Question 4.E.7.16: Let I be the identity operator on an n -dimensional space V.......
Let I be the identity operator on an n -dimensional space \mathcal{V}.
(a) Explain why
[I]_{\mathcal{B}} = \begin{pmatrix}1 &0& · · ·& 0\\0 &1& · · · &0\\\vdots&\vdots&\ddots&\vdots\\0 &0& · · ·& 1\end{pmatrix}
regardless of the choice of basis \mathcal{B}.
(b) Let \mathcal{B} = \{x_i\}^{n}_{i=1} and \mathcal{B}^{′} = \{y_i\}^{n}_{i=1} be two different bases for \mathcal{V}, and let T be the linear operator on \mathcal{V} that maps vectors from \mathcal{B}^{′} to vectors in \mathcal{B} according to the rule T(y_i) = x_i for i = 1, 2, . . . , n. Explain why
[I]_{\mathcal{B}\mathcal{B}^{′}} = [T]_{\mathcal{B}} = [T]_{\mathcal{B}^{′}} = ([x_1]_{\mathcal{B}^{′}} | [x_2]_{\mathcal{B}^{′}} | · · · | [x_n]_{\mathcal{B}^{′}}).
(c) When \mathcal{V} = ℜ^3, determine [I]_{\mathcal{B}\mathcal{B}^{′}} for
\mathcal{B} = \begin{Bmatrix}\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\1\\0\end{pmatrix}, \begin{pmatrix}0\\0\\1 \end{pmatrix} \end{Bmatrix}, \mathcal{B}^{′} = \begin{Bmatrix}\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix} 1\\1\\0\end{pmatrix}, \begin{pmatrix}1\\1\\1\end{pmatrix}\end{Bmatrix}.(a) If \mathcal{B} = \{x_i\}^{n}_{i=1} is a basis, then I(x_j) = 0x_1 + 0x_2 + · · · + 1x_j + · · · + 0x_n so that the j^{th} column in [I]_{\mathcal{B}} is just the j^{th} unit column.
(b) Suppose x_j = \sum_{i}β_{ij}y_{i} so that [x_j]_{\mathcal{B}^{′}} = \begin{pmatrix}β_{1j}\\\vdots\\β_{nj}\end{pmatrix}. Then
I(x_j) = x_j = \sum\limits_{i}β_{ij}y_{i} \Longrightarrow [I]_{\mathcal{B}\mathcal{B}^{′}} = [β_{ij}] = ([x_1]_{\mathcal{B}^{′}} | [x_2]_{\mathcal{B}^{′}} | · · · | [x_n]_{\mathcal{B}^{′}}).Furthermore, T(y_j) = x_j = \sum_{i}β_{ij}y_{i} \Longrightarrow [T]_{\mathcal{B}^{′}} = [β_{ij}], and
T(x_j) = T (\sum\limits_{i}β_{ij}y_{i}) = \sum\limits_{i}β_{ij}T(y_{i}) = \sum\limits_{i}β_{ij}x_{i} \Longrightarrow [T]_{\mathcal{B}} = [β_{ij}].(c) \begin{pmatrix}1 &−1& 0\\0 &1& −1\\0 &0& 1\end{pmatrix}