- 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.
7. $\quad$ A linear transformation $T: V_3 \rightarrow V_2$ maps the basis vectors as follows:
\begin{align*}
T(\mathbf{i}) &= (0, 0),
\quad
T(\mathbf{j}) = (1, 1),
\quad
T(\mathbf{k}) = (1, -1)
\end{align*}
(a) $\quad$ Compute $T(4\mathbf{i} - \mathbf{j} + \mathbf{k})$ and determine the nullity and rank of $T.$
(b) $\quad$ Determine the matrix of $T.$
(c) $\quad$ Use the basis $(\mathbf{i}, \mathbf{j}, \mathbf{k})$ in $V_2$ and the basis $(w_1, w_2)$ in $V_2$ where $w_1 = (1, 1),$ $w_2 = (1, 2).$ Determine the matrix of $T$ relative to these bases
(d) $\quad$ Find bases $(e_1, e_2, e_3)$ for $V_2$ and $(w_1, w_2)$ for $V_2$ relative to which the matrix of $T$ will be in diagonal form.
Solution. $\quad$
(a) $\quad$ Compute $T(4\mathbf{i} - \mathbf{j} + \mathbf{k})$ and determine the nullity and rank of $T.$
\begin{align*}
T(4\mathbf{i} - \mathbf{j} + \mathbf{k}) &= 4T(\mathbf{i}) - T(\mathbf{j}) + T(\mathbf{k})
\\
&= 4(0, 0) - (1, 1) + (1, -1)
\\
&= (0, -2)
\\
&= 0(\mathbf{i}) - 2(\mathbf{j})
\end{align*}
To compute the nullity and rank of $T,$ we will first find the null space of $T.$ To do so, we must find the set of $x = c_1\mathbf{i} + c_2\mathbf{j} + c_3\mathbf{k}$ in $V_3$ such that $$T(x) = c_1T(\mathbf{i}) + c_2T(\mathbf{j}) + c_3T(\mathbf{k}) = O$$ where $c_1, c_2, c_3$ are real scalars. Using the given equations for $T(\mathbf{i}),$ $T(\mathbf{j}),$ and $T(\mathbf{k}),$ we get
\begin{align*}
T(x) &= c_1T(\mathbf{i}) + c_2T(\mathbf{j}) + c_3T(\mathbf{k})
\\
&= c_1(0, 0) + c_2(1, 1) + c_3(1, -1)
\end{align*}
Setting this equal to zero, we get the following system of equations for $c_1, c_2,$ and $c_3:$
\begin{align*}
0(c_1) + c_2 + c_3 &= 0
\\
0(c_1) + c_2 - c_3 &= 0
\end{align*}
This system is satisfied when $c_2 = c_3 = 0,$ with arbitrary $c_1.$ In other words, $N(T)$ is the set of $x$ in $V_3$ of the form $x = c_1\mathbf{i}.$ Moreover, because $T(x) = O$ implies $c_2 = c_3 = 0,$ this means that $T(\mathbf{j})$ and $T(\mathbf{k})$ are independent elements of $T(V_3).$ As such, $T$ has nullity $1$ and rank $\geq 2.$ But by the nullity plus rank theorem, we know that $\dim N(T) + \dim T(V_3)= \dim V_3 = 3,$ giving us $$ \text{nullity 1} \quad \text{rank 2}. \quad \blacksquare$$
(b) $\quad$ Determine the matrix of $T.$
$\quad$ Setting $(\mathbf{i}, \mathbf{j})$ as the basis of $T(V_3),$ the transformations of $\mathbf{i},$ $\mathbf{j},$ and $\mathbf{k}$ can be rewritten as follows:
\begin{align*}
T(\mathbf{i}) &= (0, 0)
\\
&= 0\mathbf{i} + 0\mathbf{j}
\\
\\
T(\mathbf{j}) &= (1, 1)
\\
&= \mathbf{i} + \mathbf{j}
\\
\\
T(\mathbf{k}) &= (1, -1)
\\
&= \mathbf{i} - \mathbf{k}
\end{align*}
Then, the matrix of $T$ is the $2 \times 3$ array whose elements $t_{ik}$ are the scalars multiplying the $i^{th}$ component of $T(e_k),$ where $e_k$ is the $k^{th}$ basis element of $V_3.$ In other words, we have
\begin{align*}
(t_{ik}) &=
\begin{bmatrix}
0 & 1 & 1
\\
0 & 1 & -1
\end{bmatrix}
\quad \blacksquare
\end{align*}
(c) $\quad$ Use the basis $(\mathbf{i}, \mathbf{j}, \mathbf{k})$ in $V_3$ and the basis $(w_1, w_2)$ in $V_2$ where $w_1 = (1, 1),$ $w_2 = (1, 2).$ Determine the matrix of $T$ relative to these bases.
$\quad$ Using the revised basis for $V_2,$ we have the following transformations of the basis elements of $V_3:$
\begin{align*}
T(\mathbf{i}) &= (0, 0)
\\
&= 0w_1 + 0w_2
\\
\\
T(\mathbf{j}) &= (1, 1)
\\
&= w_1 + 0w_2
\\
\\
T(\mathbf{k}) &= (1, -1)
\\
&= 3w_1 - 2w_2
\end{align*}
This gives us the following matrix for $T:$
\begin{align*}
(t_{ik}) &=
\begin{bmatrix}
0 & 1 & 3
\\
0 & 0 & -2
\end{bmatrix}
\quad \blacksquare
\end{align*}
(d) $\quad$ Find bases $(e_1, e_2, e_3)$ for $V_3$ and $(w_1, w_2)$ for $V_2$ relative to which the matrix of $T$ will be in diagonal form.
$\quad$ We wish to find bases such that the matrix $(t_{ik})$ for $T$ is given by:
\begin{align*}
(t_{ik}) &=
\begin{bmatrix}
1 & 0 & 0
\\
0 & 1 & 0
\end{bmatrix}
\end{align*}
This matrix corresponds to the transformations:
\begin{align*}
T(e_1) &= 1(w_1) + 0(w_2)
\\
\\
T(e_2) &= 0(w_1) + 1(w_2)
\\
\\
T(e_3) &= 0(w_1) + 0(w_2)
\end{align*}
These equations are satisfied with the following basis elements:
\begin{align*}
e_1 &= \mathbf{j},\ e_2 = \mathbf{k},\ e_3 = \mathbf{i}
\qquad
w_1 = (1, 1),\ w_2 = (1, -1).
\quad \blacksquare
\end{align*}