- 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.
1. $\quad$ Determine the matrix of each of the following linear transformations of $V_n$ into $V_n:$
(a) $\quad$ the identity transformation,
(b) $\quad$ the zero transformation,
(c) $\quad$ multiplication by a fixed scalar $c.$
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*}
$\quad$ For each of the following transformations, the ordered basis for both $V$ and $W$ will be the set of unit coordinate vectors $e_1 = (1, 0, \dots, 0),\ e_2 = (0, 1, \dots, 0),\ \dots,\ e_n = (0, 0, \dots, 1),$ where it is understood that the $k^{th}$ component of $e_k$ is $1$ and the remaining components are $0.$
(a) $\quad$ The identity transformation.
$\quad$ By definition, the identity transformation $I_V$ is that which maps each element $x$ in $V$ onto itself. In other words, $T(x) = x.$ Moreover, for each basis element of $V_n,$ we have $T(e_k) = e_k.$ Thus, if we describe the transformation as the linear combination of the basis elements of $W,$ we get
\begin{align*}
T(e_k) &= \sum_{i=1}^m t_{ik}w_i
\end{align*}
But because $V = W = V_n,$ we have $m = n$ and $w_i = e_i,$ giving us
\begin{align*}
T(e_k) &= \sum_{i=1}^n t_{ik}e_i
\\
&= e_k
\end{align*}
This transformation can be achieved by setting $t_{ik} = 1$ when $i = k$ and $0$ otherwise, giving us the matrix representation:
\begin{align*}
\begin{bmatrix}
1 & 0 & \cdots & 0
\\
0 & 1 & \cdots & 0
\\
\vdots & \vdots & & \vdots
\\
0 & 0 & 0 & 1
\end{bmatrix}
\end{align*}
where the diagonal elements are $1$ and the rest are $0. \quad \blacksquare$
(b) $\quad$ The zero transformation.
$\quad$ By definition, the zero transformation $O$ maps each element $x$ in $V$ onto the zero element. In other words, $T(x) = O.$ If we describe the transformation as the linear combination of the basis elements of $W = V_n,$ we get
\begin{align*}
T(e_k) &= \sum_{i=1}^n t_{ik}e_i
\\
&= O
\end{align*}
This transformation can be achieved by setting $t_{ik} = 0$ for all $i, k,$ giving us the matrix representation:
\begin{align*}
\begin{bmatrix}
0 & 0 & \cdots & 0
\\
0 & 0 & \cdots & 0
\\
\vdots & \vdots & & \vdots
\\
0 & 0 & 0 & 0
\end{bmatrix}
\end{align*}
where every element is $0. \quad \blacksquare$
(c) $\quad$ Multiplication by a fixed scalar $c.$
$\quad$ For each element $x$ in $V,$ we have $T(x) = cx,$ where $c$ is a fixed scalar. Moreover, for each basis element, we have
\begin{align*}
T(e_k) &= \sum_{i=1}^n t_{ik}e_i
\\
&= ce_k
\end{align*}
This transformation can be achieved by setting $t_{ik} = c$ when $i = k$ and $0$ elsewhere, giving us the matrix representation:
\begin{align*}
\begin{bmatrix}
c & 0 & \cdots & 0
\\
0 & c & \cdots & 0
\\
\vdots & \vdots & & \vdots
\\
0 & 0 & 0 & c
\end{bmatrix}
\end{align*}
where the diagonal elements are $c$ and the rest are $0. \quad \blacksquare$