of the second matrix (in this case, \displaystyle \begin{align*} \frac{1}{\lambda}\mathbf{x} \end{align*}), that means that \displaystyle \begin{align*} \frac{1}{\lambda} \end{align*} must be an eigenvalue of \displaystyle \begin{align*} A^{-1} \end{align*}.
• Apr 7th 2012, 10:22 AM
renolovexoxo
Re: Eigenvalues
That was the direction I was able to complete. I was asking about the other way. Proving that for an eigenvalue of A^-1 there is a reciprocal in the set of eigenvalues for A.
• Apr 7th 2012, 11:33 AM
Haven
Re: Eigenvalues
Exact same proof, just replace A with A^{-1} and vice versa.
• Apr 8th 2012, 07:26 AM
renolovexoxo
Re: Eigenvalues
Okay, Thank you!