<Home>Teaching><MATH 115A (Spring 2023)>Worksheet 4

Worksheet 4

Problem 3

Let $V = \C$ be the vector space of complex numbers over the field $\R$ . Define the function $\func{T}{V}{V}$ by $T\p{z} = \conj{z}$ , where $\conj{z}$ is the complex conjugate of $z$ .

Prove that $T$ is additive.
Prove that $T$ is linear.
Let $\beta = \set{1, i}$ . Prove that $\beta$ is a basis of $V$ over $\R$ .
Compute $\br{T}_\beta$ .

Solution.

Complex conjugation is defined as follows: given $z \in \C$ , we can always write $z = a + bi$ , where $a, b \in \R$ . Then

\conj{z} \coloneqq a - bi.

Let $z, w \in \C$ . Then there exist $a, b, c, d \in \R$ such that $z = a + bi$ and $w = c + di$ . Recall that with this notation, addition in $\C$ is defined as $z + w = \p{a + c} + \p{b + d}i$ . Thus,
$\begin{aligned} T\p{z + w} &= \conj{z + w} \\ &= \conj{\p{a + c} + \p{b + d}i} \\ &= a + c - \p{b + d}i \\ &= \p{a - bi} + \p{c - di} \\ &= \conj{z} + \conj{w} \\ &= T\p{z} + T\p{w}. \end{aligned}$
Let $z = a + bi \in \C$ and let $c \in \R$ . Note that by definition, $cz = ca + cbi$ , and that $ca, cb \in \R$ . Thus,
$\begin{aligned} T\p{cz} &= \conj{cz} \\ &= \overline{ca + cbi} \\ &= ca - cbi \\ &= c\p{a - bi} \\ &= c \conj{z} \\ &= cT\p{z}. \end{aligned}$
By definition, $\beta$ spans $V$ (every complex number can be written in the form $a + bi$ , where $a, b \in \R$ ). We only need to show that $\beta$ is linearly independent. Assume that $z = a + bi = 0$ for some $a, b \in \R$ . Then because $T$ is linear, we have $T\p{z} = 0$ . Thus,
$\begin{cases} 0 = z + T\p{z} = \p{a + bi} + \p{a - bi} = 2a, \\ 0 = z - T\p{z} = \p{a + bi} - \p{a - bi} = 2bi. \end{cases}$
Thus, $a = b = 0$ (since $2 \neq 0$ and $i \neq 0$ in $\C$ ).
Recall that
$\br{T}_\beta = \begin{pmatrix} \vert & \vert \\ \br{T\p{1}}_\beta & \br{T\p{i}}_\beta \\ \vert & \vert \end{pmatrix},$
so we just need to write $T\p{1}$ and $T\p{i}$ as linear combinations of $\beta$ :
$\begin{aligned} T\p{1} = 1 = 1 + 0i &\implies \br{T\p{1}}_\beta = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \\ T\p{i} = -i = 0 + \p{-1}i &\implies \br{T\p{i}}_\beta = \begin{pmatrix} \phantom{-}0 \\ -1 \end{pmatrix} \end{aligned}$
Putting everything together,
$\br{T}_\beta = \begin{pmatrix} 1 & \phantom{-}0 \\ 0 & -1 \end{pmatrix}.$

Problem 5

Let $V$ be a vector space of dimension $n$ , let $\func{T}{V}{V}$ be a linear function. Suppose that $W$ is a $T$ -invariant subspace of $V$ with dimension $k$ . Show that there exists a basis $\beta$ of $V$ such that

\br{T}_\beta = \begin{pmatrix} A & B \\ O & C \end{pmatrix},

where $A$ is a $k \times k$ matrix, $B$ is a $k \times \p{n - k}$ matrix, $C$ is an $\p{n - k} \times \p{n - k}$ matrix, and $O$ is the $\p{n - k} \times k$ zero matrix.

Solution.

If $\beta = \set{v_1, v_2, \ldots, v_n}$ is a basis of $V$ , then recall that

\br{T}_\beta = \begin{pmatrix} \vert & \vert & & \vert \\ \br{T\p{v_1}}_\beta & \br{T\p{v_2}}_\beta & \cdots & \br{T\p{v_n}}_\beta \\ \vert & \vert & & \vert \\ \end{pmatrix}.

So we need to find a basis $\beta$ such that if $1 \leq i \leq k$ , then

\br{T\p{v_i}}_\beta = \begin{pmatrix} a_{1i} \\ a_{2i} \\ \vdots \\ a_{ki} \\ 0 \\ \vdots \\ 0 \end{pmatrix} \iff T\p{v_i} = a_{1i} v_1 + a_{2i} v_2 + \cdots + a_{ki} v_k + 0v_{k+1} + \cdots + 0v_n,

i.e., a basis such that $T\p{v_i}$ can be written as a linear combination of just the first $k$ basis vectors.

Let $\set{v_1, v_2, \ldots, v_k}$ be a basis of $W$ , and extend this to a basis $\beta = \set{v_1, v_2, \ldots, v_k, v_{k+1}, \ldots, v_n}$ of $V$ . Let $1 \leq i \leq k$ . Since $W$ is $T$ -invariant and $v_i \in W$ , this means that $T\p{v_i} \in W$ . But $\set{v_1, v_2, \ldots, v_k}$ is a basis of $W$ , so in particular, $W = \span\set{v_1, v_2, \ldots, v_k}$ . Thus, by definition of span, there exist scalars $a_{1i}, a_{2i}, \ldots, a_{ki} \in \F$ such that

\begin{aligned} T\p{v_i} &= a_{1i} v_1 + a_{2i} v_2 + \cdots + a_{ki} v_k \\ &= a_{1i} v_1 + a_{2i} v_2 + \cdots + a_{ki} v_k + 0v_{k+1} + \cdots + 0v_n. \end{aligned}

Hence, the claim holds with $A$ defined by $A_{ij} = a_{ij}$ .

Note that we can always write $T$ as an $n \times n$ matrix, so we don't need to deal with $B$ or $C$ since we didn't make any assumptions on the last $n - k$ columns of $\br{T}_\beta$ . We only needed to deal with the first $k$ columns and show that they have the special form above.