Exercises 1C — Axler, Linear Algebra Done Right, 4th Edition
Exercise 1. For each of the following subsets of \(\mathbb{F}^3\), determine whether it is a subspace of \(\mathbb{F}^3\).
(a) \(S_1 = \{(x_1,x_2,x_3)\in\mathbb{F}^3 : x_1+2x_2+3x_3=0\}\)
Solution. Yes, \(S_1\) is a subspace.
Additive identity. Taking \(x_1=x_2=x_3=0\) gives \(0+2(0)+3(0)=0\), so \((0,0,0)\in S_1\). \(\checkmark\)
Closed under addition. Let \(x=(x_1,x_2,x_3)\in S_1\) and \(y=(y_1,y_2,y_3)\in S_1\), so \(x_1+2x_2+3x_3=0\) and \(y_1+2y_2+3y_3=0\). Let \(z=x+y=(x_1+y_1,x_2+y_2,x_3+y_3)\). Then:
\[ \begin{aligned} z_1+2z_2+3z_3 &=(x_1+y_1)+2(x_2+y_2)+3(x_3+y_3) \\ &=(x_1+2x_2+3x_3)+(y_1+2y_2+3y_3) \\ &=0+0=0, \end{aligned} \]
so \(z\in S_1\). \(\checkmark\)
Closed under scalar multiplication. Let \(x=(x_1,x_2,x_3)\in S_1\) and \(c\in\mathbb{F}\). Let \(z=cx=(cx_1,cx_2,cx_3)\). Then:
\[ \begin{aligned} z_1+2z_2+3z_3 &=cx_1+2cx_2+3cx_3 \\ &=c(x_1+2x_2+3x_3) \\ &=c\cdot 0=0, \end{aligned} \]
so \(z\in S_1\). \(\checkmark\)
\[ \therefore\ S_1 \text{ is a subspace of } \mathbb{F}^3. \qquad \square \]
(b) \(S_2 = \{(x_1,x_2,x_3)\in\mathbb{F}^3 : x_1+2x_2+3x_3=4\}\)
Solution. No, \(S_2\) is not a subspace.
The additive identity \((0,0,0)\) satisfies \(0+2(0)+3(0)=0\neq 4\), so \((0,0,0)\notin S_2\). A subspace must contain the additive identity. \(\square\)
(c) \(S_3 = \{(x_1,x_2,x_3)\in\mathbb{F}^3 : x_1x_2x_3=0\}\)
Solution. No, \(S_3\) is not a subspace; it is not closed under addition.
Consider \((1,0,1),(0,1,0)\in\mathbb{F}^3\). Both satisfy \(x_1x_2x_3=0\) (products are \(1\cdot0\cdot1=0\) and \(0\cdot1\cdot0=0\)), so both are in \(S_3\). However, their sum is \((1,0,1)+(0,1,0)=(1,1,1)\), and \(1\cdot1\cdot1=1\neq 0\), so \((1,1,1)\notin S_3\). \(\square\)
(d) \(S_4 = \{(x_1,x_2,x_3)\in\mathbb{F}^3 : x_1=5x_3\}\)
Solution. Yes, \(S_4\) is a subspace. The proof is analogous to part (a). \(\square\)
Exercise 3. Show that the set of differentiable real-valued functions \(f\) on the interval \((-4,4)\) such that \(f'(-1)=3f(2)\) is a subspace of \(\mathbb{R}^{(-4,4)}\).
Proof.
Let \(D^{(-4,4)}\) denote the set in question. We verify the three subspace conditions.
