- 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.
9. $\quad$ Solve Exercise 8 if $T(\mathbf{i}) = (1, 0, 1)$ and $T(\mathbf{j}) = (1, 1, 1).$
Solution. $\quad$
(a) $\quad$ Compute $T(2\mathbf{i} - 3\mathbf{j})$ and determine the nullity and rank of $T.$
\begin{align*}
T(2\mathbf{i} - 3\mathbf{j}) &= 2T(\mathbf{i}) - 3T(\mathbf{j})
\\
&= 2(1, 0, 1) - 3(1, 1, 1)
\\
&= (-1, -3, -1)
\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 $x = c_1\mathbf{i} + c_2\mathbf{j}$ in $V_2$ such that $$T(x) = c_1T(\mathbf{i}) + c_2T(\mathbf{j}) = O$$ where $c_1, c_2$ are real scalars. Using the given equations for $T(\mathbf{i})$ and $T(\mathbf{j}),$ we get
\begin{align*}
T(x) &= c_1T(\mathbf{i}) + c_2T(\mathbf{j})
\\
&= c_1(1, 0, 1) + c_2(1, 1, 1)
\end{align*}
Setting this equal to zero, we get the following system of equations for $c_1$ and $c_2:$
\begin{align*}
c_1 + c_2 &= 0
\\
c_2 &= 0
\end{align*}
As we can see, this system is satisfied when $c_1 = c_2 = 0.$ This means that the null space of $T$ is the set of $x = c_1\mathbf{i} + c_1\mathbf{j}$ where $c_1 = c_2 = 0.$ In other words, $N(T) = \{O\}.$ Moreover, because $T(x) = O$ implies $c_1 = c_2 = 0,$ this means that $T(\mathbf{i})$ and $T(\mathbf{j})$ are independent elements of $T(V_3).$ As such, $T$ has nullity $0$ and rank $\geq 2.$ But by the nullity plus rank theorem, we know that $\dim N(T) + \dim T(V_2)= \dim V_2 = 2,$ giving us $$ \text{nullity 0} \quad \text{rank 2}. \quad \blacksquare$$
(b) $\quad$ Determine the matrix of $T.$
$\quad$ Setting $(\mathbf{i}, \mathbf{j}, \mathbf{k})$ as the basis of $T(V_2),$ the transformations of $\mathbf{i}$ and $\mathbf{j}$ can be rewritten as follows:
\begin{align*}
T(\mathbf{i}) &= (1, 0, 1)
\\
&= \mathbf{i} + 0(\mathbf{j}) + \mathbf{k}
\\
\\
T(\mathbf{j}) &= (1, 1, 1)
\\
&= \mathbf{i} + \mathbf{j} + \mathbf{k}
\end{align*}
Then, the matrix of $T$ is the $3 \times 2$ array whose elements $t_{ik}$ are the scalars multiplying the $i^{th}$ component of $T(e_k),$ where $e_k$ the $k^{th}$ basis element of $V_2.$ In other words, we have
\begin{align*}
(t_{ik}) &=
\begin{bmatrix}
1 & 1
\\
0 & 1
\\
1 & 1
\end{bmatrix}
\quad \blacksquare
\end{align*}
(c) $\quad$ Find bases $(e_1, e_2)$ for $V_2$ and $(w_1, w_2, w_3)$ for $V_3$ 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 & 1
\\
0 & 0
\end{bmatrix}
\end{align*}
This matrix corresponds to the transformations:
\begin{align*}
T(e_1) &= 1(w_1) + 0(w_2) + 0(w_3)
\\
\\
T(e_2) &= 0(w_1) + 1(w_2) + 0(w_3)
\end{align*}
These equations are satisfied with the following basis elements:
\begin{align*}
e_1 &= \mathbf{i},\ e_2 = \mathbf{j} - \mathbf{i}
\qquad
w_1 = (1, 0, 1),\ w_2 = (0, 1, 0),\ w_3 = (0, 0, 1).
\quad \blacksquare
\end{align*}