- 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.
3. $\quad$ A linear transformation $T: V_2 \rightarrow V_2$ maps the basis vectors $\mathbf{i}$ and $\mathbf{j}$ as follows:
\begin{align*}
T(\mathbf{i}) &= \mathbf{i} + \mathbf{j}, \quad T(\mathbf{j}) = \mathbf{2i} - \mathbf{j}
\end{align*}
(a) $\quad$ Compute $T(3\mathbf{i} - 4\mathbf{j})$ and $T^2(3\mathbf{i} - 4\mathbf{j})$ in terms of $\mathbf{i}$ and $\mathbf{j}.$
(b) $\quad$ Determine the matrix of $T$ and of $T^2$.
(c) $\quad$ Solve part (b) if the basis $(\mathbf{i}, \mathbf{j})$ is replaced by $(e_1, e_2),$ where $e_1 = \mathbf{i} - \mathbf{j},$ $e_2 = 3\mathbf{i} + \mathbf{j}.$
Solution. $\quad$
(a) $\quad$ Compute $T(3\mathbf{i} - 4\mathbf{j})$ and $T^2(3\mathbf{i} - 4\mathbf{j})$ in terms of $\mathbf{i}$ and $\mathbf{j}.$
$\quad$ Because $T$ is a linear transformation, we have
\begin{align*}
T(3\mathbf{i} - 4\mathbf{j}) &= 3T(\mathbf{i}) - 4T(\mathbf{j})
\\
&= 3(\mathbf{i} + \mathbf{j}) - 4(2\mathbf{i} - \mathbf{j})
\\
&= 7\mathbf{j} - 5\mathbf{i}
\\
\\
T^2(3\mathbf{i} - 4\mathbf{j}) &= T[T(3\mathbf{i} - 4\mathbf{j})]
\\
&= T(7\mathbf{j} - 5\mathbf{i})
\\
&= 7T(\mathbf{j}) - 5T(\mathbf{i})
\\
&= 7(\mathbf{2i} - \mathbf{j}) - 5(\mathbf{i} + \mathbf{j})
\\
&= 9\mathbf{i} - 12\mathbf{j} \quad \blacksquare
\end{align*}
(b) $\quad$ Determine the matrix of $T$ and of $T^2$.
$\quad$ If we set $(e_1, e_2)$ and $(w_1, w_2)$ as the respective bases of $V$ and $W,$ with $e_1 = w_1 = \mathbf{i}$ and $e_2 = w_2 = \mathbf{j},$ then, denoting the matrix of $T$ by $(t_{ik}),$we have
\begin{align*}
T(e_1) &= t_{11}w_1 + t_{21}w_2
\\
&= w_1 + w_2
\\
T(e_2) &= t_{12}w_1 + t_{22}w_2
\\
&= 2w_1 - w_2
\end{align*}
From this, we can see that the matrix of $T$ is
\begin{bmatrix}
1 & 2
\\
1 & -1
\end{bmatrix}
Then, if we set $u_1 = \mathbf{i}$ and $u_2 = \mathbf{j}$ as the basis elements of $T^2,$ applying $T$ once more, we get:
\begin{align*}
\\
T^2(e_1) &= T[T(e_1)]
\\
&= T[w_1 + w_2]
\\
&= T(w_1) + T(w_2)
\\
&= T(\mathbf{i}) + T(\mathbf{j})
\\
&= (\mathbf{i} + \mathbf{j}) + (2\mathbf{i} - \mathbf{j})
\\
&= 3u_1 + 0u_2
\\
&= v_{11}u_1 + v_{21}u_2
\\
\\
T^2(e_2) &= T[T(e_2)]
\\
&= T[2w_1 - w_2]
\\
&= 2T(w_1) - T(w_2)
\\
&= 2T(\mathbf{i}) - T(\mathbf{j})
\\
&= 2(\mathbf{i} + \mathbf{j}) - (2\mathbf{i} - \mathbf{j})
\\
&= 0u_1 + 3u_2
\\
&= v_{12}u_1 + v_{22}u_2
\end{align*}
where $(v_{ik})$ denotes the matrix of $T^2,$ giving us the following matrix representation:
\begin{align*}
\begin{bmatrix}
3 & 0
\\
0 & 3
\end{bmatrix}
\quad \blacksquare
\end{align*}
