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Suppose $A\in\mathbb R^{n\times n}$, $a,b\in\mathbb R$, $a\ne b$ and $$(A+aI_n)(A+bI_n)=0$$ where $I_n$ denotes the identity. Prove
1). $\text{rank}(A+aI_n)+\text{rank}(A+bI_n)=n$
2). $A$ is similar to a diagonal matrix.

Since I'm still struggling with the first, I haven't yet seen the relation between the two.
Here is my attempt for the first:
Clearly, $$\mathscr C[(A+bI_n)]\subset \text{Ker}(A+aI_n)$$, where $\mathscr C(\cdot)$ denotes column space. And note that $$\begin{align} \dim\text{Ker}(A+aI_n)+\dim\text{Im}(A+aI_n) &=\dim\text{Ker}(A+aI_n)+\dim\mathscr C[(A+aI_n)] \\&=\dim\text{Ker}(A+aI_n)+\text{rank}(A+aI_n) \\&=n \end{align}$$ Then all I need to do is to show $$\dim\mathscr C[(A+bI_n)]=\dim \text{Ker}(A+aI_n)$$ or equivalently, $$\mathscr C[(A+bI_n)]= \text{Ker}(A+aI_n)$$ and that's where I get stuck.
Any help or hint will be appreciated. Best regards!

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  • $\begingroup$ Do you know about eigenvalues? Jordan form? Minimal polynomials? $\endgroup$ – Omnomnomnom Apr 9 '15 at 17:14
  • $\begingroup$ @Omnomnomnom I know a little. But just a little. And I think maybe there are simpler ways without those advanced skills to solve it, since the problem appear in the exam for elementary LA, we students who took it had only learnt "real-number" LA. $\endgroup$ – Vim Apr 9 '15 at 17:20
  • $\begingroup$ I'll try not to lean on those, then. $\endgroup$ – Omnomnomnom Apr 9 '15 at 17:21
  • $\begingroup$ @Omnomnomnom and months before a friend of mine showed me his solution, which was extremely clever, and did not even involve the eigenvalues. But the pity is I forget it. $\endgroup$ – Vim Apr 9 '15 at 17:22
  • $\begingroup$ @Omnomnomnom Thanks! That's very kind of you. $\endgroup$ – Vim Apr 9 '15 at 17:22
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We start by noting $$ \mathscr C[(A+bI)]\subset \text{Ker}(A+aI) $$ as you say. It follows that $$ \operatorname{rank}(A + bI) \leq \dim\ker(A + aI) $$ By the rank-nullity theorem, we have

$$ \operatorname{rank}(A + bI) \leq n - \operatorname{rank}(A + aI) \implies\\ \operatorname{rank}(A + bI) + \operatorname{rank}(A + aI) \leq n $$ Up to some slight differences, this is basically what you have so far

It now remains to be shown that $$ \operatorname{rank}(A + bI) + \operatorname{rank}(A + aI) \geq n $$ In order to do this, we note that $$ \ker(A + aI) \cap \ker(A + bI) = \{0\} $$ I will leave the proof of this fact to you (at least give it a shot). From there, it follows that $$ \dim \ker(A + aI) + \dim \ker(A + bI) \leq n \implies\\ (n - \operatorname{rank}(A + aI)) + (n - \operatorname{rank}(A + bI)) \leq n $$ and I will let you take it from there.

We now have enough to answer the first part of the question.


First, remember $(A + aI)$ and $(A + bI)$ have disjoint kernels whose direct sum is $\Bbb R^n$.

Let $\{v_1,\dots,v_k\}$ denote a basis for $\ker(A + aI)$. Let $\{v_{k+1},\dots,v_n\}$ be a basis for $\ker(A + bI)$. The vectors $\{v_1,\dots,v_n\}$ form a basis of $\Bbb R^n$, and the matrix of $A$ relative to this basis is diagonal.


Since you have vague knowledge of minimal polynomials, here is the quick-and-easy approach to part $2$. Let $p$ denote the polynomial $p(x) = (x+a)(x+b)$. Because $p(A) = 0$, the minimal polynomial of $A$ divides $p$. It follows that the minimal polynomial of $A$ has no repeated factors, which means that $A$ diagonalizable.


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  • $\begingroup$ I will try to answer the second part. In the mean time, I thought this was worth posting. Feel free to ask for clarification. $\endgroup$ – Omnomnomnom Apr 9 '15 at 17:34
  • $\begingroup$ Thanks! But I don't quite understand why $$\ker(A + aI) \cap \ker(A + bI) = \{0\}$$ $\endgroup$ – Vim Apr 9 '15 at 17:36
  • $\begingroup$ Note that any $x$ in this intersection would satisfy $(A + aI)x = (A + bI)x[= 0]$. Remember that $a \neq b$. $\endgroup$ – Omnomnomnom Apr 9 '15 at 17:44
  • $\begingroup$ Also, be sure to read my latest edit. Let me know if anything else in there needs clarification. $\endgroup$ – Omnomnomnom Apr 9 '15 at 17:46
  • $\begingroup$ Perfect. I love this answer! $\endgroup$ – Vim Apr 9 '15 at 17:50

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