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Given matrix $A$ over $\mathbb R$, its characteristic polynomial is $p(x)=(x+3)^2(x-1)(x-5)$ and $\operatorname{rank}(A+2I)+\operatorname{rank}(A+3I)+\operatorname{rank}(A-5I)=9$ prove that $A$ is diagonalizable.

The polynomial can be factored into linear terms then we only need to prove that the corresponding algebraic and geometric multiplicities are equal. For $1$ and $5$ the multiplicities are equal.

Also:

  1. $\operatorname{rank}(A+2I)=\operatorname{rank}(-(-A-2I)=4$ because $-2$ is not an eigenvalue of $A$;
  2. $\operatorname{rank}(A-5I)=\operatorname{rank}(-(5I-A))=3$ because geometric multiplicity of $5$ is $1$.

From the given and from 1. and 2. we can conclude that: $$ \operatorname{rank}(A+3I)=2\Rightarrow \operatorname{rank}(-3I-A)=2. $$ Therefore the geometric multiplicity of $3$ is $2$ and $A$ is diagonalizable.

I'm aware that $A$ is $4\times 4$ matrix yet I cannot grasp the points 1), 2) and the conclusion. Please explain those.

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    $\begingroup$ I think there are typos in the solution you have. For (1) it should be rank of $A+2I$ is $4$ because $-2$ is not an eigen value. Similarly in (2) it should be rank of $A-5I$ and so on. $\endgroup$ – Anurag A Jul 11 '17 at 19:02
  • $\begingroup$ @AnuragA I think you're right, it was a mistake in the solution. But then we have that $\operatorname{rank}(A+3I)=3$ but it has to be $2$ in order to for the geometric multiplicity to equal the algebraic. $\endgroup$ – Yos Jul 11 '17 at 19:18
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One way to show that $A$ is diagonalizable is to prove that there exists a basis of eigenvectors of $A$. So your argument shows that there exists two linearly independent eigenvectors associated to the eigenvalue $-3$ (since $\operatorname{rank} (A+3I) = 2$, we also have $\operatorname{dim ker} (A+3I) = 2$), and one for each of the other two. So you get 4 linearly independent eigenvectors and then $A$ is diagonalizable.

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  • $\begingroup$ but why $\operatorname{rank} (A+3I) = 2$? $\endgroup$ – Yos Jul 11 '17 at 19:24
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    $\begingroup$ You can use the information that $\operatorname{rank}(A+2I)+\operatorname{rank}(A+3I)+\operatorname{rank}(A−5I)=9$ to get that $4 + \operatorname{rank} ⁡(A+3I) + 3=9.$ $\endgroup$ – Diego Marcon Jul 11 '17 at 19:25

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