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

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.

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 $$\lambda I-A=\lambda I-SDS^{-1}=\lambda SS^{-1}-SDS^{-1}=S(\lambda I-D)S^{-1}.$$

Observe that the problem for diagonalization comes from the eigenvalues $\;0,\,-3\;$ , for which it isn't sure their geometric multiplicity = algebraic multiplicity.

If we denote by $\;V_\lambda\;$ the eigenspace corresponding to the eigenvalue $\;\lambda\;$ ,then it must be that

$$\;\dim V_0=4\,,\,\,\dim V_{-3}=2\iff \dim V_0+\dim V_{-3}=6$$

the last double arrow and inequality following from the almost trivial fact that $\;V_0\cap V_{-3}=\{0\}\;$

But by the dimensions theorem we get

$$\dim V_0=\dim\ker A=4\implies\text{rank}\,A=7-\dim\ker A=7-\dim V_0=3$$ and also

$$\dim V_{-3}=\dim\ker(A+3I)\le2\implies 5\le\text{rank}(A+3I)\le6$$

where we used in the last inequality that $\;A+3I\;$ is singular, so now do your math since

$$\text{rank}\,A+\text{rank}\,(A+3I)=8\implies \text{rank}\,(A+3I)=5\implies\ldots$$