5.11.1.1E: Examples and Basic Properties Exercises

Exercises for 1

solutions

2

Let $V$ denote the set of ordered triples $(x, y, z)$ and define addition in $V$ as in $\mathbb{R}^3$. For each of the following definitions of scalar multiplication, decide whether $V$ is a vector space.

1. $a(x, y, z) = (ax, y, az)$
2. $a(x, y, z) = (ax, 0, az)$
3. $a(x, y, z) = (0, 0, 0)$
4. $a(x, y, z) = (2ax, 2ay, 2az)$

2. No; S5 fails.
3. No; S4 and S5 fail.

Are the following sets vector spaces with the indicated operations? If not, why not?

1. The set $V$ of nonnegative real numbers; ordinary addition and scalar multiplication.
2. The set $V$ of all polynomials of degree $\geq 3$, together with $0$; operations of $\mathbf{P}$.
3. The set of all polynomials of degree $\leq 3$; operations of $\mathbf{P}$.
4. The set $\{1, x, x^{2}, \dots\}$; operations of $\mathbf{P}$.
5. The set $V$ of all $2 \times 2$ matrices of the form $\left[ \begin{array}{cc} a & b \\ 0 & c \end{array} \right]$; operations of $\mathbf{M}_{22}$.
6. The set $V$ of $2 \times 2$ matrices with equal column sums; operations of $\mathbf{M}_{22}$.
7. The set $V$ of $2 \times 2$ matrices with zero determinant; usual matrix operations.
8. The set $V$ of real numbers; usual operations.
9. The set $V$ of complex numbers; usual addition and multiplication by a real number.
10. The set $V$ of all ordered pairs $(x, y)$ with the addition of $\mathbb{R}^2$, but using scalar multiplication $a(x, y) = (ax, -ay)$.
11. The set $V$ of all ordered pairs $(x, y)$ with the addition of $\mathbb{R}^2$, but using scalar multiplication $a(x, y) = (x, y)$ for all $a$ in $\mathbb{R}$.
12. The set $V$ of all functions $f$ : $\mathbb{R} \to \mathbb{R}$ with pointwise addition, but scalar multiplication defined by $(af)(x) = f(ax)$.
13. The set $V$ of all $2 \times 2$ matrices whose entries sum to $0$; operations of $\mathbf{M}_{22}$.
14. The set $V$ of all $2 \times 2$ matrices with the addition of $\mathbf{M}_{22}$ but scalar multiplication $*$ defined by $a * X = aX^{T}$.

2. No; only A1 fails.
3. No.
4. Yes.
5. Yes.
6. No.
7. No; only S3 fails.
8. No; only S4 and S5 fail.

[ex:ex6_1_3] Let $V$ be the set of positive real numbers with vector addition being ordinary multiplication, and scalar multiplication being $a \cdot v = v^{a}$. Show that $V$ is a vector space.

[ex:ex6_1_4] If $V$ is the set of ordered pairs $(x, y)$ of real numbers, show that it is a vector space with addition $(x, y) + (x_{1}, y_{1}) = (x + x_{1}, y + y_{1} + 1)$ and scalar multiplication $a(x, y) = (ax, ay + a - 1)$. What is the zero vector in $V$?

The zero vector is $(0, -1)$; the negative of $(x, y)$ is $(-x, -2 - y)$.

Find $\mathbf{x}$ and $\mathbf{y}$ (in terms of $\mathbf{u}$ and $\mathbf{v}$) such that:

$\begin{array}[t]{rlrcr} 2\mathbf{x} & + & \mathbf{y} & = & \mathbf{u} \\ 5\mathbf{x} & + & 3\mathbf{y} & = & \mathbf{v} \\ \end{array}$ $\begin{array}[t]{rlrcr} 3\mathbf{x} & - & 2\mathbf{y} & = & \mathbf{u} \\ 4\mathbf{x} & - & 5\mathbf{y} & = & \mathbf{v} \\ \end{array}$

2. $\mathbf{x} = \frac{1}{7}(5\mathbf{u} - 2\mathbf{v}), \mathbf{y} = \frac{1}{7}(4\mathbf{u} - 3\mathbf{v})$

In each case show that the condition $a\mathbf{u} + b\mathbf{v} + c\mathbf{w} = \mathbf{0}$ in $V$ implies that $a = b = c = 0$.

