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I read that a Matrix $A$ has the eigenvalue $0$ if and only if $\ker(A) \neq \{0\}$.

Why so?

Edit Okay actually I figured it out myself. If $0$ is an eigenvalue of a matrix $A$ then $\det(A)=0$ and then $A$ is not invertible, therefore the rows are not linearly independent: $\ker(A) \neq0$.

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  • $\begingroup$ Here is a related question. This shows that $A \text{ has eigenvalue } 0 \implies A \text{ is not invertible}$ and from this, we can deduce $\ker(A) \neq \{\mathbf 0\}$. $\endgroup$
    – Noble Mushtak
    Commented Jun 18, 2016 at 12:58
  • $\begingroup$ Yep I just needed a little longer to figure that out, long time since I did some linear algebra..;D $\endgroup$
    – Tesla
    Commented Jun 18, 2016 at 12:59

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$A$ has eigenvalue $0$ $\iff$ $\exists v \neq0$ such that $Av=0v=0$ $\Rightarrow 0\neq v\in \ker (A) \Rightarrow \ker (A) \neq\{0\} $ .

On the other hand : $\ker (A) \neq\{0\} \Rightarrow \exists v\neq0$ such that $v \in \ker (A)\Rightarrow \exists v\neq0$ such that $Av=0=0v$. So $0$ is eigenvalue of $A$.

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Because there exists non-trivial vector $x$ such that $Ax = 0$, so then $Ax = 0 *x$ and $0$ is an eigenvalue.

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