#math/linear-algebra
**Problem 1** For each of the following subsets of $F^3$, determine whether it is a subspace of $F^3$:
1. ${(x_{1},x_{2},x_{3}) \in F^3 : x_{1} + 2x_{2} + 3x_{3} = 0}$
2. ${(x_{1},x_{2},x_{3}) \in F^3 : x_{1} + 2x_{2} + 3x_{3} = 4}$
3. ${(x_{1},x_{2},x_{3}) \in F^3 : x_{1}x_{2}x_{3} = 0}$
4. ${(x_{1},x_{2},x_{3}) \in F^3 : x_{1} = 5x_{3}}$
**Problem 1.1** $U = \{(x_{1},x_{2},x_{3}) \in F^3 : x_{1} + 2x_{2} + 3x_{3} = 0\}$
1. $0 \in{} U$
2. Let $x, y \in{} U$. Then if we have $x + y$
$
\begin{align}
(x+y)&= (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{align}
$
3. Let $u \in{} U$ and $c \in{} F$. Then if we have $cu$
$
\begin{align}
c(u_{1}+2u_{2} + 3u_{3}) & = c(0) \\
& = 0
\end{align}
$
$\therefore U$ is a subspace of $F^3$.
**Problem 1.2** $U = \{(x_{1},x_{2},x_{3}) \in F^3 : x_{1} + 2x_{2} + 3x_{3} = 4\}$
1. $0 \not\in U$. $(0,0,0)$ can not satisfy the constraint.
$\therefore U$ is not a subspace of $F^3$.
**Problem 1.3** $U = \{(x_{1},x_{2},x_{3}) \in F^3 : x_{1}x_{2}x_{3} = 0\}$
1. $0 \in{} F^3$.
2. Let $u, v \in{} F^3$.
$
\begin{align}
u+v & = (u_{1}u_{2}u_{3}) + (v_{1}v_{2}v_{3}) \\
& = 0 + 0 \\
& = 0
\end{align}
$
3. Let $u \in{} F^3$ and $c\in{}F$.
$
\begin{align}
cu & = c(u_{1}u_{2}u_{3}) \\
& = c0\\
& = 0
\end{align}
$
So $cu\in U$
$\therefore U$ is a subspace of $F^3$.
**Problem 1.4** $U = \{(x_{1},x_{2},x_{3}) \in F^3 : x_{1} = 5x_{3}\}$
Before even fully answering... we know that the condition is linear and [[Homogenous Equation|Homogenous]] representing a plane or complex plane if $F=\mathbb{R}$ or $F=\mathbb{C}$, respectively.
1. $0 \in{} U$
2. Let $u=(5u_{3},u_{2},u_{3}), v=(5v_{3},v_{2},v_{3}) \in{} U$
$
\begin{align}
u+v & = (5u_{3},u_{2},u_{3}) + (5v_{3},v_{2},v_{3}) \\
& = (5(u_{3} + v_{3}), (u_{2}+v_{2}), (u_{3} + v_{3}))
\end{align}
$
3. Let $c \in{} F, u \in{} U$.
$
\begin{align}
cu & = c(5u_{3}, u_{2}, u_{3}) \\
& = (5(cu_{3}),cu_{2}, cu_{3})
\end{align}
$
$\therefore U$ is a subspace of $F^3$.
---
**Problem 2** Verify all the assertions in Example 1.35.
(a) If $b \in{} F$, then
$
\{(x_{1},x_{2},x_{3},x_{4}) \in{} F^4 : x_{3} = 5x_{4}+b\}
$
is a subspace of $F^4$ if and only if $b=0$.
$(\Rightarrow)$Let $b=0$.
$U = \{(x_{1},x_{2},5x_{4},x_{4}) \in{} F^4\}$
(1) $(0,0,0,0) \in{} U$
(2) Let $u = (a, b, 5d, d) : a,b,d \in{} F$ and $w=(x,y,5z, z) : x,y,z \in{} F$.
