3.4.2

Consider the standard basis $$\textbf{E} = (e_1,e_2,\dots,e_n) $$ of $$\mathbb{F}^n $$, and the new basis $$\textbf{E'} = (e_n,e_{n-1},\dots,e_1) $$. By observing that $$ e_1 = 0e_n + 0e_{n-1} + \dots + 1e_1$$, and $$ e_2 = 0e_n + 0e_{n-1} + \dots +1e_2 + 0e_1$$, and so on, we have that $$

(e_n,e_{n-1},\dots,e_1) \begin{bmatrix}

0 & 0 & ... & 0 & 1\\

0 & 0 & ... & 1 & 0\\

...&...&...&...&...\\

0&1&...&0&0\\

1&0&...&0&0\\

\end{bmatrix} = (e_1,e_2,\dots,e_n).

$$

Thus the change of basis matrix $$P$$ when going from the standard basis $$\textbf E$$ to the new basis $$\textbf{E'}$$ is the matrix above, which has 1's on the antidiagonal and 0's everywhere else.