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Let $A\in M_{10}(\Bbb{R})$ be singular matrix such that $\mathrm{rank}(I-A)=4 \ \mathrm{and} \ \mathrm{rank}(3I-A)=7$. Show that $A$ is diagonalizable over $\Bbb{R}$. Which diagonal matrix $A$ is similar to?

My try:

We know that $(I-A)\in M_{10}(\Bbb{R})$, and $\mathrm{rank}(I-A)=4$, hence $\mathrm{det}(I-A)=0$, so $1$ is an eigenvalue of $A$. For the same reasoning, $3$ is an eigenvalue of $A$.

Also, $A$ is singular, hence $0$ is an eigenvalue of $A$.

Unfortunately, I don't know how to find the multiplicity of any eigenvalue and how t show that is diagonalizable.

Please help, thank you!

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Since $\mathrm{rank}(I-A)=4$, we get that eigenspace for eigenvalue $1$ has dimension $6$, i.e. $1$ has geometric multiplicity $6$; $\gamma_A(1)=6$. Similarly, $\gamma_A(3)=3$. Also $\gamma_A(0)\geq 1$.

Since algebraic multiplicity is always greater than geometric, $\mu_A(\lambda)\geq\gamma_A(\lambda)$, for $\lambda\in\{0,1,3\}$, and $10\geq\mu_A(0)+\mu_A(1)+\mu_A(3)\geq\gamma_A(0)+\gamma_A(1)+\gamma_A(3)\geq 1+6+3=10$, we get that algebraic and geometric multiplicities are equal.

Since their sum is $10$ we conclude that $0$, $1$ and $3$ are the only eigenvalues. Since all algebraic and geometric multiplicities are equal, we conclude that $A$ is diagonalizable. And diagonal matrix similar to $A$ has one $0$, six $1$ and three $3$ on the diagonal.

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  • $\begingroup$ You're welcome. $\endgroup$ – SMM Jan 14 '15 at 18:19
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Hint: Note that for each eigenvalue $\lambda$ one has $\text{null}(\lambda I-A)=10-\text{rank}(\lambda I-A)$ and recall that $\text{null}(\lambda I-A)$ is the geometric multiplicity of $\lambda$.

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  • $\begingroup$ If I recall correctly, you mean that $1$ has geometric multiplicity of $6$, $3$ has geometric multiplicity of $3$ and $0$ has geometric multiplicity of $1$ and therefore the algebraic multiplicity of any of them equals the geometric multiplicity, thus A is diagonalizable. Correct? $\endgroup$ – Galc127 Jan 14 '15 at 18:05
  • $\begingroup$ @Galc127 Yes, but why does the geometric multiplicity equal the algebraic multiplicity? Missing that justification might cost you some points in an exam. $\endgroup$ – Git Gud Jan 14 '15 at 18:13
  • $\begingroup$ I know how to justify it, thank you for your answer. $\endgroup$ – Galc127 Jan 14 '15 at 18:15
  • $\begingroup$ @Galc127 You're welcome. $\endgroup$ – Git Gud Jan 14 '15 at 18:16

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