$
\begin{align}
u + w & = (a, b, 5d, d) + (x,y,5z,z) \\
& = (a + x, b + y, 5(d + z), d+z)
\end{align}
$
(3) Let $u = (a,b,5d,d)$ and $\alpha \in{} F$.
$
\begin{align}
\alpha u & = (\alpha a, \alpha b, 5(\alpha d), \alpha d)
\end{align}
$
$\therefore U$ is a subspace of $F^4$.
$(\Leftarrow)$ Let $U$ be a subspace of $F^4$.
$0 \in{} U$. Setting all the free variables to $0$, $(x_{1},x_{2},5x_{4}+b,x_{4}) = (0,0,b,0)$.
$\therefore b = 0$
****
(b) The set of continuous real-valued functions on the interval $[0,1]$ is a subspace of $R^{[0,1]}$.
Let $U = \{f \in{} R^{[0,1]}: f \text{ is continuous }\}$ .
(1) $0 \in{} U$ since $f(x) = 0$ is continuous.
(2) The sum of continuous functions, $(f + g)(x)$ is continuous.
(3) Scaling a continuous function by $c \in{} \mathbb{R}$, $cf(x)$, is continuous.
****
(c) The set of differentiable real-valued functions on $\mathbb{R}$ is a subspace of $\mathbb{R}^\mathbb{R}$.
Let $U = \{f \in{} R^{R}: f \text{ is differentiable }\}$ .
(1) $0 \in{} U$ since $f(x) = 0$ is differentiable. $f'(x) = 0$.
(2) If $f \in{} U$ and $g \in{} U$, then $(f+g)(x) \in{} U$ is differentiable. $(f+g)'(x) = f'(x) + g'(x)$.
(3) If $c\in{}\mathbb{R}$ and $f \in{} U$ then $cf \in{} U$ since $(cf)'(x) = cf'(x)$.
****
(d) The set of differentiable real-valued functions $f$ on the interval (0, 3) such that $f'(2) = b$ is a subspace of $\mathbb{R}^{(0,3)}$ if and only if $b=0$.
Let $U = \{f \in{} R^{(0,3)} : f\text{ is differentiable and }f'(2) = b \in{}\mathbb{R}\}$.
Let $b = 0$ and $f'(2) = 0$.
- (1) $0 \in{} U$, since $0'(x) = 0$.
- (2) $f,g \in{} U$ then $(f + g)'(x) = f'(x) + g'(x)$. and $(f+g)'(2) = f'(2) + g'(2) = 0 + 0 = 0$. So $(f+g)(x) \in{} U$.
- (3) $f \in{} U$ and $c \in{} \mathbb{\mathbb{R}}$. Then $(cf)(x) = cf(x)$ and $(cf)'(2) = cf'(2) = c(0) = 0$. So $cf(x) \in{} U$. Thus if $b=0$ then $U$ is a subspace of $R^{(0,3)}$
Let $U$ be a subspace of $\mathbb{R}^{(0,3)}$. Let $f,g \in{} U$ then $(f + g)'(x) = f'(x) + g'(x)$. and $(f+g)'(2) = f'(2) + g'(2) = b + b = 2b$. Thus
$
\begin{align}
b & =2b \\
\end{align}
$
So $b=0$.
$\therefore U \text{ is a subspace of } \mathbb{R}^{(0,3)} \iff b=0$.
Without any proofs we know if we add two functions with b not equaling 0, then we'd just end up with 2b. Which means it wouldn't hold that two functions in the space added together are also in the space.
****
(e) The set of all sequences of complex numbers with limit $0$ is a subspace of $C^{\infty}$.
Let $U = \{(x_{1}, x_{2}, \dots) : x_{i} \in{} C \text{ for } i = 1, 2, \dots \text{ and } \lim_{ i \to \infty } x_{i} = 0\}$.
(1) $0 \in{} U$ since $\lim_{ i \to \infty } 0_{i} = 0$.
