I'm asked to prove, among other properties, that $(f\circ g)^t = g^t\circ f^t$, where $f^t$ stands for the transpose of the linear map $f$, though the definition of $f^t$ is itself problematic to me. The fact that I'm not able to proceed makes me think I got something wrong about the theory, so I'm here to ask you what did I got wrong.

Following an introduction to linear maps and dual spaces, this is what I undestand: the set of linear forms of a vector space $V$ over $\mathbb{F}$, i.e. $\mathcal{L}(V,\mathbb{F})$, is called the dual space of $V$ and noted $V^*$. Let $f\in\mathcal{L}(V,W)$, for $W$ another vector space over $\mathbb{F}$; we define $f^t: W^*\to V^*: \phi\mapsto \phi \circ f$, for all $\phi \in\mathcal{L}(W,\mathbb{F})$.

Now here is where the problem starts. Doesn't $f^t= \phi\circ f$ mean that $f^t(\phi)= (\phi\circ f)(\phi)$? If that's the case, how is it evaluated? The domain of definition of $f$ is $V$, not $W^*$. If $f^t= \phi\circ f$ doesn't mean that $f^t(\phi)= (\phi\circ f)(\phi)$, what is the argument for $f$ in $\phi\circ f=\phi(f(\cdot))$? If the argument was a vector from $V$, then $f^t: V \to \mathbb{F}$, which is contradictory to its definition: $f^t$ is a map between two vector spaces whose elements are themselves functions.

I have the feeling I'm measing something very basic that I cannot see.

EDIT: of course, I've googled about it, read Wikipedia's article about dual spaces and transposes of linear maps, even checked some related questions on Stackexchange, and still didn't got it. That's the main reason why I suspect what I'm missing is fundamental.

  • 1
    $\begingroup$ $f^t(\phi)= \phi\circ f$ not $f^t(\phi)= (\phi\circ f)(\phi)$. $\endgroup$
    – cqfd
    Jan 20, 2020 at 11:07
  • $\begingroup$ Well, then what does $\phi\circ f$ mean? What is the output of such a function? A function itself? If that is the case, a then understand than the function $\phi\circ f\in W^*$. However, I still don't see how to prove the aforementioned property. $\endgroup$
    – Albert
    Jan 20, 2020 at 11:08
  • 2
    $\begingroup$ If $v\in V$, then $f^t(\phi)v= (\phi\circ f)(v)$. Note that $(\phi\circ f)$ is a linear functional from $V$ to $\mathbb F$. $\endgroup$
    – cqfd
    Jan 20, 2020 at 11:11
  • $\begingroup$ What about the property? $\endgroup$
    – Albert
    Jan 20, 2020 at 11:15
  • 1
    $\begingroup$ Yes. If $\phi\in W^*$, then $(f\circ g)^t\phi=\phi\circ(f\circ g)=(\phi\circ f)\circ g=g^t(\phi\circ f)=\cdots$ $\endgroup$
    – cqfd
    Jan 20, 2020 at 11:52

1 Answer 1


Let me first recall some theory of dual spaces, which might be confusing. It's all a matter of writing down the definitions.

We have two vector spaces $V$ and $W$ over a field $\mathbb{K}$. We denote $\mathcal{L}(V,W)$ the set of all $\mathbb{K}$-linear maps from $V$ to $W$.

Let's say we have $f\in \mathcal{L}(V,W)$ and a linear functional $\alpha\in W^*$.

Question: How can we now create a linear functional in $V^*$?

Simple: we compose: $\alpha\circ f\in V^*$. Why? If we take $v\in V$, then $f(v)\in W$ ($f$ is a way to get from $V$ to $W$) and $\alpha\circ f(v)=\alpha(f(v))\in \mathbb{K}$. We denote $\alpha\circ f=:f^t(\alpha)$.

This operation is called "transposing". More precisely, we have proven that we have a map

$$\boxed{\cdot^t:\mathcal{L}(V,W)\to\mathcal{L}(W^*,V^*),f\mapsto f^t}$$

(Look at this and meditate.) It is easy to see that this is even a linear map, but that doesn't matter for now.

Let's now say we have $f:V\to W$ and $g:W\to Z$. We want to prove that $(g\circ f)^t=f^t\circ g^t$. Recall that this is an equality of linear maps in $\mathcal{L}(W^*,V^*)$.

(Why? We have $g\circ f:V\to Z$, so $(g\circ f)^t\in \mathcal{L}(Z^*,V^*)$ takes a linear functional $\alpha\in Z^*$ and gives a linear functional $(g\circ f)^t(\alpha)\in V^*$.)

Let us therefore take $\alpha\in Z^*$ and see what happens. We have

$$(g\circ f)^t(\alpha)=\alpha\circ (g\circ f),$$

and the other hand

$$(f^t\circ g^t)(\alpha)=f^t(g^t(\alpha))=f^t(\alpha\circ g)=(\alpha\circ g)\circ f.$$

By associativity of composition of maps, we see that indeed $(g\circ f)^t(\alpha)=(f^t\circ g^t)(\alpha)$ for all $\alpha\in Z^*$, so $(g\circ f)^t=f^t\circ g^t$.

  • $\begingroup$ You made me understand more than what I originally sought. $\endgroup$
    – Albert
    Jan 20, 2020 at 13:10
  • $\begingroup$ Glad I could be of help! :) $\endgroup$
    – rae306
    Jan 20, 2020 at 14:16

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