Exercise 2.2.4

Answers

Since

\begin{array}{l} T (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) & = 1 + 0 x + 0 x^{2} \\ T (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}) & = 1 + 0 x + 1 x^{2} \\ T (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) & = 0 + 0 x + 0 x^{2} \\ T (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) & = 0 + 2 x + 0 x^{2} \end{array}

we have

{[T]}_{β}^{γ} = (\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 \end{matrix}) .

jephianlin

2011-06-27 00:00