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I read some book which says:

$$A\vec{v} = \lambda\vec{v} \iff (A-\lambda I)\vec{v} = 0$$

and it claims such a vector $\vec{v}$ exists iff $(A-\lambda I)\vec{v}$ is singular which means $\operatorname{det}(A-\lambda I) = 0$

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It's bog-standard that $Bx=0$ has a nonzero solution if and only if $\det B=0$. – Gerry Myerson Mar 2 '13 at 11:28
@GerryMyerson how can one show that? – 0x90 Mar 2 '13 at 11:33
By using elementary row operations to reduce to row-echelon form. – Gerry Myerson Mar 2 '13 at 11:36
up vote 2 down vote accepted

Definition: a non-zero vector $\,v\,$ is an eigenvector of $\,A\,$ iff

$$Av=\lambda v\,\,\,,\,\,\lambda\in\Bbb F\Longleftrightarrow (A-\lambda I)v=0\Longleftrightarrow$$

the operator (or matrix) $A-\lambda I\,$ is singular as

$$\,0\neq v\in\ker A-\lambda I\Longrightarrow \ker (A-\lambda I)\neq\{0\}\Longleftrightarrow \det(A-\lambda I)=0$$

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When $\det(A-\lambda I)=0$ it means there exists an $v\neq 0$ so that $(A-\lambda I)v=0$, if not, $(A-\lambda I)$ would be injective and hence bijective, so $\det (A-\lambda I)\neq 0$.

So we know $$(A-\lambda I) v= A v - \lambda v=0 \iff Av=\lambda v$$

On the other hand if $Av=\lambda v$ with $v\neq 0$ you know that $$(A-\lambda I)v=0$$ and hence the kernel isn't trivial, so $\det(A-\lambda I)=0$

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