Additive identity. The zero function \(f(x)=0\) is real-valued and differentiable on \((-4,4)\), with \(f'(x)=0\) for all \(x\). Thus \(f'(-1)=0=3\cdot 0=3f(2)\), so \(0\in D^{(-4,4)}\). \(\checkmark\)
Closed under addition. Let \(f,g\in D^{(-4,4)}\), so \(f'(-1)=3f(2)\) and \(g'(-1)=3g(2)\). Since \(f\) and \(g\) are differentiable on \((-4,4)\), their sum \(f+g\) is also differentiable on \((-4,4)\), with \((f+g)'=f'+g'\) and \([f+g](x)=f(x)+g(x)\). Thus:
\[ \begin{aligned} (f+g)'(-1) &=f'(-1)+g'(-1) \\ &=3f(2)+3g(2) \\ &=3\bigl(f(2)+g(2)\bigr) \\ &=3\,[f+g](2). \end{aligned} \]
Therefore \((f+g)\in D^{(-4,4)}\). \(\checkmark\)
Closed under scalar multiplication. Let \(f\in D^{(-4,4)}\) and \(c\in\mathbb{F}\). The function \(cf\) is differentiable on \((-4,4)\) with \((cf)'(x)=cf'(x)\) and \([cf](x)=cf(x)\). Thus:
\[ \begin{aligned} (cf)'(-1) &=c\,f'(-1) \\ &=c\cdot 3f(2) \\ &=3\,cf(2) \\ &=3\,[cf](2). \end{aligned} \]
Therefore \(cf\in D^{(-4,4)}\). \(\checkmark\)
\[ \therefore\ D^{(-4,4)} \text{ is a subspace of } \mathbb{R}^{(-4,4)}. \qquad \square \]
Exercise 4. Suppose \(b\in\mathbb{R}\). Show that the set \(S\) of continuous real-valued functions on the interval \([0,1]\) such that \(\int_0^1 f = b\) is a subspace of \(\mathbb{R}^{[0,1]}\) if and only if \(b=0\).
Proof.
Forward direction (\(\Rightarrow\)). Suppose \(S\) is a subspace of \(\mathbb{R}^{[0,1]}\). Suppose for contradiction that \(b\neq 0\). Since \(S\) is a subspace, it must contain the additive identity, i.e. the zero function \(f(x)=0\). But \(\int_0^1 0 = 0 \neq b\), so the zero function is not in \(S\) — a contradiction. Therefore \(b=0\).
\[ \therefore\ b=0. \qquad \square \]
Reverse direction (\(\Leftarrow\)). Suppose \(b=0\). We show \(S\) satisfies the three subspace conditions.
Additive identity. The zero function \(f(x)=0\) is continuous on \([0,1]\) and \(\int_0^1 0 = 0 = b\), so \(0\in S\). \(\checkmark\)
Closed under addition. Let \(f,g\in S\), so \(\int_0^1 f = 0\) and \(\int_0^1 g = 0\). Then:
\[ \int_0^1(f+g) = \int_0^1 f + \int_0^1 g = 0+0 = 0, \]
so \(f+g\in S\). \(\checkmark\)
Closed under scalar multiplication. Let \(f\in S\) and \(c\in\mathbb{R}\). Then:
\[ \int_0^1 cf = c\int_0^1 f = c(0) = 0, \]
so \(cf\in S\). \(\checkmark\)
\[ \therefore\ \text{if } b=0 \text{ then } S \text{ is a subspace of } \mathbb{R}^{[0,1]}. \qquad \square \]
Exercise 19. Suppose \(V_1, V_2, U\) are subspaces of \(V\) such that \(V_1 + U = V_2 + U\). Does it follow that \(V_1 = V_2\)?
Solution. No. Let \(V_1 = \{0\}\), \(V_2 = U\), and suppose \(U \neq \{0\}\). Then
\[V_1 + U = \{0\} + U = U = U + U = V_2 + U,\]
but \(V_1 = \{0\} \neq U = V_2\). \(\square\)
Exercise 20. Suppose \(U = \{(x,x,y,y) \in \mathbb{F}^4 : x,y \in \mathbb{F}\}\). Find a subspace \(W\) of \(\mathbb{F}^4\) such that \(\mathbb{F}^4 = U \oplus W\).
Solution. Let \(W = \{(0,w_1,0,w_2) \in \mathbb{F}^4 : w_1,w_2 \in \mathbb{F}\}\).
\(W\) is a subspace of \(\mathbb{F}^4\): it contains \((0,0,0,0)\), and is closed under addition and scalar multiplication.
\(U \cap W = \{0\}\). Suppose \((a,a,c,c) \in U \cap W\). Since the element is also in \(W\), its first and third coordinates are \(0\), so \(a = 0\) and \(c = 0\). Thus \(U \cap W = \{0\}\).
