01. 벡터공간Vector Space

Solution.

(a) $M_{13} = 3$.

(b) $$M+N = \begin{pmatrix} 2+4 & 5+(-2) & 3+5 \\ 1+(-5) & 0+3 & -7+2 \end{pmatrix} = \begin{pmatrix} 6 & 3 & 8 \\ -4 & 3 & -5 \end{pmatrix}$$

(c) $$5M = \begin{pmatrix} 10 & 25 & 15 \\ 5 & 0 & -35 \end{pmatrix}$$

Solution. Adding entrywise gives the following table.

Solution. The zero vector of $M_{3\times 4}(F)$ is the $3\times 4$ matrix all of whose entries are $0$.

$$O = \begin{pmatrix} 0&0&0&0\\0&0&0&0\\0&0&0&0 \end{pmatrix}$$

Solution. $$A+B = \begin{pmatrix} 2+8 & -1+(-3) \\ 0+0 & 4+1 \end{pmatrix} = \begin{pmatrix} 10 & -4 \\ 0 & 5 \end{pmatrix}$$

For $A+B=C$ we need $\begin{pmatrix}10&-4\\0&5\end{pmatrix} = \begin{pmatrix}10&x\\0&y\end{pmatrix}$, so $x=-4$ and $y=5$.

Proof. Define addition and scalar multiplication on $V$ by $(f+g)(t)=f(t)+g(t)$ and $(cf)(t)=cf(t)$ for $f,g\in V$, $c\in\mathbb{R}$, $t\in\mathbb{R}$.

(Closure.) The sum of two differentiable functions is differentiable with $(f+g)'=f'+g'$, and a constant multiple of a differentiable function is differentiable with $(cf)'=cf'$. Hence $f+g\in V$ and $cf\in V$ whenever $f,g\in V$ and $c\in\mathbb{R}$.

The function $0$ defined by $0(t)=0$ for all $t\in\mathbb{R}$ is a constant function, hence differentiable, so $0\in V$; and for any $f\in V$, $(f+0)(t)=f(t)+0=f(t)$, so $f+0=f$. Thus $0$ is the zero vector of $V$.

For any $f\in V$, define $(-f)(t):=-f(t)$; since $-f$ is a $-1$ multiple of $f$, it is differentiable and $-f\in V$, and $(f+(-f))(t)=f(t)-f(t)=0=0(t)$, so $f+(-f)=0$.

The remaining axioms — commutativity and associativity of addition, $c(f+g)=cf+cg$, $(c+d)f=cf+df$, $c(df)=(cd)f$, and $1f=f$ — follow pointwise, for each $t\in\mathbb{R}$, from the corresponding field axioms of $\mathbb{R}$ applied to the values $f(t),g(t)$. For instance, $(f+g)(t)=f(t)+g(t)=g(t)+f(t)=(g+f)(t)$ for every $t$, so $f+g=g+f$; the other axioms follow similarly. Hence $V$ is a vector space.

Proof. Let $V=\{v_0\}$. Define addition and scalar multiplication by $v_0+v_0:=v_0$ and $cv_0:=v_0$ for $c\in F$.

Since $V$ has only the single element $v_0$, both operations always take the value $v_0$, so $V$ is closed under them, and commutativity and associativity hold trivially since every term equals $v_0$. Since $v_0+v_0=v_0$, $v_0$ is the zero vector, and since $v_0+v_0=v_0$, $v_0$ is its own additive inverse.

For scalar multiplication, $c(v_0+v_0)=cv_0=v_0$ and $cv_0+cv_0=v_0+v_0=v_0$, so $c(v_0+v_0)=cv_0+cv_0$; similarly $(c+d)v_0=v_0=cv_0+dv_0$, $c(dv_0)=v_0=(cd)v_0$, and $1v_0=v_0$ all hold. Hence $V$ is a vector space under these operations, called the zero vector space.

Solution.

(a) Not a vector space. If a zero vector $(z_1,z_2)$ existed, then $(a_1,a_2)+(z_1,z_2)=(a_1+z_1,a_2z_2)=(a_1,a_2)$ for all $(a_1,a_2)$ forces $z_1=0$, $z_2=1$, i.e., the zero vector would be $(0,1)$. But an additive inverse $(b_1,b_2)$ of $(1,0)$ would need to satisfy $(1+b_1,0\cdot b_2)=(0,1)$, and $0\cdot b_2=0\neq1$ for any $b_2$. Hence $(1,0)$ has no additive inverse, so the axiom requiring existence of additive inverses fails.

(b) Not a vector space. $(1,0)+(0,1) = (1+2\cdot0, 0+3\cdot1) = (1,3)$, while $(0,1)+(1,0) = (0+2\cdot1, 1+3\cdot0) = (2,1)$, so $(1,0)+(0,1)\neq(0,1)+(1,0)$. Hence addition is not commutative, so this is not a vector space.

