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This question already has an answer here:

Given $A \in \mathbb{R}^{n \times n}$ and $\text{rank}(A) = 1$. By working only on real field, show that $A$ is diagonalizable if and only if $\text{tr}(A) \neq 0$. Here, $\text{tr}(A)$ is the sum of all eigenvalues of $A$.

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marked as duplicate by Marc van Leeuwen linear-algebra Apr 25 '15 at 11:55

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  • $\begingroup$ What are your thoughts on the problem? $\endgroup$ – Servaes May 23 '14 at 15:14
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    $\begingroup$ @Servaes Prove that 0 is an eigenvalue of $A$ and it has geometric multiplicity of $n-1$. If $A$ is diagonalizable, $0$ also has algebraic multiplicity of $n-1$, so there is another nonzero eigenvalue, so tr(A) is not zero. On the other hand, if $A$ is not diagonalizable, then $0$ must have algebraic multiplicity of $n$, so that tr(A) must be zero. Is it true? $\endgroup$ – Kevin Limanta May 23 '14 at 15:19
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    $\begingroup$ Seems fine to me. $\endgroup$ – Servaes May 23 '14 at 15:24