Let $T$ be a linear operator on a finite dimensional inner product space $V$. If $T$ has an eigenvector, then so does $T^*$.

Proof. Suppose that $v$ is an eigenvector of $T$ with corresponding eigenvalue $\lambda$. Then for any $x \in V$,
$$ 0 = \langle0,x\rangle = \langle(T-\lambda I)v,x\rangle = \langle v, (T-\lambda I)^*x\rangle = \langle v,(T^*-\bar{\lambda} I)x\rangle $$ This means that $(T^*-\bar{\lambda} I)$ is not onto. WHY?
(Of course the proof is not completed in here)

  • $\begingroup$ I think there is a mistake in your proof. Specifically, the term after the third equal to sign is wrong. You do not need that, the term after the second equal to sign is equal to the last term by the definition of adjoint. $\endgroup$ May 28, 2013 at 4:22
  • $\begingroup$ Actually it's not mine. It's from the friedberg's book. $\endgroup$
    – noname
    May 28, 2013 at 4:24
  • 2
    $\begingroup$ This is correct. If it was onto, you would get $(v,y)=0$ for every $y\in V$, whence $v=0$. $\endgroup$
    – Julien
    May 28, 2013 at 4:24
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    $\begingroup$ If $T^* - \lambda I$ were onto, then wouldnt you be able to find an $x$ such that $(T^* - \lambda I)x = v$? Then what can you say about $v$? $\endgroup$ May 28, 2013 at 4:25
  • $\begingroup$ Oops, its correct. I did not see the star at the top! $\endgroup$ May 28, 2013 at 4:27

2 Answers 2


Assume $T^*-\overline{\lambda}I$ is onto and take $v$ such that $(T-\lambda I)v=0$.

By surjectivity, there exists $x\in V$ such that $(T^*-\overline{\lambda}I)x=v$. For this specific $x$, we get $$ 0=(v,(T^*-\overline{\lambda}I)x)=(v,v)=\|v\|^2\quad \Rightarrow \quad v=0. $$ Therefore $T^*-\overline{\lambda}I$ surjective implies $T-\lambda I$ injective.

By rank-nullity and symmetry, it follows that $T-\lambda I$ is not injective if and only if $T^*-\overline{\lambda}I$ is not injective. That is $\lambda $ is an eigenvalue for $T$ if and only if $\overline{\lambda}$ is an eigenvalue for $T^*$.


If you want to know why $T^{*}$ is not onto if $T$ is not onto, just observe that the matrix of $T^{*}$ is just the conjugate transpose of the matrix of $T$ and we know that a matrix and its conjugate transpose have the same rank (which is equal to the dimension of the range space).


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