(c) Not a vector space. Let $x=(0,1)$ and $c=d=1$. Then $(c+d)x = 2x = (0,1/2)$, while $cx+dx = x+x = (0,1)+(0,1) = (0,2)$, so $(c+d)x\neq cx+dx$. Hence the distributive law for scalar addition fails, so this is not a vector space.

Proof. $V=\mathbb{C}^n$ carries the usual entrywise addition and scalar multiplication over $\mathbb{C}$, which makes it a vector space over $\mathbb{C}$. Regard $\mathbb{R}$ as a subfield of $\mathbb{C}$, and define scalar multiplication of $V$ by real scalars as the restriction of the scalar multiplication over $\mathbb{C}$ to $\mathbb{R}\subset\mathbb{C}$: for $a\in\mathbb{R}\subset\mathbb{C}$ and $x\in V$, $ax$ is the same value as given by the already-defined scalar multiplication over $\mathbb{C}$.

The axioms concerning addition (commutativity, associativity, existence of the zero vector $0=(0,\dots,0)$, existence of $-x$ for each $x$) hold as they already hold for $V$ as a vector space over $\mathbb{C}$. The scalar multiplication axioms $a(x+y)=ax+ay$, $(a+b)x=ax+bx$, $a(bx)=(ab)x$, $1x=x$ already hold for all $a,b\in\mathbb{C}$, hence in particular for all $a,b\in\mathbb{R}\subset\mathbb{C}$. Therefore $V=\mathbb{C}^n$ is a vector space over $\mathbb{R}$ with this scalar multiplication.

Proof.

(a) For any scalar $a\in F$, $a0 = a(0+0) = a0+a0$, and adding the additive inverse of $a0$ to both sides gives $0=a0$. That is, $a0=0$ for every $a\in F$.

Now suppose $ax=0$. If $a=0$ there is nothing to prove, so assume $a\neq0$. Then $a$ has a multiplicative inverse $a^{-1}$ in $F$, and

$$x = 1x = (a^{-1}a)x = a^{-1}(ax) = a^{-1}0 = 0.$$

Hence if $ax=0$, then $a=0$ or $x=0$.

(b) No. If $a=0$, then $ax=0=ay$ for all $x,y\in V$, but $x$ need not equal $y$. For example, take $V=F$, $x=1$, $y=0$: then $0\cdot1=0=0\cdot0$ but $x\neq y$. So it is not true in general that $ax=ay$ implies $x=y$. (However, applying (a) to $a(x-y)=ax-ay=0$ shows that when $a\neq0$, $x=y$ does hold.)

Proof. We use the distributive laws $(a+b)v=av+bv$ and $c(x+y)=cx+cy$, together with commutativity and associativity of vector addition.

$$(a+b)(x+y) = (a+b)x+(a+b)y = (ax+bx)+(ay+by)$$

where the first equality applies $c(x+y)=cx+cy$ with $c=a+b$, and the second applies $(a+b)v=av+bv$ to $v=x$ and $v=y$. Rearranging the terms using commutativity and associativity of addition,

$$(ax+bx)+(ay+by) = ax+ay+bx+by.$$

Hence $(a+b)(x+y)=ax+ay+bx+by$.

Proof. (Existence) Since $a\neq0$, $a^{-1}$ exists in $F$. Let $y:=a^{-1}(-x)=-a^{-1}x$. Then

$$ay+x = a(-a^{-1}x)+x = -(aa^{-1})x+x = -1x+x = -x+x = 0,$$

so this $y$ satisfies $ay+x=0$.

(Uniqueness) Suppose $y_1,y_2\in V$ both satisfy $ay_1+x=0$ and $ay_2+x=0$. Then $ay_1+x=ay_2+x$, and adding $-x$ to both sides gives $ay_1=ay_2$. Multiplying both sides by $a^{-1}$,

$$y_1 = 1y_1=(a^{-1}a)y_1=a^{-1}(ay_1)=a^{-1}(ay_2)=(a^{-1}a)y_2=1y_2=y_2,$$

so $y_1=y_2$. Hence there is a unique $y\in V$ with $ay+x=0$.

Solution.

(a) $$A^t = \begin{pmatrix} -4 & 5 \\ 2 & -1 \end{pmatrix}$$

(b) $\operatorname{tr}(A) = -4+(-1) = -5$.

Proof.

(a) Let $A\in M_{m\times n}(F)$. By the definition of transpose, $((A^t)^t)_{ij}=(A^t)_{ji}=A_{ij}$ for all $i,j$. Hence $(A^t)^t=A$.

(b) Let $A\in M_{n\times n}(F)$ and $B:=A+A^t$. Then $B_{ij}=A_{ij}+(A^t)_{ij}=A_{ij}+A_{ji}$ and $B_{ji}=A_{ji}+A_{ij}$, so $B_{ij}=B_{ji}$ for all $i,j$. Hence $A+A^t$ is symmetric.

