- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.12 Exercises
$\quad$ In all exercises involving the vector space $V_n,$ the usual basis of unit coordinate vectors is to be chosen, unless another basis is specifically mentioned. In exercises concerned with the matrix of a linear transformation $T: V \rightarrow W$ where $V = W,$ we take the same basis in both $V$ and $W$ unless another choice is indicated.
2. $\quad$ Determine the matrix for each of the following projections.
(a) $\quad T: V_3 \rightarrow V_2,$ where $T(x_1, x_2, x_3) = (x_1, x_2).$
(b) $\quad T: V_3 \rightarrow V_2,$ where $T(x_1, x_2, x_3) = (x_2, x_3).$
(c) $\quad T: V_5 \rightarrow V_3,$ where $T(x_1, x_2, x_3, x_4, x_5) = (x_2, x_3, x_4).$
Solution. $\quad$ In Section 2.10 we saw that if $T: V \to W$ is a linear transformation, where $V$ and $W$ are finite linear spaces with respective bases $(e_1, \dots, e_n)$ and $(w_1, \dots, w_m),$ we can express the transformation of the $k^{th}$ basis element of $V,$ $T(e_k)$ as the linear combination of the basis elements of $W:$ \begin{align*} T(e_k) &= \sum_{i=1}^m t_{ik}w_i \end{align*} where $t_{1k}, \dots, t_{mk}$ are the components of $T(e_k)$ with respect to the ordered basis $(w_1, \dots, w_m).$ These components make up the $k^{th}$ column of the matrix representation of $T:$ \begin{align*} \begin{bmatrix} t_{1k} \\ t_{2k} \\ \vdots \\ t_{mk} \end{bmatrix} \end{align*} Then, placing the column matrices for each of the $n$ elements $T(e_1), \dots, T(e_n)$ side by side, we get the component matrix for $T:$ \begin{align*} \begin{bmatrix} t_{11} & t_{12} & \cdots & t_{1n} \\ t_{21} & t_{22} & \cdots & t_{2n} \\ \vdots & \vdots & & \vdots \\ t_{m1} & t_{m2} & \cdots & t_{mn} \end{bmatrix} \end{align*}
(a) $\quad T: V_3 \rightarrow V_2,$ where $T(x_1, x_2, x_3) = (x_1, x_2).$
$\quad$ In this transformation, $T$ maps $x = x_1e_1 + x_2e_2 + x_3e_3$ onto $w = x_1w_1 + x_2w_2,$ where $(e_1, e_2, e_3)$ and $(w_1, w_2)$ are the respective unit coordinate bases for $V$ and $W.$ Thus, if
\begin{align*}
T(1, 0, 0) &= \sum_{i=1}^2 t_{i1}w_i
\\
&= (1, 0)
\\
T(0, 1, 0) &= \sum_{i=1}^2 t_{i2}w_i
\\
&= (0, 1)
\\
T(0, 0, 1) &= \sum_{i=1}^2 t_{i3}w_i
\\
&= (0, 0)
\end{align*}
then $(t_{ik})$ is given by
\begin{align*}
\begin{bmatrix}
1 & 0 & 0
\\
0 & 1 & 0
\end{bmatrix}
\quad \blacksquare
\end{align*}
(b) $\quad T: V_3 \rightarrow V_2,$ where $T(x_1, x_2, x_3) = (x_2, x_3).$
$\quad$ In this transformation, $T$ maps $x = x_1e_1 + x_2e_2 + x_3e_3$ onto $w = x_2w_1 + x_3w_2,$ where $(e_1, e_2, e_3)$ and $(w_1, w_2)$ are the respective unit coordinate bases for $V$ and $W.$ Thus, if
\begin{align*}
T(1, 0, 0) &= \sum_{i=1}^2 t_{i1}w_i
\\
&= (0, 0)
\\
T(0, 1, 0) &= \sum_{i=1}^2 t_{i2}w_i
\\
&= (1, 0)
\\
T(0, 0, 1) &= \sum_{i=1}^2 t_{i3}w_i
\\
&= (0, 1)
\end{align*}
then $(t_{ik})$ is given by
\begin{align*}
\begin{bmatrix}
0 & 1 & 0
\\
0 & 0 & 1
\end{bmatrix}
\quad \blacksquare
\end{align*}
(c) $\quad T: V_5 \rightarrow V_3,$ where $T(x_1, x_2, x_3, x_4, x_5) = (x_2, x_3, x_4).$
$\quad$ In this transformation, $T$ maps $x = x_1e_1 + x_2e_2 + x_3e_3 + x_4e_4 + x_5e_5$ onto $w = x_2w_1 + x_3w_2 + x_4w_3$ where $(e_1, e_2, e_3, e_4, e_5)$ and $(w_1, w_2, w_3)$ are the respective unit coordinate bases for $V$ and $W.$ Thus, if
\begin{align*}
T(1, 0, 0, 0, 0) &= \sum_{i=1}^3 t_{i1}w_i
\\
&= (0, 0, 0)
\\
T(0, 1, 0, 0, 0) &= \sum_{i=1}^3 t_{i2}w_i
\\
&= (1, 0, 0)
\\
T(0, 0, 1, 0, 0) &= \sum_{i=1}^3 t_{i3}w_i
\\
&= (0, 1, 0)
\\
T(0, 0, 0, 1, 0) &= \sum_{i=1}^3 t_{i4}w_i
\\
&= (0, 0, 1)
\\
T(0, 0, 0, 0, 1) &= \sum_{i=1}^3 t_{i5}w_i
\\
&= (0, 0, 0)
\end{align*}
then $(t_{ik})$ is given by
\begin{align*}
\begin{bmatrix}
0 & 1 & 0 & 0 & 0
\\
0 & 0 & 1 & 0 & 0
\\
0 & 0 & 0 & 1 & 0
\\
\end{bmatrix}
\quad \blacksquare
\end{align*}