4.2.5

Let A be an nxn matrix, let V = $$\mathbb{F}^n $$ denote the space of row vectors. Consider the linear operator $$T:V \rightarrow V $$ defined by $$Tv = vA $$, let its matrix with respect to the standard basis E of V be B. Observe that with $$e_1 = (1,0,...,0) $$, $$ Te_1 = e_1A = A_1 = A_{11}e_1 + A_{12}e_2 + ... + A_{1n}e_n$$, where $$A_1 $$ denotes row 1 of A, and $$A_{ij} $$ denotes the entry in row i and column j of A. Thus the first column of B is $$B_{j1} = A_{1j} $$, for j = 1,2,..n, showing that the first column of B is in fact the first row of A. By considering the action of T on the other standard basis elements $$e_i $$ we find that $$ B_{ji} = A_{ij}$$ in general, so that $$B = A^T $$, and the matrix of T with respect to the standard basis E is the transpose of A, $$ A^T$$.