- 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.
5. $\quad$ Let $T: V_3 \rightarrow V_3$ be a linear transformation such that
\begin{align*}
T(\mathbf{k}) &= 2\mathbf{i} + 3\mathbf{j} + 5\mathbf{k},
\quad
T(\mathbf{j} + \mathbf{k}) = \mathbf{i},
\quad
T(\mathbf{i} + \mathbf{j} + \mathbf{k}) = \mathbf{j} - \mathbf{k}.
\end{align*}
(a) $\quad$ Compute $T(\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})$ and determine the nullity and rank of $T.$
(b) $\quad$ Determine the matrix of $T.$
Solution. $\quad$
(a) $\quad$ Compute $T(\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})$ and determine the nullity and rank of $T.$
$\quad$ We can use the linearity of $T$ to compute $T(\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}):$
\begin{align*}
T(\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) &= T(\mathbf{i} + \mathbf{j} + \mathbf{k}) + T(\mathbf{j} + \mathbf{k}) + T(\mathbf{k})
\\
&= (\mathbf{j} - \mathbf{k}) + (\mathbf{i}) + (2\mathbf{i} + 3\mathbf{j} + 5\mathbf{k})
\\
&= 3\mathbf{i} + 4\mathbf{j} + 4\mathbf{k}
\end{align*}
To determine the nullity of $T,$ we will first find $N(T).$ In other words, we wish to find the set of all elements $x = c_1\,\mathbf{i} + c_2\,\mathbf{j} + c_3\,\mathbf{k}$ in $V_3$ such that $T(x) = O.$ Using the above-defined equations, we can apply the linearity of $T$ to get
\begin{align*}
T(\mathbf{i}) &= T(\mathbf{i} + \mathbf{j} + \mathbf{k}) - T(\mathbf{j} + \mathbf{k})
\\
&= -\mathbf{i} + \mathbf{j} - \mathbf{k}
\\
&=(-1, 1, -1)
\\
\\
T(\mathbf{j}) &= T(\mathbf{j} + \mathbf{k}) - T(\mathbf{k})
\\
&= -(\mathbf{i} + 3\mathbf{j} + 5\mathbf{k})
\\
&= (-1, -3, -5)
\\
\\
T(\mathbf{k}) &= 2\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}
\\
&= (2, 3, 5)
\end{align*}
If $T(x) = O$ for $x = c_1\,\mathbf{i} + c_2\,\mathbf{j} + c_3\,\mathbf{k},$ then we have $c_1T(\mathbf{i}) + c_2T(\mathbf{j}) + c_3T(\mathbf{k}) = O,$ giving us the following system of equations for the scalars $c_1, c_2, c_3:$
\begin{align*}
-c_1 - c_2 + 2c_3 &= 0
\\
c_1 - 3c_2 + 3c_3 &= 0
\\
-c_1 -5c_2 + 5c_3 &= 0
\end{align*}
From this, we can see that this system of equations is satisfied only when $c_1 = c_2 = c_3 = 0,$ which means that $T(x) = O$ implies $x = O,$ giving us $N(T) = \{O\}.$ Moreover, we find that the elements $T(\mathbf{i}),$ $T(\mathbf{j}),$ and $T(\mathbf{k})$ are independent. Thus, as shown in the nullity plus rank theorem, $T$ has nullity $0$ and rank $3. \quad \blacksquare$
(b) $\quad$ Determine the matrix of $T.$
$\quad$ To determine the matrix representation of $T,$ we must find the coefficients $t_{ik}$ such that for the $k^{th}$ basis element of $V,$ $e_k,$ we have $T(e_k) = \sum_{i=1}^m t_{ik}w_i,$ where $m$ is the dimension of $T(V)$ and $w_i$ is the $i^{th}$ basis element of $T(V).$ From (a), we know that $\dim T(V) = 3,$ and if we set $e_1 = w_1 = \mathbf{i},$ $e_2 = w_2 = \mathbf{j},$ and $e_3 = w_3 = \mathbf{k},$ we get
\begin{align*}
T(\mathbf{i}) &= t_{11}\mathbf{i} + t_{21}\mathbf{j} + t_{31}\mathbf{k}
\\
\\
T(\mathbf{j}) &= t_{12}\mathbf{i} + t_{22}\mathbf{j} + t_{32}\mathbf{k}
\\
\\
T(\mathbf{k}) &= t_{13}\mathbf{i} + t_{23}\mathbf{j} + t_{33}\mathbf{k}
\end{align*}
Applying our result from (a), placing component $t_{ik}$ in the $i^{th}$ row and $k^{th}$ column, we get the following matrix representation of $T:$
\begin{align*}
(t_{ik}) &=
\begin{bmatrix}
-1 & -1 & 2
\\
1 & -3 & 3
\\
-1 & -5 & 5
\end{bmatrix}
\quad \blacksquare
\end{align*}