# Thread: relationship between eigenvalue and inverse

1. ## [SOLVED]relationship between eigenvalue and inverse

Let $\lambda$ be an eigenvalue of the nonsingular matrix A with associated eigenvector $\vec x$. Show that $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$ with associated eigenvector $\vec x$.

Things I thought about: roots of $det(\lambda I_n-A)$ are eigenvalues, inverse is calculated by rref[A|In]

2. Originally Posted by Jskid
Let $\lambda$ be an eigenvalue of the nonsingular matrix A with associated eigenvector $\vec x$. Show that $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$ with associated eigenvector $\vec x$.

Things I thought about: roots of $det(\lambda I_n-A)$ are eigenvalues, inverse is calculated by rref[A|In]
you know that there exists some $x$ such that $Ax=\lambda x$ so that $x=A^{-1}\lambda x$ or $\frac{1}{\lambda}x=A^{-1}x$ (where we used the fact that $A$ is invertible thus $\lambda\in R^\times$)

3. Originally Posted by Drexel28
$x=A^{-1}\lambda x$ or $\frac{1}{\lambda}x=A^{-1}x$
But you can't divide by a vector or matrices, in this case $\lambda$? Are you allowed to multiply by its reciprocal?

4. Originally Posted by Jskid
But you can't divide by a vector or matrices, in this case $\lambda$? Are you allowed to multiply by its reciprocal?
Huh? I'm sorry I must be misuderstanding you. Isn't $\lambda$ a scalar?

5. Originally Posted by Drexel28
Huh? I'm sorry I must be misuderstanding you. Isn't $\lambda$ a scalar?
I thought it was a vector but it turned out I was confused