\(\mathbb{F}^4 = U + W\). Let \(z = (a,b,c,d) \in \mathbb{F}^4\). Set \(u = (a,a,c,c) \in U\) and \(w = (0,b-a,0,d-c) \in W\). Then
\[u + w = (a,\; a+(b-a),\; c,\; c+(d-c)) = (a,b,c,d) = z.\]
Since every \(z \in \mathbb{F}^4\) decomposes as \(u + w\) with \(u \in U\) and \(w \in W\), and since \(U \cap W = \{0\}\),
\[\mathbb{F}^4 = U \oplus W. \qquad \square\]
Exercise 23. Prove or give a counterexample: if \(V_1, V_2, U\) are subspaces of \(V\) such that \(V = V_1 \oplus U\) and \(V = V_2 \oplus U\), then \(V_1 = V_2\).
Solution. False. Let \(V = \mathbb{R}^2\), and let
\[U = \{(0,y) : y \in \mathbb{R}\}, \quad V_1 = \{(x,0) : x \in \mathbb{R}\}, \quad V_2 = \{(x,x) : x \in \mathbb{R}\}.\]
\(U\), \(V_1\), and \(V_2\) are subspaces of \(\mathbb{R}^2\). Moreover \(V_1 \cap U = \{0\}\) and \(V_2 \cap U = \{0\}\).
Let \((x,y) \in \mathbb{R}^2\). Then
\[ \begin{aligned} (x,y) &= (x,0) + (0,y), && (x,0) \in V_1,\ (0,y) \in U, \\ (x,y) &= (x,x) + (0,y-x), && (x,x) \in V_2,\ (0,y-x) \in U. \end{aligned} \]
Therefore \(V = V_1 \oplus U\) and \(V = V_2 \oplus U\), but \(V_1 \neq V_2\). \(\square\)
Exercise 24. Let \(V_e\) denote the set of real-valued even functions on \(\mathbb{R}\), and let \(V_o\) denote the set of real-valued odd functions on \(\mathbb{R}\). Show that \(\mathbb{R}^{\mathbb{R}} = V_e \oplus V_o\).
Proof.
\(V_e\) and \(V_o\) are subspaces of \(\mathbb{R}^{\mathbb{R}}\). The zero function belongs to both \(V_e\) and \(V_o\). For \(f, g \in V_e\), \(h, m \in V_o\), and \(c, d \in \mathbb{R}\):
\[ \begin{aligned} (f+g)(-x) &= f(-x)+g(-x) = f(x)+g(x) = (f+g)(x), \\ (h+m)(-x) &= h(-x)+m(-x) = -h(x)+(-m(x)) = -(h+m)(x), \end{aligned} \]
so \(V_e\) and \(V_o\) are each closed under addition. Similarly,
\[ \begin{aligned} (cf)(-x) &= cf(-x) = cf(x), \\ (dh)(-x) &= dh(-x) = -dh(x), \end{aligned} \]
so \(V_e\) and \(V_o\) are each closed under scalar multiplication. Hence both are subspaces of \(\mathbb{R}^{\mathbb{R}}\).
\(V_e \cap V_o = \{0\}\). Only the zero function is simultaneously even and odd: if \(f \in V_e \cap V_o\) then \(f(x) = f(-x) = -f(x)\) for all \(x\), so \(f(x) = 0\) for all \(x\).
Every \(z \in \mathbb{R}^{\mathbb{R}}\) decomposes into an even part and an odd part. Let \(z \in \mathbb{R}^{\mathbb{R}}\). Define
\[z_1(x) = \frac{z(x)+z(-x)}{2}, \qquad z_2(x) = \frac{z(x)-z(-x)}{2}.\]
Then \(z = z_1 + z_2\). Moreover,
\[z_1(-x) = \frac{z(-x)+z(x)}{2} = z_1(x), \quad \text{so } z_1 \in V_e,\]
\[z_2(-x) = \frac{z(-x)-z(x)}{2} = -\!\left[\frac{z(x)-z(-x)}{2}\right] = -z_2(x), \quad \text{so } z_2 \in V_o.\]
\[\therefore\ \mathbb{R}^{\mathbb{R}} = V_e \oplus V_o. \qquad \square\]