If $p_A(x) = x^4 (x+3)^2 (x-4)$ then $A$ is diagonalizable iff $\operatorname{Rank}(A) + \operatorname{Rank}(-3I-A) = 8$

Question

Given the Characteristic polynomial of a matrix $A$ is
$$
p(x) = x^4 (x+3)^2 (x-4),
$$
show that $A$ is diagonalizable if and only if
$$
\operatorname{Rank}(A) + \operatorname{Rank}(-3I-A) = 8.
$$

---

Given: $A$ is diagonalizable
Prove: $\operatorname{Rank}(A) + \operatorname{Rank}(-3I-A) = 8$
(Help is needed in the other way around)
The geometric multiplicity equals the algebraic multiplicity, hence the diagonal matrix $D$ will look like
$$
D
= \begin{pmatrix}
-3 & & & & & & \\
& -3 & & & & & \\
& & 4 & & & & \\
& & & 0 & & & \\
& & & & 0 & & \\
& & & & & 0 & \\
& & & & & & 0
\end{pmatrix}
$$
Since $D$ and $A$ are similar matrices, their rank must be the same:
$$
\operatorname{Rank}(A) = \operatorname{Rank}(D) = 3.
$$
Also
$$
-3I – A
= \begin{pmatrix}
0 & & & & & & \\
& 0 & & & & & \\
& & -7 & & & & \\
& & & 0 & & & \\
& & & & -3 & & \\
& & & & & -3 & \\
& & & & & & -3
\end{pmatrix}
$$
and therefore
$$
\operatorname{Rank}(A) + \operatorname{Rank}(-3I-A)
= 3 + 5
= 8.
$$
Perfect.
(Is it okay to place $D$ instead of $A$ in $\operatorname{Rank}(-3I-A)$? The rank will be the same, is it not?)

---

Given: $\operatorname{Rank}(A) + \operatorname{Rank}(-3I-A) = 8$
Prove: $A$ is diagonalizable

No clue really…
I was thinking since geometric multiplicity is less or equal to algebraic multiplicity, I was thinking of finding all possible $D$'s and show that the one that makes the statement $\operatorname{Rank}(A) + \operatorname{Rank}(-3I-A) = 8$ hold makes $A$ diagonal.

Any hint is appreciated.

Answer 1

Look at the possible Jordan form of $A$ $$ \begin{bmatrix} -3 & * & 0 & 0 & 0 & 0 & 0\\ 0 & -3 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 4 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & * & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & * & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & *\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$ (* elements are $0$ or $1$) to deduce that

1. $\operatorname{rank}A\ge 3$ and $\operatorname{rank}A=3$ $\iff$ the block with eigenvalue $0$ is diagonal.
2. $\operatorname{rank}(-3I-A)\ge 5$ and $\operatorname{rank}(-3I-A)=5$ $\iff$ the block with eigenvalue $-3$ is diagonal.

Therefore, $\operatorname{rank}A+\operatorname{rank}(-3I-A)\ge 8$ and $\operatorname{rank}A+\operatorname{rank}(-3I-A)=8$ $\iff$ $\operatorname{rank}A=3$, $\operatorname{rank}(-3I-A)=5$. Then...

P.S. To answer your question "is it ok to place $D$ instead of $A$ in $\operatorname{rank}(-3I-A)$": yes, it is ok, since $-3I-A$ and $-3I-D$ are similar as well $$ \lambda I-A=\lambda I-SDS^{-1}=\lambda SS^{-1}-SDS^{-1}=S(\lambda I-D)S^{-1}. $$