Suppose that \(A=\left[\begin{array}{rr} 2 \amp 1 \\ 1\amp 2 \\ \end{array}\right]\) and

\begin{equation*} \vvec_1=\twovec{1}{1}, \vvec_2=\twovec{1}{-1} \text{.} \end{equation*}
  1. Explain why \(\bcal=\{\vvec_1,\vvec_2\}\) is a basis for \(\real^2\text{.}\)

  2. Find \(A\vvec_1\) and \(A\vvec_2\text{.}\)

  3. Use what you found in the previous part of this problem to find \(\coords{A\vvec_1}{\bcal}\) and \(\coords{A\vvec_2}{\bcal}\text{.}\)

  4. If \(\coords{\xvec}{\bcal} = \twovec{1}{-5}\text{,}\) find \(\coords{A\xvec}{\bcal} \text{.}\)

  5. Find a matrix \(D\) such that \(\coords{A\xvec}{\bcal} = D\coords{\xvec}{\bcal}\text{.}\)

You should find that the matrix \(D\) is a very simple matrix, which means that this basis \(\bcal\) is well suited to study the effect of multiplication by \(A\text{.}\) This observation is the central idea of the next chapter.