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True of False: If $A$ is an $n\times n$ diagonalizable matrix, then $0$ can not be in eigenvalue of $A$

I believe this is false, because you can have a diagonalizable matrix with eigenvalues of zero. Diagonalization can fail if there are repeated eigenvalues. I am not able to think of a subtle counter-example or theorem that would make it more direct.

my counter-example:

$3\times 3$ matrix with $\lambda = 0,1$ for $p(\lambda) = \lambda(\lambda^2-1)=0$

$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$

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    $\begingroup$ Just take any diagonal matrix with a zero in the diagonal. $\endgroup$ – Student Feb 6 '17 at 7:30
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    $\begingroup$ I would just take the null matrix. The characteristic polynomial for your example is incorrect. $\endgroup$ – Git Gud Feb 6 '17 at 7:36
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Take the Null matrix as an example

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Take any vector space V with basis $v_1,v_2..v_n$. Now define $ A : V \to V$ to be $A(v_1) = 0 , A(v_i)= v_i \forall i \neq 1 , i \le n$.Now the matrix of A shall be with 1 0 and all 1 ones in the diagonal which is possible.

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