(2) If $x, y \in{} U$, then $\lim_{ i \to \infty }(x_{i}+y_{i}) = \lim_{ i \to \infty }x_{i} + \lim_{ i \to \infty } y_{i} = 0 + 0 = 0$
(3) If $x \in{} U$ and $c\in{}C$, then $\lim_{ i \to \infty }(cx_{i}) = c\left(\lim_{ i \to \infty }x_{i}\right) = c0 = 0$.
Thus $U$ is a subspace of $C^{\infty}$.
****
**Problem 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 $\mathbb{R}^{(-4,4)}$.
Let $U = \{f\in{}\mathbb{R}^{(-4,4)} : f \text{ is differentiable and } f'(-1)=3f(2)\}$
1. $0 \in U$. Since $0'(-1) = 0 = 3(0(2))$.
2. Let $u, v \in{} U$. Then .
$
\begin{align}
(u+v)'(-1) & = u'(-1) + v'(-1) & \text{ Definition for Sum of Functions}\\
& = 3u(2) + 3v(2) \\
& = 3(u+v)(2) & \text{ Definition for Sum of Functions}
\end{align}
$
3. Let $c \in{} \mathbb{R}, u \in{} U$. Then .
$
\begin{align}
(cu)(-1) & = c(u(-1)) & \text{ Definition for Product with Function} \\
& = c(3u(2)) \\
& 3((cu)(2))
\end{align}
$
$\therefore$ the set of differentiable real-valued functions $f$ on the interval $(-4,4)$ such that $f'(-1) = 3f(2)$ is a subspace of $\mathbb{R}^{(-4,4)}$.
---
**Problem 4** Suppose $b \in \mathbb{R}$. Show that the set of continuous real-valued functions $f$ on the interval $[0,1]$ such that $\int_0^1 f = b$ is a subspace of $\mathbb{R}^{[0,1]}$ if and only if $b = 0$.
Let $U = \{f \in{} \mathbb{R}^{[0,1]} : f \text{ is continuous and }\int_{0}^{1} f\, = b\}$.
$\Rightarrow$ Let $U$ be a subspace of $F^{[0,1]}$ and $u \in U, c \in \mathbb{R}$. So, $\int_{0}^{1} \,cu = c\int_{0}^{1} \, u = cb$. For $cb = b$ to be true, it must be that $b = 0$.
$\Leftarrow$ Let $b=0$.
1. $0 \in U$
2. Let $u, v \in U$
$
\begin{align}
\int_{0}^{1} \, (u+v) & = \int_{0}^{1} \, u + \int_{0}^{1} \, v & \text{Definition of Addition of Integrals}\\
& =b + b \\
& = 0 + 0 \\
& = 0
\end{align}
$
3. Let $c \in \mathbb{R}, u \in U$.
$
\begin{align}
\int_{0}^{1} \, cu & = c \int_{0}^{1} \, u & \text{ Definition of Mul with Integral}\\
& =c (b) \\
& = c(0) \\
& =0
\end{align}
$
So, $U$ is a subspace of $\mathbb{R}^{[0,1]}$.
$\therefore$ The set of continuous real-valued functions $f$ on the interval $[0,1]$ such that $\int_0^1 f = b$ where $b \in \mathbb{R}$ is a subspace of $\mathbb{R}^{[0,1]}$ if and only if $b = 0$.
---
**Problem 5** Is $\mathbb{R}^2$ a subspace of the complex vector space $\mathbb{C}^2$?
No. If $c=i \in \mathbb{C}, x=(1,1)\in \mathbb{R}^2$, then $cx = (i,i) \not\in \mathbb{R}^2$
