5
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.