Transpose of composite functions

Question

I want to prove that $(gof)^T$=$f^Tog^T$ where $f,g$ are linear maps.

I know that I can just use definitions but I don't know exactly how. Can anyone point me in the right way?

Answer 1

If $\alpha$ is a linear form, and $v$ a vector (on/in appropriate vector spaces) then
$$ (g\circ f)^T(\alpha)(v) =\alpha((g\circ f)(v)) =\alpha(g(f(v))) =g^T(\alpha)(f(v)) =f^T(g^T(\alpha))(v) =(f^T\circ g^T)(\alpha)(v), $$ and since this holds for all $\alpha,v$, one has $(g\circ f)^T=f^T\circ g^T$.

Answer 2

Let $f : V \to W$, $g : W \to U$, then $f^T : W^* \to V^*$, $g^T : U^* \to W^*$.

$f^T$ is defined by $$\langle f^T x,y\rangle = \langle x, f y\rangle$$ for all $x,y$, where $\langle \cdot, \cdot\rangle$ denotes the appropriate duality pairing, and I will leave it implicit that $x \in W^*, y \in V$, etc., so that the LHS is the duality in $V$, the RHS is the duality in $W$, etc.

Try starting by writing $\langle (g\circ f)^T x,y \rangle$, and then applying the definitions to show this is equal to $ \langle f^Tg^Tx,y\rangle$. Since you do this for arbitrary $x,y$, this proves what you want.

Explicitly:

$$\langle (g\circ f)^T x,y \rangle = \langle x,(g \circ f)y\rangle = \langle g^Tx,fy\rangle = \langle f^Tg^Tx,y\rangle$$