---
**Problem 6**
**(a)** Is ${(a,b,c) \in \mathbb{R}^3 : a^3 = b^3}$ a subspace of $\mathbb{R}^3$?
**(b)** Is ${(a,b,c) \in \mathbb{C}^3 : a^3 = b^3}$ a subspace of $\mathbb{C}^3$?
**Problem 6A** Is $U = \{(a,b,c) \in \mathbb{R}^3 : a^3 = b^3\}$ a subspace of $\mathbb{R}^3$?
1. $0 \in U$
2. Let $x,y \in U$. Then
$
\begin{align}
(x+y) & = (x_{1}+y_{1}, x_{2}+y_{2},x_{3}+y_{3}) \\
& = (x_{1}+y_{1}, x_{1}+y_{1},x_{3}+y_{3}) & \text{Since } x \mapsto x^{3} \text{ is injective on R }\\
& \in U
\end{align}
$
3. Let $x \in U, a\in \mathbb{R}$. Then
$
\begin{align}
ax & =(ax_{1},ax_{2},ax_{3}) \\
& =(ax_{1},ax_{1},ax_{3}) & \text{ Since } x \mapsto x^{3} \text{ is injective on R }\\
& \in U & \text{ Since } (ax_{1})^{3}=(ax_{2})^{3}
\end{align}
$
$\therefore U$ is a subspace of $\mathbb{R}^{3}$
**Problem 6B** Is $U=\{(a,b,c) \in \mathbb{C}^3 : a^3 = b^3\}$ a subspace of $\mathbb{C}^3$?
We hinged the previous proof on the fact that $f(x)=x^{3}$ is an [[Injective Function]] allowing us to conclude that if $a^3=b^3$ then $a=b$ . This is not always the case when $a,b\in \mathbb{C}$. So, to disprove, we want to find a value of $a$ and $b$ such that $a^{3}=b^{3}$ but $a\neq b$. Then when we add it with another term $u \in U$, we'll _probably_ end up with something $\not\in U$. So lets look at the [[Roots of Polynomials|roots]] of the function.
$
\begin{align}
a^{3} - b^{3} & =0 \\
(a-b)(a^2+ab+b^2) & =0 \\
\end{align}
$
For the first term, we get $a=b$, which is not we're looking for so lets rule that out. The roots of the second term, treating $b$ as the constant, are
$
\begin{align}
a & = \frac{-b\pm \sqrt{b^2-4b^2}}{2} \\
& = \frac{-b\pm \sqrt{-3b^2}}{2} \\
& = \frac{-b\pm (\sqrt{3})bi}{2}\\
& = b\frac{-1\pm (\sqrt{3})i}{2}
\end{align}
$
This is a generic formula for values of a relative to b which satisfy $a^{3}=b^{3}$. For a concrete counter example, choose $b=1$, then
$
a = \frac{-1\pm (\sqrt{3})i}{2}
$
Clearly, $a\neq b$ and $a^{3}=b^{3}$. Now, (choosing the positive radical root), let
$u = (\frac{-1 + (\sqrt{3})i}{2}, 1, 0) \in U \quad\text{ Since }u_{1}^3=u_{2}^3$
and
$
v = \left( \frac{1}{2}, -1, 0 \right)
$
Now $u+v = \left( \frac{\sqrt{3}}{2}i, 0 , 0\right)$, but $\left( \frac{\sqrt{3}}{2}i \right)^{3}=-\frac{3\sqrt{3}i}{8} \neq \left(0\right)^3$ So $(u+v)\not\in U$.
$\therefore U$ is not a subspace of $\mathbb{R}^{3}$
---
**Problem 7** Give an example of a nonempty subset $U$ of $\mathbb{R}^2$ such that $U$ is closed under addition and under taking additive inverses (meaning $-u \in U$ whenever $u \in U$), but $U$ is not a subspace of $\mathbb{R}^2$.
Since the inverse implies $0\in U$, we'd want an example where $\lambda \in R, u\in U, \text{ but }\lambda u\not\in U$. Let
$
U = \{(-1,1), (1, -1), (0,0)\}
$
Evidently each element has an additive inverse. For closure under addition we have
$(-1,1) + (1,-1) = (0,0) \in U$
$(-1,1) + (0,0) = (-1, 1)\in U$
$(1,-1) + (0,0) = (1, -1)\in U$
But, $3(-1,1) = (-3, 3) \not\in U$. So $U$ is not a subspace of $\mathbb{R}^2$.
