Isomorphism from $T\rightarrow T^*$

Question

I don't know how to do this problem. Please help me..

Let $V$ be a finite dimensional vector space over field $F$. Show that $T\rightarrow T^*$ is an isomorphism of $L(V,V)$ onto $L(V^*,V^*)$.

I have one more thing to clarify: Am I right in assuming $T^*$ as the transpose of $T$ and $V^*$as the Dual space of $V$ here?

Answer 1

$V^{*}$ is surely the dual space. And $T^{*}$ is defined by $$ T^{*}: V^{*} \to V^{*} : f \mapsto (v \mapsto f(T(v))) $$ i.e., $T^{*}(f)$, for a typical covector $f$, is a new covector, whose value on a typical element $v \in V$ is simply $f(T(v))$.

Answer 2

Let $\dim V:= n=\dim V^*$(Since $V$ and $V^*$ have the same dimension)

Thus $\dim L(V,V)=\dim (V^*,V^*)=n^2.$

Hence if we show that the mapping $\zeta : L(V,V)\to (V^*,V^*)$ where $\zeta(T)=T^t$ is $1:1,$ we are done.

But we have to verify that $\zeta$ indeed is a function(trivial) and a linear transformation.

• Let $c\in F,T_1\&T_2\in L(V,V).$ Then for a given $f\in V^*$ and any $\alpha\in V,$ \begin{align} \big[(cT_1+T_2)^tf\big](\alpha)&=f\big[(cT_1+T_2)(\alpha)\big]\\ &=f\big[cT_1(\alpha)+T_2(\alpha)\big]\\ &=cf[T_1(\alpha)]+f[T_2(\alpha)]\\ &=cT_1^tf(\alpha)+T_2^tf(\alpha). \end{align}

Thus $\zeta (cT_1+T_2)=(cT_1+T_2)^t=cT_1^t+T_2^t=c\zeta (T_1) +\zeta (T_2)$ so that $\zeta$ is linear.

• $\zeta$ is $1:1$

For any given $T\in L(V,V), T^t=0\iff T^t(f)=0$ for all $f\in V^*$
$\hspace{6.6cm}$i.e., $f[T(\alpha)]=0$ for all $f\in V^*,\alpha\in V$
$\hspace{6.5cm} \iff$ Range$(T)\subseteq \ker f$ for all $f\in V^*$
$\hspace{6.5cm} \iff$ Range$(T)\subseteq \bigcap_{f\in V^*} \ker f$ $\hspace{6.5cm} \iff$ Range$(T)=0$
$\hspace{6.5cm} \iff T=0$

Thus $\zeta (T)=0\iff T^t=0 \iff T=0$ so that $\zeta$ is $1:1$ and we have the desired result.