1. Hard eigenvector/eigenvalue proof

Struggling with this:

I started by saying A is invertible if and only if Ax = 0 has a trivial solution.

So there exists x = 0 where Ax = 0

0 = lamda. 0 = 0

I'm not quite sure this proves that lamda is not equal to zero though?!

2. $\displaystyle \displaystyle Ax=\lambda x$

Multiple both sides by A inverse

$\displaystyle \displaystyle A^{-1}Ax=A^{-1}\lambda x\rightarrow x=A^{-1}\lambda x\rightarrow \frac{1}{\lambda}x=A^{-1}x$

3. Originally Posted by chr91

Struggling with this:

I started by saying A is invertible if and only if Ax = 0 has a trivial solution.
You've pretty much got everything backwards here. Ax= 0 for x= 0 for any A. What is true is that A is invertible if and only if there is NO non-trivial solution. That is if x= 0 is the only x such that Ax= 0.

Saying that $\displaystyle \lambda$ is an eigenvalue of A means there exist non-zero x such that $\displaystyle Ax= \lambda x$. Use the fact that A is invertible to multiply both sides by $\displaystyle A^{-1}$.

So there exists x = 0 where Ax = 0

0 = lamda. 0 = 0

I'm not quite sure this proves that lamda is not equal to zero though?!

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