WTS: A scalar $\lambda$ is an eigenvalue of a matrix $A$ $\iff$ $\det(\lambda I-A)=0$
My proof: Assume $\lambda$ is an eigenvalue of A. So $Av=\lambda v$ for a non-zero vector, v.This is equivalent to $(\lambda I-A)v=0$. However, since v is non-zero, $\det(\lambda I-A) = 0$. Assume $\det(\lambda I-A)=0$ that means that for any vector $b$, the equation $(\lambda I-A)v=b$ has no unique solution, therefore there is a non-zero solution.
May please tell me if the proof is correct? Does it make sense? How can I improve it?
Best Answer
The first part ($\lambda$ eigenvalue $\implies$ $\det(\lambda\operatorname{Id}-A)=0$) is correct. For the other implication, you should not have introduced that vector $b$. Just say that$$\det(\lambda\operatorname{Id}-A)=0\implies\text{ there is a non-zero vector $v$ such that }(\lambda\operatorname{Id}-A)v=0.$$