---
**Problem 8** Give an example of a nonempty subset $U$ of $\mathbb{R}^2$ such that $U$ is closed under scalar multiplication, but $U$ is not a subspace of $\mathbb{R}^2$.
Scalar multiplication probably means we need an infinitely large set.
Let
$
U = \{(a,b) \in \mathbb{R}^2 : a=0 \text{ or } b= 0\}
$
Evidently $U$ is closed under scalar multiplication $(0, \lambda b) \text{ and } (\lambda a, 0) \in U.$ Then, $(0, 1) \in U \text{ and } (1,0) \in U$, but $(0,1) + (1,0) = (1,1) \not\in U$. So, $U$ is not a subspace.
---
**Problem 9** A function $f : \mathbb{R} \to \mathbb{R}$ is called periodic if there exists a positive number $p$ such that $f(x) = f(x + p)$ for all $x \in \mathbb{R}$. Is the set of periodic functions from $\mathbb{R}$ to $\mathbb{R}$ a subspace of $\mathbb{R}^\mathbb{R}$? Explain.
Intuition says yes because frequency waves when added together are still periodic.
Let
$
U = \{f \in \mathbb{R}^\mathbb{R} : f(x) = f(x+p) \}
$
1. $0 \in U$
2. Let $f \in U, \lambda \in \mathbb{R}$. Then $(\lambda f)(x) = \lambda(f(x)) = \lambda(f)(x+p) = (\lambda f)(x+p)$
3. Let $f, g \in U$. Then $f(x) = f(x+p) \text{ and } g(x) = g(x+p).$ $(f+g)(x) = f(x)+g(x).$
$= f(x+p) + g(x+p) = (f+g)(x+p).$
So $U$ is a subspace of $\mathbb{R}^\mathbb{R}$.
---
**Problem 10** Suppose $U_{1}$ and $U_{2}$ are subspaces of $V$. Prove that the intersection $U_{1} \cap U_{2}$ is a subspace of $V$.
Let $I = U_{1}\cap U_{2}$
1. Since $0\in U_{1}$ and $0\in U_{2}$, $0 \in I$.
2. Let $i,j \in I$. Then $i,j\in U_{1}$ and $i,j\in U_{2}$ and $U_{1}, U_{2}$ are closed under addition, so $(i+j) \in U_{1} \text{ and } (i+j) \in U_{2} \implies (i+j)\in I$.
3. Let $i\in I, \lambda \in F$. Since $i\in U_{1} \implies \lambda i \in U_{1}$ and $i\in U_{2}\implies \lambda i\in U_{2}$.
$\therefore i\in I.$
So $I$ is a subspace of $V.$
---
**Problem 11** Prove that the intersection of every collection of subspaces of $V$ is a subspace of $V$.
Let
$
I = \bigcap_{\alpha \in A} U_{\alpha} \text{ where } A \text{ is an arbitrary indexing set}$
1. Since $0 \in U_{\alpha}$ for every $\alpha \in A$, $0 \in I$.
2. Let $x, y \in I$. Then $x, y \in U_{\alpha}$ for every $\alpha \in A$. Since each $U_{\alpha}$ is a subspace, $x + y \in U_{\alpha}$ for every $\alpha \in A$. $\therefore (x+y) \in I$.
3. Let $x \in I, \lambda \in F$. Since $x \in U_{\alpha} \implies \lambda x \in U_{\alpha}$ for every $\alpha \in A$. $\therefore \lambda x \in I.$ So $I$ is a subspace of $V.$
---
**Problem 12** Prove that the union of two subspaces of $V$ is a subspace of $V$ if and only if one of the subspaces is contained in the other.
