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Let $A$ be a square non-Hermitian matrix and $c$ be an eigenvalue of $A$ with algebraic multiplicity $1$. Let $Ax = cx$ and $y^{H}A = cy^{H}$ where $y^{H}$ is a conjugate transpose of $y$. Prove that $y^{H}x \neq 0$.

Please give me a hint. Thanks.

fiverules
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2 Answers2

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EDIT:@me10240 pointed out a flaw in the porevious proof. So i am giving a new proof.

Let say $y^Hx=0$. Then $x\in \mathcal{R}(A-cI)$ since $y\in \mathcal{N}(A^H-c^*I)$. Then, $\exists u\ne 0$ such that $x=(A-cI)u\Rightarrow (A-cI)^2u=0$ Now, since $c$ has algebraic multiplicity $1$, we have $\dim(\mathcal{N}(A-cI)^2)=\dim(\mathcal{N}(A-cI))=1$. Hence, $u=\alpha x$ for some $\alpha\ne 0\Rightarrow x=(A-cI)\alpha x=0$ which is a contradiction.

Community
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Samrat Mukhopadhyay
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    This proof is not correct. You cannot conclude that $y$ is an eigenvector of $A$. Please fix it if possible. Here is a counterexample: Let $A =\begin{pmatrix} 5 & -1\ -2 & 6\end{pmatrix}$. Then 4 is a simple eigenvalue with right eigenvector $x = (1,1)^T$ and left eigenvector $y = (2,1)^T$. Clearly, $y \neq k x$ for any $k$. – me10240 Jun 07 '14 at 21:09
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    @me10240 Thank you for pointing out the flaw in the previous proof. I have given a new proof. Please see if this correct. – Samrat Mukhopadhyay Jun 08 '14 at 09:54
  • This is almost correct, but in proving $u = \alpha x$ you have made an assumption that the eigenvectors form a basis. This is not given in the problem, and is not required either. To complete it, recall that you obtained $u \in Null(A -cI)^2$, that is $u$ is a generalized eigenvector. But the algebraic and geometric multiplicity of $c$ is 1, which implies $dim(Null(A -cI)) = dim(Null(A -cI)^2) = 1$, hence $u = \alpha x$. If you can modify your proof to clean up this bit, I can remove my comment. – me10240 Jun 09 '14 at 05:51
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I suppose you forgot to say that $x,y$ are nonzero, otherwise this is not true of course. On the other hand the hypothesis about being non-Hermitian is a red herring, and the conjugate-transpose in $y^H$ serves no purpose: $y^H$ is just a complicated name for an arbitrary $1\times n$ matrix, a linear form on $V=\Bbb C^n$.

Choose a basis such that the linear form $y^H$ is the final coordinate function, i.e., start with choosing a basis $[e_1,\ldots,e_{n-1}]$ of $W=\ker(y^H)$ and complete with a final vector $e_n$ for which $y^He_n=1$. Now the fact that $$y^HAe_i=cy^He_i=c\delta_{i,n}$$ shows that after change of basis $A$ to this new basis $[e_1,\ldots,e_n]$, the final row of the new matrix $A'$ is $\begin{pmatrix}0&\ldots&0&c\end{pmatrix}$ (since that final row is the result of composing the final coordinate function $y^H$ with $A$, and then evaluating it on all basis vectors $e_1,\ldots,e_n$). Then the characteristic polynomial$~\chi$ of $A'$ (and therefore of $A$) is the product of the characteristic polynomial$~\chi'$ of the top-left $(n-1)\times(n-1)$ submatrix$~B$ of $A'$ (which gives the restriction of $A$ to the subspace$~W$) and a factor $X-c$ (for the bottom right $1\times1$ submatrix of$~A'$). Since $c$ is a simple root of$~\chi$, it is not a root of $\chi'$. This means there is no eigenvector of$~A$ for$~c$ in the subspace$~W$. Since $x$ is an eigenvector of$~A$ for$~c$, one has $x\notin W$ as desired.

I've included quite a bit of detail, but if you are used to less matrix-centred reasoning, you can formulate this argument in a single sentence: by hypothesis $c$ is a simple eigenvalue of the operator that $A$ defines on$~V$, and it is also the eigenvalue of the operator it defines on the $1$-dimensional quotient module $V/W$, but then it cannot also be an eigenvalue of the restriction to$~W$.

Marc van Leeuwen
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