Let $A$ and $B$ in $M_n(\mathbb C)$ with $n$ being odd.
Suppose $AB + BA = A$. Prove that $A$ and $B$ have a common eigenvector.
(It is from an oral exam.)
I ask for a hint for this, not a complete solution. Thank you.
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If $x\in\ker{A}$ then $Bx\in\ker{A}$ and this kernel is not trivial. – marwalix May 18 '18 at 15:30
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@marwalix: Are you sure? I get the same result with a slip. – May 18 '18 at 15:38
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$Ax=0$ leads to $ABx+0=0$... – marwalix May 18 '18 at 15:39
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I speak about "this kernel is not trivial". – May 18 '18 at 15:40
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True. The kernel may be trivial – marwalix May 18 '18 at 15:43
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Show that $A^2B=BA^2$ and use this answer: https://math.stackexchange.com/a/1227219/58818 – Luiz Cordeiro May 18 '18 at 16:12
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Hint: consider first $A^2$ and $B$. – ancient mathematician May 18 '18 at 15:38
3 Answers
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Observe that $$AB -\frac{1}{2}A +BA-\frac{1}{2}A=0$$ $$A(B -\frac{1}{2} I) =-(B -\frac{1}{2} I)A$$ Therefore matrices $A$ and $C=B -\frac{1}{2} I$ satisfies the identity $$AC=-CA.$$ Suposse that $\lambda$ is an eigenvalue of $A$ and $x_{\lambda}$ is a corresponding eigenvector. Then $$CAx_{\lambda} =\lambda Cx_{\lambda}=-ACx_{\lambda}$$ hence $-Cx_{\lambda}$ is an eigenvector of $A$ corresponding to the eingenvalue $\lambda.$ If we assume that one of eigenspaces of $A$ is one dimensional then $-Cx_{\lambda}$ must be a multiple of $x_{\lambda}$ hence $$-Cx_{\lambda}=\mu x_{\lambda}$$ $$(\frac{1}{2}I -B)x_{\lambda}=\mu x_{\lambda}$$ $$Bx_{\lambda}=\left(\frac{1}{2} -\mu \right) x_{\lambda}$$ Thus $A$ and $B$ have a common eigenvector $x_{\lambda}.$
user26857
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The key observation, as pointed out in another answer, is that the equation can be rewritten as $AC=-CA$, where $C=B-\frac12 I$ have the same eigenvectors as $B$.
Since $AC=-CA$ and $n$ is odd, if you take determinants on both sides, you see that one of $A,C$ must be singular.
Suppose $C$ be singular and $0\ne v\in\ker C$. Then $v,Av,A^2v,\ldots\in\ker C$. You may continue from here.
user1551
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$\ker C$ is stable by $A$ hence there exists an eigenvector $x$ of $A$ in $\ker C$ (since matrices are complex). And $Cx = 0 \implies Bx = \frac{1}{2}x$, so $x$ fits. – May 18 '18 at 17:55
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@Nûr Yes. The true question itself ($n$ is odd and $AC=-CA$ imply that $A,C$ have a common eigenvector) is fairly easy, but the $C=B-\frac12 I$ part is more like a trick question. If I were taking that oral exam, I don't think I would be able to spot that under pressure. – user1551 May 18 '18 at 18:03
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If $Bv=cv$, then $BAv=(1-c)Av$. If $Av=0$, then we are done. If not, then $Av$ is eigenvector of $B$.
Note that $A^{2i}v$ is eigenvector wrt $c$ and $A^{2i+1}v$ is eigenvector wrt $1-c$.
i) $c=\frac{1}{2}$ : So assume that by subsequence, $\frac{A^{i} v}{|A^{i} v|}\rightarrow X$. And $BX=cX$. And $$ AX = \lim\ \frac{A^{i+1}v}{|A^{i+1}v|} \frac{|A^{i+1}v|}{|A^{i} v|} =\lambda X$$
ii) $c\neq \frac{1}{2}$ : By subsequence, $\frac{A^{2i} v}{|A^{2i} v|}\rightarrow X$. And $BX=cX$. And $$ A^2X = \lim\ \frac{A^{2i+2}v}{|A^{2i+2}v|} \frac{|A^{2i+2}v|}{|A^{2i} v|} =\lambda X$$
And $Y=AX,\ BY=(1-c)Y$. Further, $AY=A^2X=\lambda X$.
Hence we do the same on subspace orthogonal to ${\rm span}(X,Y)$.
HK Lee
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