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 $A^*$?

  • $\begingroup$ I take it you mean that $A(x) = x$ for some particular $x$ (that is, not for any $x$)? $\endgroup$
    – Muphrid
    Commented Feb 10, 2013 at 16:44
  • $\begingroup$ 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. $\endgroup$
    – GaryMak
    Commented Sep 22, 2013 at 10:07

2 Answers 2


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 $A^*x=(1,1)^T\neq x$.


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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