1. $V = \mathbb{R}^4$; $\mathbf{u} = (2, 1, 0, 2)$, $\mathbf{v} = (1, 1, -1, 0)$, $\mathbf{w} = (0, 1, 2, 1)$
2. $V =\|{M}_{22}$; $\mathbf{u} = \left[ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right]$, $\mathbf{v} = \left[ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right]$, $\mathbf{w} = \left[ \begin{array}{rr} 1 & 1 \\ 1 & -1 \end{array} \right]$
3. $V =\|{P}$; $\mathbf{u} = x^{3} + x$, $\mathbf{v} = x^{2} + 1$, $\mathbf{w} = x^{3} - x^{2} + x + 1$
4. $V =\|{F}[0, \pi]$; $\mathbf{u} = \sin x$, $\mathbf{v} = \cos x$, $\mathbf{w} = 1$—the constant function

2. Equating entries gives $a + c = 0$, $b + c = 0$, $b + c = 0$, $a - c = 0$. The solution is $a = b = c = 0$.
3. If $a \sin x + b \cos y + c = 0$ in $\mathbf{F}[0, \pi]$, then this must hold for every $x$ in $[0, \pi]$. Taking $x = 0, \frac{\pi}{2}$, and $\pi$, respectively, gives $b + c = 0$, $a + c = 0$, $-b + c = 0$ whence, $a = b = c = 0$.

Simplify each of the following.

1. $3[2(\mathbf{u} - 2\mathbf{v} - \mathbf{w}) + 3(\mathbf{w} - \mathbf{v})] - 7(\mathbf{u} - 3\mathbf{v} - \mathbf{w})$
2. $4(3\mathbf{u} - \mathbf{v} + \mathbf{w}) - 2[(3\mathbf{u} - 2\mathbf{v}) - 3(\mathbf{v} - \mathbf{w})] \newline + 6(\mathbf{w} - \mathbf{u} - \mathbf{v})$

2. $4\mathbf{w}$

Show that $\mathbf{x} = \mathbf{v}$ is the only solution to the equation $\mathbf{x} + \mathbf{x} = 2\mathbf{v}$ in a vector space $V$. Cite all axioms used.

Show that $-\mathbf{0} = \mathbf{0}$ in any vector space. Cite all axioms used.

[ex:6_1_10] Show that the zero vector $\mathbf{0}$ is uniquely determined by the property in axiom A4.

If $\mathbf{z} + \mathbf{v} = \mathbf{v}$ for all $\mathbf{v}$, then $\mathbf{z} + \mathbf{v} = \mathbf{0} + \mathbf{v}$, so $\mathbf{z} = \mathbf{0}$ by cancellation.

[ex:6_1_11] Given a vector $\mathbf{v}$, show that its negative $-\mathbf{v}$ is uniquely determined by the property in axiom A5.

[ex:6_1_12]

1. Prove (2) of Theorem [thm:017797]. [Hint: Axiom S2.]
2. Prove that $(-a)\mathbf{v} = -(a\mathbf{v})$ in Theorem [thm:017797] by first computing $(-a)\mathbf{v} + a\mathbf{v}$. Then do it using (4) of Theorem [thm:017797] and axiom S4.
3. Prove that $a(-\mathbf{v}) = -(a\mathbf{v})$ in Theorem [thm:017797] in two ways, as in part (b).

2. $(-a)\mathbf{v} + a\mathbf{v} = (-a + a)\mathbf{v} = 0\mathbf{v} = \mathbf{0}$ by Theorem [thm:017797]. Because also $-(a\mathbf{v}) + a\mathbf{v} = \mathbf{0}$ (by the definition of $-(a\mathbf{v})$ in axiom A5), this means that $(-a)\mathbf{v} = -(a\mathbf{v})$ by cancellation. Alternatively, use Theorem [thm:017797](4) to give $(-a)\mathbf{v} = [(-1)a]\mathbf{v} = (-1)(a\mathbf{v}) = -(a\mathbf{v})$.

[ex:6_1_13] Let $\mathbf{v}, \mathbf{v}_{1}, \dots, \mathbf{v}_{n}$ denote vectors in a vector space $V$ and let $a, a_{1}, \dots, a_{n}$ denote numbers. Use induction on $n$ to prove each of the following.