(c) $\quad$ Solve part (b) if the basis $(\mathbf{i}, \mathbf{j})$ is replaced by $(e_1, e_2),$ where $e_1 = \mathbf{i} - \mathbf{j},$ $e_2 = 3\mathbf{i} + \mathbf{j}.$
$\quad$ If we replace the basis elements of $V$ and $W$ by $e_1$ and $e_2$ as described, keeping the transformation the same, we get:
\begin{align*}
T(e_1) &= T(\mathbf{i} - \mathbf{j})
\\
&= T(\mathbf{i}) - T(\mathbf{j})
\\
&= (\mathbf{i} + \mathbf{j}) - (\mathbf{2i} - \mathbf{j})
\\
&= -\mathbf{i} + 2\mathbf{j}
\\
&= t_{11}(\mathbf{i} - \mathbf{j}) + t_{21}(3\mathbf{i} + \mathbf{j})
\\
\\
T(e_2) &= T(3\mathbf{i} + \mathbf{j})
\\
&= 3T(\mathbf{i}) + T(\mathbf{j})
\\
&= 3(\mathbf{i} + \mathbf{j}) + (\mathbf{2i} - \mathbf{j})
\\
&= 5\mathbf{i} + 2\mathbf{j}
\\
&= t_{12}(\mathbf{i} - \mathbf{j}) + t_{22}(3\mathbf{i} + \mathbf{j})
\end{align*}
which gives us the following systems of equations for the components of $(t_{ik}):$
\begin{align*}
t_{11} + 3t_{21} &= -1
\\
t_{21} - t_{11} &= 2
\\
\\
t_{12} + 3t_{22} &= 5
\\
t_{22} - t_{12} &= 2
\end{align*}
To find the matrix of $T$ relative to the ordered basis $(e_1, e_2),$ we solve for the components $t_{ik}:$
\begin{align*}
t_{11} &= -\frac{7}{4}, \quad t_{21} = \frac{1}{4}, \quad t_{12} = -\frac{1}{4}, \quad t_{22} = \frac{7}{4}
\end{align*}
which gives us the matrix:
\begin{align*}
(t_{ik}) &=
\begin{bmatrix}
-\frac{7}{4} & -\frac{1}{4}
\\
\frac{1}{4} & \frac{7}{4}
\end{bmatrix}
\end{align*}
Applying $T$ once more, maintaining the same basis elements $e_1$ and $e_2,$ we get:
\begin{align*}
T^2(e_1) &= T[T(e_1)]
\\
&= T(-\mathbf{i} + 2\mathbf{j})
\\
&= -T(\mathbf{i}) + 2T(\mathbf{j})
\\
&= -(\mathbf{i} + \mathbf{j}) + 2(\mathbf{2i} - \mathbf{j})
\\
&= 3\mathbf{i} - 3\mathbf{j}
\\
&= v_{11}(\mathbf{i} - \mathbf{j}) + v_{21}(3\mathbf{i} + \mathbf{j})
\\
\\
T^2(e_2) &= T[T(e_2)]
\\
&= T(5\mathbf{i} + 2\mathbf{j})
\\
&= 5T(\mathbf{i}) + 2T(\mathbf{j})
\\
&= 5(\mathbf{i} + \mathbf{j}) + 2(\mathbf{2i} - \mathbf{j})
\\
&= 9\mathbf{i} + 3\mathbf{j}
\\
&= v_{12}(\mathbf{i} - \mathbf{j}) + v_{22}(3\mathbf{i} + \mathbf{j})
\end{align*}
which gives the following systems of equations for the components $v_{ik}$ of the matrix representation of $T^2:$
\begin{align*}
v_{11} + 3v_{21} &= 3
\\
v_{21} - v_{11} &= -3
\\
\\
v_{12} + 3v_{22} &= 9
\\
v_{22} - v_{12} &= 3
\end{align*}
Solving the systems gives us the components:
\begin{align*}
v_{11} &= 3, \quad v_{21} = 0, \quad v_{12} = 0, \quad v_{22} = 3
\end{align*}
giving us the matrix representation of $T^2:$
\begin{align*}
(v_{ik}) &=
\begin{bmatrix}
3 & 0
\\
0 & 3
\end{bmatrix}
\quad \blacksquare
\end{align*}