prove that matrix is diagonal by matrix rank and eigenvalue rank

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$A$ is matrix $9\times9$ with rank of $5$, there is rank$(A-3I)=5$, the matrix has another eigenvalue of 5. I need to prove that $A$ is diagonal and find the similar diagonal matrix of $A$. I'm stuck, I thing that the geometry multiplicity of $(A-3I)$ is $4$, because $9$ - rank$(A-3I) = 4$. Is there any use of Jordan blocks here maybe? but I don't know if its true and how to continue from here.

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Firstly, you can't show that $A$ is diagonal just looking at ranks, but I think you mean diagonalizable.

The fact that $A$ has rank $5$ means that it has an eigenvalue of $0$ with geometric mulitiplicity $4$. Since $(A-3I)$ has rank $5$, that means the eigenvalue $3$ has geometric multiplicity $4$. You know that $5$ is an eigenvalue, and it must have multiplicity $1$ since the sum of the multiplicities must equal $9$. When the sum of the geometric multiplicities is maximal, the matrix is diagonalizable.

I think this is all easiest to see by first putting $A$ in Jordan canonical form as you suggest. Then the fact that rank is $5$ immediately shows you there's a block of $0$s, and that $A-3I$ similarly has a block of zeros, which means the Jordan block for eigenvalue $3$ is diagonal. At that point, you only have one dimension left for eigenvalue $5$, so its block is diagonal as well.