1. $a(\mathbf{v}_{1} + \mathbf{v}_{2} + \dots + \mathbf{v}_{n}) = a\mathbf{v}_{1} + a\mathbf{v}_{2} + \dots + a\mathbf{v}_{n}$
2. $(a_{1} + a_{2} + \dots + a_{n})\mathbf{v} = a_{1}\mathbf{v} + a_{2}\mathbf{v} + \dots + a_{n}\mathbf{v}$

2. The case $n = 1$ is clear, and $n = 2$ is axiom S3. If $n > 2$, then $(a_{1} + a_{2} + \dots + a_{n})\mathbf{v} = [a_{1} + (a_{2} + \dots + a_{n})]\mathbf{v} = a_{1}\mathbf{v} + (a_{2} + \dots + a_{n})\mathbf{v} = a_{1}\mathbf{v} + (a_{2}\mathbf{v} + \dots + a_{n}\mathbf{v})$ using the induction hypothesis; so it holds for all $n$.

[ex:6_1_14] Verify axioms A2—A5 and S2—S5 for the space $\mathbf{F}[a, b]$ of functions on $[a, b]$ (Example [exa:017760]).

Prove each of the following for vectors $\mathbf{u}$ and $\mathbf{v}$ and scalars $a$ and $b$.

1. If $a\mathbf{v} = \mathbf{0}$, then $a = 0$ or $\mathbf{v} = \mathbf{0}$.
2. If $a\mathbf{v} = b\mathbf{v}$ and $\mathbf{v} \neq \mathbf{0}$, then $a = b$.
3. If $a\mathbf{v} = a\mathbf{w}$ and $a \neq 0$, then $\mathbf{v} = \mathbf{w}$.

3. If $a\mathbf{v} = a\mathbf{w}$, then $\mathbf{v} = 1\mathbf{v} = (a^{-1}a)\mathbf{v} = a^{-1}(a\mathbf{v}) = a^{-1}(a\mathbf{w}) = (a^{-1}a)\mathbf{w} = 1\mathbf{w} = \mathbf{w}$.

By calculating $(1 + 1)(\mathbf{v} + \mathbf{w})$ in two ways (using axioms S2 and S3), show that axiom A2 follows from the other axioms.

[ex:ex6_1_17] Let $V$ be a vector space, and define $V^{n}$ to be the set of all $n$-tuples $(\mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n})$ of $n$ vectors $\mathbf{v}_{i}$, each belonging to $V$. Define addition and scalar multiplication in $V^{n}$ as follows:

$$\begin{gathered} (\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n) + (\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n) \\ \quad = (\mathbf{u}_1 + \mathbf{v}_1, \mathbf{u}_2 + \mathbf{v}_2, \dots, \mathbf{u}_n + \mathbf{v}_n) \\ a(\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n) = (a\mathbf{v}_1, a\mathbf{v}_2, \dots, a\mathbf{v}_n)\end{gathered} \nonumber$$

Show that $V^{n}$ is a vector space.

[ex:6_1_18] Let $V^{n}$ be the vector space of $n$-tuples from the preceding exercise, written as columns. If $A$ is an $m \times n$ matrix, and $X$ is in $V^{n}$, define $AX$ in $V^{m}$ by matrix multiplication. More precisely, if

$$A = \left[ a_{ij} \right] \mbox{ and } X = \left[ \begin{array}{c} \mathbf{v}_1 \\ \vdots \\ \mathbf{v}_n \end{array} \right], \mbox{ let } AX = \left[ \begin{array}{c} \mathbf{u}_1 \\ \vdots \\ \mathbf{u}_n \end{array} \right] \nonumber$$

where $\mathbf{u}_{i} = a_{i1}\mathbf{v}_{1} + a_{i2}\mathbf{v}_{2} + \dots + a_{in}\mathbf{v}_{n}$ for each $i$. Prove that:

1. $B(AX) = (BA)X$
2. $(A + A_{1})X = AX + A_{1}X$
3. $A(X + X_{1}) = AX + AX_{1}$
4. $(kA)X = k(AX) = A(kX)$ if $k$ is any number
5. $IX = X$ if $I$ is the $n \times n$ identity matrix
6. Let $E$ be an elementary matrix obtained by performing a row operation on the rows of $I_{n}$ (see Section [sec:2_5]). Show that $EX$ is the column resulting from performing that same row operation on the vectors (call them rows) of $X$. [Hint: Lemma [lem:005213].]