eigenvector of A+I

• Dec 9th 2009, 02:33 PM
ecc5
eigenvector of A+I
I need to either prove or give a counterexample of the following: If x is an eigenvector of A, then x is also an eigenvector of A+I. I tried several different matrices, and this proposition does seem to be true. However, I have no idea how to prove it.

Help!
• Dec 9th 2009, 03:11 PM
Defunkt
Suppose x is an eigenvector of $\displaystyle A$, ie. $\displaystyle \exists \lambda \in \mathbb{R} \ : \ Ax = \lambda x \Rightarrow (A + I)x = Ax + x = ...$
And finish.
• Dec 9th 2009, 03:17 PM
kjchauhan
If $\displaystyle x$ is an eigen vector of a matrix $\displaystyle A$ corresponding to an eigen value $\displaystyle \lambda$

$\displaystyle \therefore Ax=\lambda x$

$\displaystyle \therefore Ax+x=\lambda x+x$

$\displaystyle \therefore (A+I)x=(\lambda +1)x$

$\displaystyle \therefore (A+I)x=\lambda_1 x$, where $\displaystyle \lambda_1=\lambda+1$

$\displaystyle \therefore x$ is an eigen vector of a matrix $\displaystyle A+I$ corresponding to an eigen value $\displaystyle \lambda_1=\lambda+1$.

Proved.