$$\newcommand{\mc}{\mathcal}$$ Let $V$ be an $n$-dimensional vector space over a field $F$. (We use $\mc L(V)$ to denote $End(V)$).

For each $v\in V$, define $\Theta_v:\mc L(V)\to V$ as $\Theta_v(T)=Tv$ for all $T\in \mc L(V)$.

Then each linear functional on $\mc L(V)$ can be written as a composition $\varepsilon \circ \Theta_v$, for some $\varepsilon\in V^*$ and $v\in V$.

If we choose a basis for we and transform the problem in terms of matrices, then the above fact is clear: Operating an $n\times n$ matrix on $(0, \ldots, 0, 1, 0\ldots, 0)^T$, where $1$ is atthe $i$-th position, gives us the $i$-th column of the matrix. Now the $j$-th projection function yields the $i,j$-th entry of the matrix.

I am wondering how to do this without using bases.

A natural idea is to define a function $f:V^*\times V\to \mc L(V)^*$ as $$f(\varepsilon, v)=\varepsilon\circ \Theta_v$$ We note that $f$ is bilinear. Thus $f$ factors through the tensor product $V^*\otimes V$.

But this doesn't seem to lead anywhere.

Can somebody help?


I apologize for posting a wrong question. This has been made clear to me by the user whose answer I have accepted. Thanks to him/her.

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    $\begingroup$ Is $\mathcal L(V)=\operatorname{End}(V)$? $\endgroup$ – Hagen von Eitzen May 14 '15 at 20:25
  • $\begingroup$ Yes. I thought this is standard. Let me specify int he question. $\endgroup$ – caffeinemachine May 14 '15 at 20:26
  • $\begingroup$ How would you write the trace in the form above? $\endgroup$ – copper.hat May 14 '15 at 20:31
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    $\begingroup$ The trace is a linear functional on the space of square matrices, but cannot be written in the form $A \mapsto u^T A v$ for any $u,v$. $\endgroup$ – copper.hat May 14 '15 at 20:36
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    $\begingroup$ There is no need to apologise. This is the way it works. $\endgroup$ – copper.hat May 14 '15 at 20:47

I don't think what you say is true.

$\mathcal{L}(V) = \text{Hom}(V, V) = V \otimes V^*$. Your map $\Theta_v: V \otimes V^* \rightarrow V$ is given by tensoring the evaluation map $ev_v: V^* \rightarrow k$ with $V$.

A functional on $\text{Hom}(V, V)$ is the same as an element of $\text{Hom}(V, V)$, the relationship being taking the dual of an endomorphism of $V$. I find it easier to work here. Here, $ev_v$ is identified with the map $1 \mapsto v$. Then, what you ask is that every endomorphism of $V$ is given by $T(x) = f(x) v$ for a functional $f$. But this isn't true. What is true is that every $T$ can be written as a sum of such things.

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  • $\begingroup$ Thank you. And apologies for the wrong question. $\endgroup$ – caffeinemachine May 14 '15 at 20:34
  • $\begingroup$ No need to apologize :) $\endgroup$ – user148177 May 14 '15 at 20:34

A different reason why it can't be true is that there just aren't enough $(\varepsilon,v)$ pairs to represent all the functionals when $F$ is finite.

Suppose for example $F=\mathbb F_p$ and $n=3$.

$\mathcal L(V)$ is a 9-dimensional vector space and therefore there are $p^9$ different linear functionals on $\mathcal L(V)$.

On the other hand there are only $p^3$ vectors in $V$ and $p^3$ vectors in $V^*$, so you can't express more than at most $p^3p^3=p^6$ different functionals in the form $\varepsilon \circ \Theta_v$.

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  • $\begingroup$ Nice! $\phantom{xxxxxx}$ $\endgroup$ – caffeinemachine May 14 '15 at 20:39

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