- 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.
10. $\quad$ Let $V$ and $W$ be linear spaces, each with dimension 2 and each with basis $(e_1, e_2).$ Let $T: V \to W$ be a linear transformation such that
\begin{align*}
T(e_1 + e_2) &= 3e_1 + 9e_2, \qquad T(3e_1 + 2e_2) = 7e_1 + 23e_2.
\end{align*}
(a) $\quad$ Compute $T(e_2 - e_1)$ and determine the nullity and rank of $T.$
(b) $\quad$ Determine the matrix of $T$ relative to the given basis of $T.$
(c) $\quad$ Use the basis $(e_1, e_2)$ for $V$ and find a new basis of the form $(e_1 + ae_2, 2e_1 + be_2)$ for $W,$ relative to which the matrix of $T$ will be in diagonal form.
Solution. $\quad$
(a) $\quad$ Compute $T(e_2 - e_1)$ and determine the nullity and rank of $T.$
$\quad$ We can find $T(e_2 - e_1)$ by using the linearity of $T$ with our given equations for $T(e_1 + e_2)$ and $T(3e_1 + 2e_2):$
\begin{align*}
T(e_2 - e_1) &= T(e_1 + e_2) - 2T(e_1)
\\
&= T(e_1 + e_2) - 2[T(3e_1 + 2e_2) - 2T(e_1 + e_2)]
\\
&= (3e_1 + 9e_2) - 2[(7e_1 + 23e_2) - 2(3e_1 + 9e_2)]
\\
&= (3e_1 + 9e_2) - 2(e_1 + 5e_2)
\\
&= e_1 - e_2
\end{align*}
To determine the nullity and rank of $T,$ we will first find the null space of $T.$ Thus, we wish to find the set of all $x = c_1e_1 + c_2e_2$ in $V$ such that $T(x) = O,$ where $c_1, c_2$ are scalars. Since we know that $T(e_2 - e_1) = e_1 - e_2$ and $T(e_1) = e_1 + 5e_2,$ $T(x)$ is given by:
\begin{align*}
T(x) &= c_1T(e_1) + c_2T(e_2)
\\
&= c_1T(e_1) + c_2[T(e_2 - e_1) + T(e_1)]
\\
&= c_1(e_1 + 5e_2) + c_2[(e_1 - e_2) + (e_1 + 5e_2)]
\\
&= (c_1 + 2c_2)e_1 + (5c_1 + 4c_2)e_2
\end{align*}
Because $e_1$ and $e_2$ are independent elements of $W,$ setting $T(x) = O$ implies that
\begin{align*}
c_1 + 2c_2 &= 0
\\
5c_1 + 4c_2 &= 0
\end{align*}
Solving this system of equations, we find that $c_1 = -2c_2$ and $5c_1 + 4c_2 = -6c_2 = 0,$ giving us $c_1 = c_2 = 0.$ As such, the set of all $x = c_1e_1 + c_2e_2$ satisfying $T(x) = O$ is the set containing only the zero element. In other words, $N(T) = \{O\}.$ And because $T(x) = O$ implies $c_1 = c_2 = 0,$ this means that $T(e_1)$ and $T(e_2)$ are independent elements of $T(V).$ This means that $T$ has nullity $0$ and rank $\geq 2.$ But by the nullity plus rank theorem, because $\dim V = 2,$ we know that $\dim T(V)$ must be $\leq 2,$ giving us
\begin{align*}
\text{nullity 0} \quad \text{rank 2} \quad \blacksquare
\end{align*}
(b) $\quad$ Determine the matrix of $T$ relative to the given basis of $T.$
$\quad$ To determine the matrix of $T$ relative to the given basis $(e_1, e_2),$ we must first find the transformation of each basis element in terms of the basis components of $W:$
\begin{align*}
T(e_1) &= e_1 + 5e_2
\\
T(e_2) &= 2e_1 + 4e_2
\end{align*}
Then, the scalar multiplying the $i^{th}$ component of $T(e_k)$ goes in the $i^{th}$ row and $k^{th}$ column of a $2 \times 2$ array to give us:
\begin{align*}
(t_{ik}) &=
\begin{bmatrix}
1 & 2
\\
5 & 4
\end{bmatrix}
\quad \blacksquare
\end{align*}
(c) $\quad$ Use the basis $(e_1, e_2)$ for $V$ and find a new basis of the form $(e_1 + ae_2, 2e_1 + be_2)$ for $W,$ relative to which the matrix of $T$ will be in diagonal form.
$\quad$ For the matrix relative to the new basis to be diagonal, we need to find scalars $a$ and $b$ such that
\begin{align*}
T(e_1) &= 1(e_1 + ae_2) + 0(2e_1 + be_2)
\\
&= e_1 + ae_2
\\
\\
T(e_2) &= 0(e_1 + ae_2) + 1(2e_1 + be_2)
\\
&= 2e_1 + be_2
\end{align*}
But as we saw in (b), this is satisfied when $a = 5$ and $b = 4. \quad \blacksquare$