Exercise 3.F.33

Proof. Let $t\in L({𝔽}^{m,n},{𝔽}^{n,m})$ denote the linear map that takes a matrix to its transpose. For this exercise, assume $1\le k\le m$ and $1\le j\le n$.

Suppose $A,C\in {𝔽}^{m,n}$. Then

Let $\lambda \in 𝔽$. We have

Therefore $t$ is indeed a linear map.

Since $\dim <!--\; FUNCTION\; APPLICATION\; -->({𝔽}^{m,n})=\dim <!--\; FUNCTION\; APPLICATION\; -->({𝔽}^{n,m})$, to prove that $t$ is invertible we only need to show that it is injective.

Suppose $t(A)=0$ for some $A\in F^{m,n}$ (here 0 denotes a matrix in ${𝔽}^{n,m})$ with 0 in all entries). We have that

Because $A$ has zero in all its entries, it follows that $A=0$ and, therefore, $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->t=0$, which implies that $t$ is injective. □