Prove that a matrix with a given characteristic polynomial is diagonalizable

Question

Matrix $A$ is defined over real number.

Characteristic polynomial : $p(x)=(x+3)^2(x-1)(x-5)$

It also known that :

$$\text{rank}(A+2I)+\text{rank}(A+3I)+\text{rank}(A-5I)=9$$

• prove $A$ diagonalize.

My solution

• $-3,1,5$ are eigenvalues, using the characteristic polynomial we can conclude that matrix $A$ is $4 \times 4$

• Since eigenvalue $1,5$ has algebraic multiplicity of $1$, we can conclude that geometric multiplicity is also $1$ hence:

$$\text{rank}(I+A)=3$$

$$\text{rank}(5I-A)=3$$

I don't find a way to continue from here.

Answer 1

Picking up from where you left off.

From $\text{rank}(A-5I)=3$ and from $\text{rank}(A+2I)+\text{rank}(A+3I)+\text{rank}(A-5I)=9$ you get $$\text{rank}(A+2I)+\text{rank}(A+3I)=6.$$

Now prove that $A+2I$ is invertible. What does that tell you about the rank of $A+2I$?

Infer that $\text{rank}(A+3I)=2$. What does that tell you about the geometric multiplicity of $-3$?

Conclude.

Answer 2

Hint: with such a characteristic polynomial, the matrix is congruent to either $$ \left( \begin{array}{ccc} -3 & 1 & 0 &0\\ 0 & -3 & 0 &0\\ 0 & 0 & 1 &0\\ 0 & 0 & 0 &5\end{array} \right) $$ or $$ \left( \begin{array}{ccc} -3 & 0 & 0 &0\\ 0 & -3 & 0 &0\\ 0 & 0 & 1 &0\\ 0 & 0 & 0 &5\end{array} \right)$$

Then in each case, compute $P(A)$ in this new base.

Answer 3

Hint: The rank of $A - \lambda I$ is $n$ minus (the number of Jordan blocks of $A$ that are associated with $\lambda$).