Let $U_{1},U_{2}$ be subspaces of $V$ and $U=U_{1}\cup U_{2}$
($\Rightarrow$) Assume $U$ is a subspace. Assume towards a contradiction, that $U_{1} \not\subseteq U_{2}$ and $U_{2} \not\subseteq U_{1}$. Since $U_{1} \not\subseteq U_{2}$ there exists a $u\in U_{1}$ with $u\not\in U_{2}$. And since $U_{2} \not\subseteq U_{1}$ and there exists $w\in U_{2}$ with $w\not\in U_{1}$. Since $U$ is a subspace, $(u+w) \in U$. Then either $(u+w) \in U_{1}$ or $(u+w)\in U_{2}$. Without loss of generality, $(u+w) \in U_{1}$, then $-u + (u + w) \in U_{1} \implies w\in U_{1}$. Which is a contradiction.
$\therefore U_{1}\subseteq U_{2} \text{ or } U_{2}\subseteq U_{1}$
$(\Leftarrow)$ Assume without loss of generality that $U_{1}\subseteq U_{2}$. Then $U_{1}\cup U_{2}=U_{2}$.
$\therefore U_{1}\cup U_{2}$ is a subspace.
The union of two subspaces of $V$ is a subspace of $V$ if and only if one of the subspaces is contained in the other.
>[!Example]
>Consider $U_{1}=\{(x,0) \in \mathbb{R}^2\}$ and $U_{2}=\{(0,x)\in \mathbb{R}^2\}$.
---
**Problem 13** Prove that the union of three subspaces of $V$ is a subspace of $V$ if and only if one of the subspaces contains the other two.
Let $U = U_{1}\cup U_{2}\cup U_{3}$
$(\Rightarrow)$ Assume $U$ is a subspace of $V$.
**Case A:** WLOG, assume $U_{1}\cup U_{2}$ is not a subspace, (by problem 12 we know neither is a subset of the other). Then there exists $u\in U_{1}-U_{2}$ and $w\in U_{2}-U_{1}$. Then $(u+w) \not\in U_{1}\cup U_{2}$. (Proof: Suppose $u+w\in U_{1}$, then $w=u+w+(-u)\in U_{1}$ contradicting $w\in U_{2}-U_{1}$. Suppose $u+w\in U_{2}$, then $u=u+w+(-w)\in U_{2}$ contradicting $u\in U_{1}-U_{2}$).
WLOG, Let $u'\in U_{1}$.
$\text{If }(u'\in U_{2})$ then consider $v=u'+w+u$.
- $v\not\in U_{1}$ : ATC $v\in U_{1}\implies(-u')+v=(-u')+u+w+u=w+u\in U_{1}$. But $w+u\not\in U_{1}\implies v\not\in U_{1}$
- $v\not\in U_{2}$ : ATC $v\in U_{2}\implies(-u')+v=(-u')+u+w+u=w+u\in U_{2}$. But $w+u\not\in U_{2}\implies v\not\in U_{2}$
- Since $u'+w+u$ lies in $U$, then it must be $u'+w+u \in U_{3}$ since $U$ is a subspace.
Then $u'=-(w+u) + (u'+w+u) \in U_{3}$.
$\text{If }(u'\not\in U_{2})$ then consider $v= u'+\lambda w : \lambda \in F$.
- $v\not\in U_{1}$ : ATC $v = u' + \lambda w\in U_{1}\implies-u'+u'+\lambda w\in U_{1}\implies w\in U_{1}$. But $w\not\in U_{1}\implies v\not\in U_{1}$
- $v \not\in U_{2}$ : ATC $v = u'+\lambda w\in U_{2}\implies-\lambda w+u'+\lambda w\in U_{2}\implies u'\in U_{2}$. But $u'\not\in U_{2}\implies v\not\in U_{2}$.
- Since $v=u'+\lambda w \in U \implies u'+\lambda w\in U_{3}$ since $U$ is a subspace.
_We want to isolate $u'$ so we need to manipulate multiple $\lambda