Proof if characteristic polynomial of T splits, then the adjoint T* has at least one eigenvector

Question

I don't understand a part of this proof, hope you can help me sort it out.

Let $T$ be a linear operator on inner product space $V$ for which the characteristic polynomial splits. Then $T^*$ has at least one eigenvector.

Proof. Because $T$ splits, it has an eigenvector $v$ with eigenvalue $\lambda$ $\Rightarrow$ $(T-\lambda I)v=0$
$\Rightarrow\langle (T-\lambda I)v, x \rangle = 0$ for all $x \in V$
$\Rightarrow\langle v, (T-\lambda I)^* x \rangle = 0$
$\Rightarrow\langle v, (T^*- \overline{\lambda}I) x \rangle = 0$
$\Rightarrow v \notin R(T^*- \overline{\lambda}I)$
$\Rightarrow R(T^*- \overline{\lambda}I) \ne V$
Therefore, $(T^*- \overline{\lambda}I)$ is not invertible (why???)
Hence, there exists an engenvector of $T^*$ (with eigenvalue $\lambda$) (why???).

I don't understand the last two conclusion.

1. Why $(T^*- \overline{\lambda}I) \ne V$ implies $(T^*- \overline{\lambda}I)$ is not invertible?
2. Why $(T^*- \overline{\lambda}I)$ is not invertible implies an eigenvector of $T^*$?

Thanks for any help!

Answer 1

First of all, you seem to have written the final step incorrectly. We should state that $R(T^* - \bar \lambda I) \neq V$; this is not the same as $(T^* - \bar \lambda I) \neq V$, which doesn't make much sense.

Note that a linear operator is invertible if and only if it is both one-to-one (injective) and onto (surjective). Because $(T - \bar \lambda I)$ is not surjective, it cannot be invertible.

To put that another way: the statement $v \notin R(T - \bar \lambda I)$ is the same as saying that the equation $(T - \bar \lambda I) x = v$ has no solution (for $x$). However, if $(T - \bar \lambda I)$ had an inverse, then we could use the inverse to solve the equation and get $x = (T - \bar \lambda I)^{-1}v$.

To your second question, if $(T - \bar \lambda I)$ is not invertible, then $\bar \lambda $ must be an eigenvalue of $T^*$. One way to see that this is the case is as follows: if $(T^* - \bar \lambda I)$ is not invertible, then the equation $(T^* - \bar \lambda I)x = 0$ has a solution $x \neq 0$ (i.e. a "non-trivial" solution). Expanding the left side of this equation leads to $$ T^*x - \bar \lambda Ix = 0 \implies T^*x = \bar \lambda x. $$ So, $x$ is an eigenvector of $T^*$ associated with $\bar \lambda$.

Answer 2

1. Why $(T^*- \overline{\lambda}I) \ne V$ implies $(T^*- \overline{\lambda}I)$ is not invertible?

You have a typo there, it should say $R(T^*- \overline{\lambda}I) \ne V$. It just says that $T^*-\overline\lambda I$ is not surjective, so it cannot be invertible.

2. Why $(T^*- \overline{\lambda}I)$ is not invertible implies an eigenvector of $T^*$?

If $V$ is finite-dimensional, then an operator is surjective if and only if it is injective. So $\ker(T^*- \overline{\lambda}I) \ne\{0\}$, which means that there exists $w$ with $T^*w- \overline{\lambda}w=0$ .

Answer 3

First point, a map $\phi$ that is not surjective can’t be invertible. What would be the inverse image under $\phi$ of an element that is not in the image of $\phi$?

Second point, in a finite dimension linear space, the kernel of a non invertible endomorphism is not reduced to the zero linear subspace.