Let \(T:\mathbb{R}^2\rightarrow \mathbb{R}^2\) be given by
\(T\left(\begin{bmatrix} x_1\\x_2\end{bmatrix}\right)=
\begin{bmatrix} 2x_1-x_2 \\ x_1 - x_2\end{bmatrix}.\)
Let \(\Gamma = \left( \begin{bmatrix} 1 \\ 0\end{bmatrix},
\begin{bmatrix} 1 \\ 1\end{bmatrix}\right)\) and
\(\Omega = \left( \begin{bmatrix} 2 \\ 1\end{bmatrix},
\begin{bmatrix} 1 \\ 0\end{bmatrix}\right)\) be ordered bases for
\(\mathbb{R}^2\).
Then \([T]_\Gamma^\Omega = \begin{bmatrix} 1 & 0 \\ 0 & a\end{bmatrix}\).
What is \(a\)?
The answer is 1.
Recall that the second column of
\([T]_\Gamma^\Omega\) is given by
the tuple representation of \(T\left(\begin{bmatrix} 1 \\ 1\end{bmatrix}
\right)\) with respect to \(\Omega\).
Since \(T\left(\begin{bmatrix} 1 \\ 1\end{bmatrix} \right) = \begin{bmatrix}
2(1) - 1 \\ 1 -1 \end{bmatrix} =\begin{bmatrix} 1 \\ 0 \end{bmatrix}
= 0\begin{bmatrix} 2 \\1 \end{bmatrix} + 1 \begin{bmatrix}
1 \\ 0\end{bmatrix}\),
its tuple representation with respect to \(\Omega\)
is \(\begin{bmatrix} 0 \\ 1 \end{bmatrix}\).