eigenvector of A+I

**ecc5** · December 9th 2009, 02:33 PM

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!

**Defunkt** · December 9th 2009, 03:11 PM

Suppose x is an eigenvector of $A$ , ie. $\exists \lambda \in \mathbb{R} \ : \ Ax = \lambda x \Rightarrow (A + I)x = Ax + x = ...$
And finish.

**kjchauhan** · December 9th 2009, 03:17 PM

If $x$ is an eigen vector of a matrix $A$ corresponding to an eigen value $\lambda$

$\therefore Ax=\lambda x$

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

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

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

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

Proved.

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