# Conjugate transpose of a matrix

I have a fairly simple question. If $A$ is a matrix and $A^*$ denotes its conjugate transpose, is it true that if $Ax = x$, then $A^*x = x$? The matrix $A^*$ will certainly have $1$ as an eigenvalue, but will it be with the same eigenvector? And if not, what is the relation between the eigenvector of $A$ and the one of $A^*$?

Thanks!

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I take it you mean that $A(x) = x$ for some particular $x$ (that is, not for any $x$)? – Muphrid Feb 10 '13 at 16:44
One simple thing that should be clarified here: if one distinguishes left and right eigenvectors then one CAN say something, ie Ax = kx is the same as xA = kx. – GaryMak Sep 22 '13 at 10:07

No, take the matrix

$$\begin{pmatrix} 1 & 1\\ 0 & -1\end{pmatrix}$$

which has $x=(1,0)^T$ as an eigenvector with eigenvalue 1. Yet $A^*x=(1,1)^T\neq x$.

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The easiest example would be to consider the rank one matrix $$A = xy^\top$$ Then $$Ax = (xy^\top) x= x(y^\top x) = x\lambda = \lambda x$$ and $$A^* x = (\bar{y} \bar{x}^\top) x =\bar{y} (\bar{x}^\top x )= \bar{y} k = k\bar{y}$$

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The eigenvectors of $A$ and $A^\ast$, in general, have no relationship with each other.

Let $e_i$ denotes the $i$-th vector in the canonical basis of $\mathbb{C}^n$. Pick any two vectors $u,v$, such that $u$ is linearly independent of $e_n$ and $v$ is linearly independent of $e_1$. Then there always exists an invertible matrix $S$ such that $Se_1 = u$ and $Sv = e_n$. Now, consider $A=SJS^{-1}$, where $J$ is an $n\times n$ Jordan block associated with the eigenvalue $1$. Then $u$ is the eigenvector of $A$ (up to scalar multiples) and $\bar{v}$ is the eigenvector of $A^\ast$. Since $u$ and $v$ are picked arbitrarily, they have no relationship with each other whatsoever.

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