# Proving

• May 6th 2009, 03:38 AM
Roam
Proving
(a) Show that if P is a non-singular nxn matrix then $\displaystyle \lambda$ is an eigenvalue of A if and only if, $\displaystyle \lambda$ is an eigenvalue of $\displaystyle PAP^{-1}$.

(b) Show that if P is a non-singular nxn matrix such that $\displaystyle A^3=o_{n}$. Show that the only possible eigenvalue for A is $\displaystyle \lambda = 0$. Here $\displaystyle 0_{n}$ is the nxn zero matrix.

For part (a) I think I need to manipulate the expression, here's what I think:

$\displaystyle PAP^{-1} (x) = \lambda x$

$\displaystyle A(PxP^{-1}) = \lambda x$

$\displaystyle A(x)PP^{-1} = A(x) I_{n} = A(x) = \lambda x$

Can anyone help me with part (b)?
I know that lambda must satisfy $\displaystyle Ax=\lambda x$

we have: $\displaystyle A.A.A (x) = \lambda x$

I don't know what to do with this & I'm I'm really stuck here...
• May 6th 2009, 04:29 AM
NonCommAlg
Quote:

Originally Posted by Roam
(a) Show that if P is a non-singular nxn matrix then $\displaystyle \lambda$ is an eigenvalue of A if and only if, $\displaystyle \lambda$ is an eigenvalue of $\displaystyle PAP^{-1}$.

if $\displaystyle Ax=\lambda x,$ put $\displaystyle y=Px.$ then $\displaystyle PAP^{-1}y=\lambda y.$ conversely if $\displaystyle PAP^{-1}x=\lambda x,$ then put $\displaystyle P^{-1}x=y$ and see that $\displaystyle Ay=\lambda y.$ in each case it's important to note that $\displaystyle x \neq 0$ iff $\displaystyle y \neq 0.$

Quote:

(b) Show that if P is a non-singular nxn matrix such that $\displaystyle A^3=o_{n}$. Show that the only possible eigenvalue for A is $\displaystyle \lambda = 0$. Here $\displaystyle 0_{n}$ is the nxn zero matrix.

if $\displaystyle Ax=\lambda x,$ then $\displaystyle A^2x=A\lambda x=\lambda Ax=\lambda^2 x.$ repeat this to get $\displaystyle 0=A^3x=\lambda^3x,$ and thus $\displaystyle \lambda = 0,$ because $\displaystyle x \neq 0.$