Eigenvectors in the Dual and Double Dual of T

Question

Let $T$ be a linear operator on a vector space $V$ . Prove that if $x$ is an eigenvector for $T$, then $\hat{x}$ is an eigenvector for $T ^{tt}$ .

(Where $T^{tt}$ is the double dual of $T$ and $\hat{x}: T^t\longrightarrow F $)

I'm having trouble keeping straight what exactly elements in the double dual represent and therefore how they relate to getting "scaled" under t (being an eigenvector) and therefore what that implies in $T^{tt}$. Any help is appreciated!

Answer 1

This is actually a comment on Federico Fallucca's great solution. However, it is too long to be put into the comment section.

More generally, let $\alpha:V\to V$, $\beta:W\to W$, and $\phi:V\to W$ be linear maps such that the diagram below commutes: $$\require{AMScd} \begin{CD} V @>\phi>> {W} \\ @V\alpha V V @VV\beta V\\ V @>>\phi> W\,. \end{CD} $$ (That is, $\beta\circ \phi=\phi\circ\alpha$.) If $x\in V\setminus\{0\}$ is an eigenvector of $\alpha$ corresponding to the eigenvalue $\lambda$ and $\phi(x)\neq 0$, then $\phi(x)$ is an eigenvector of $\beta$ corresponding to the eigenvalue $\lambda$.

Furthermore, if $p(s)$ is a polynomial such that $p(\alpha)(x)=0$ for some $x\in V$, then $p(\beta)\big(\phi(x)\big)=0$. Therefore, if $x\in V\setminus\{0\}$ is a generalized eigenvector of $\alpha$ with respect to the eigenvalue $\lambda$ and $x\notin\ker(\phi)$, then $\phi(x)$ is also a generalized eigenvector of $\beta$ corresponding to the same eigenvalue $\lambda$. Consequently, the minimal polynomial of $\beta|_{\text{im}(\phi)}:\text{im}(\phi)\to\text{im}(\phi)$ (if exists) divides the minimal polynomial of $\alpha$ (if exists).

In particular, if $W:=V^{**}$, $\phi:=\iota$, $\alpha:=T$, and $\beta:=T^{**}$, where $\iota:V\to V^{**}$ is the canonical injection (sending $x\in V$ to $x^{**}\in V^{**}$ which maps $f\in V^*$ to $f(x)$), then $$\require{AMScd} \begin{CD} V @>\iota>> {V^{**}} \\ @V T V V @VV T^{**} V\\ V @>>\iota> V^{**} \end{CD} $$ is a commutative diagram. Since $\iota$ is injective, $\ker(\iota)=0$. Therefore, for any $x\in V\setminus\{0\}$ that is an eigenvector of $T$, $x^{**}=\iota(x)$ is an eigenvector of $T^{**}$ (in the OP's notaion, $x^{**}=\hat{x}$ and $T^{**}=T^{tt}$). Particularly, when $V$ is finite-dimensional, $\iota$ is an isomorphism. Therefore, $T$ and $T^{**}$ have exactly the same Jordan normal form.

Answer 2

The double dual map $T^{**}$ maps every element $v\in V^{**}$ to $v\circ T^*$ so if $x$ is an eighevector of $T$ of eighenvalue of $\lambda$ then

$T^{**}(x^-)(f)=(x^-\circ T^*)(f)=$

$x^-(f\circ T)=f(T(x))=f(\lambda x)= \lambda f(x)$

$=\lambda x^-(f) $

for every $ f\in V^*$ and so

$T^{**}(x^-)=\lambda x^-$

In any case this is a little observation of an bigger important result:

For every linear injective map $L: V\to W$ if $ x$ is an eighenvector of a linear map $T: V\to V $ than $L(x) $ is an eighenvector of the linear map $L(T) : Im(L)\to W$ that maps every $L(v)$ to $ L(T(v))$.

So if you consider the canonical injection

$J: V \to V^{**} $

you have that this map is injective and so you must only apply the previous result.