A is singular. Show that A is diagonalizable over $\Bbb{R}$

Question

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!

Answer 1

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.

Answer 2

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 $\lambda$.