(c) Let $A,B\in M_{n\times n}(F)$ and $a,b\in F$. By the definition of trace,

$$\operatorname{tr}(aA+bB) = \sum_{i=1}^n (aA+bB)_{ii} = \sum_{i=1}^n (aA_{ii}+bB_{ii}) = a\sum_{i=1}^n A_{ii} + b\sum_{i=1}^n B_{ii} = a\operatorname{tr}(A)+b\operatorname{tr}(B).$$

(d) Let $A\in M_{n\times n}(F)$. Since $(A^t)_{ii}=A_{ii}$,

$$\operatorname{tr}(A^t) = \sum_{i=1}^n (A^t)_{ii} = \sum_{i=1}^n A_{ii} = \operatorname{tr}(A).$$

Solution.

(a) It is a subspace. $W=\{(3a_2,a_2,-a_2):a_2\in\mathbb{R}\}=\operatorname{span}\{(3,1,-1)\}$. We have $0=(0,0,0)\in W$ (take $a_2=0$), and for $u=(3s,s,-s)$, $v=(3t,t,-t)\in W$, $c\in\mathbb{R}$, $u+v=(3(s+t),s+t,-(s+t))\in W$ and $cu=(3(cs),cs,-(cs))\in W$. Hence $W$ is a subspace of $\mathbb{R}^3$.

(b) It is not a subspace. $0=(0,0,0)$ satisfies $0+2\cdot0-3\cdot0=0\neq1$, so $0\notin W$. Since a subspace must contain the zero vector, $W$ is not a subspace.

(c) It is not a subspace. Let $u=(\sqrt3,\sqrt5,0)$ and $v=(\sqrt3,-\sqrt5,0)$. Then $5(\sqrt3)^2-3(\sqrt5)^2+6\cdot0^2=15-15=0$, so $u,v\in W$. But for $u+v=(2\sqrt3,0,0)$, $5(2\sqrt3)^2-3\cdot0^2+6\cdot0^2=5\cdot12=60\neq0$, so $u+v\notin W$. Hence $W$ is not closed under addition and is not a subspace.

Solution.

(a) It is not a subspace (for $n\geq1$). Both $f(x)=x^n$ and $g(x)=-x^n+1$ have degree $n$, so $f,g\in W$. But $(f+g)(x)=1$ is neither the zero polynomial nor of degree $n$, so $f+g\notin W$. Hence $W$ is not closed under addition and is not a subspace of $P(F)$.

(b) It is a subspace. The zero matrix has all entries $0$, hence is upper triangular, so $O\in W$. If $A,B\in W$, i.e., $A_{ij}=B_{ij}=0$ for $i>j$, then $(A+B)_{ij}=A_{ij}+B_{ij}=0$ for $i>j$, so $A+B\in W$, and $(cA)_{ij}=cA_{ij}=0$ for $i>j$ and $c\in F$, so $cA\in W$. Hence $W$ is a subspace of $M_{m\times n}(F)$.

(c) It is a subspace. The zero function satisfies $0(s_0)=0$, so $0\in W$. If $f,g\in W$, then $(f+g)(s_0)=f(s_0)+g(s_0)=0+0=0$, so $f+g\in W$, and for $c\in F$, $(cf)(s_0)=cf(s_0)=c\cdot0=0$, so $cf\in W$. Hence $W$ is a subspace of $\mathcal{F}(S,F)$.

(d) It is a subspace. The zero function has all derivatives equal to $0$, hence continuous, so it belongs to $C^n(\mathbb{R})$. If $f,g\in C^n(\mathbb{R})$, then $(f+g)^{(n)}=f^{(n)}+g^{(n)}$ is a sum of continuous functions, hence continuous, and for $c\in\mathbb{R}$, $(cf)^{(n)}=cf^{(n)}$ is a constant multiple of a continuous function, hence continuous. Thus $f+g,cf\in C^n(\mathbb{R})$, and $C^n(\mathbb{R})$ is a subspace of $\mathcal{F}(\mathbb{R},\mathbb{R})$.

Proof. Since $F_1$ is a field, $-x\in F_1$ for every $x\in F_1$, and an even (resp. odd) function is a function $f\in\mathcal{F}(F_1,F_2)$ satisfying $f(-x)=f(x)$ (resp. $f(-x)=-f(x)$) for all $x\in F_1$. Let $W_e$ denote the set of even functions and $W_o$ the set of odd functions.

The zero function $0$ satisfies $0(-x)=0=0(x)$ and $0(-x)=0=-0(x)$ for all $x$, so $0\in W_e$ and $0\in W_o$.

Let $f,g\in W_e$ and $c\in F_2$. For all $x\in F_1$,

$$(f+g)(-x) = f(-x)+g(-x) = f(x)+g(x) = (f+g)(x), \quad (cf)(-x)=cf(-x)=cf(x)=(cf)(x),$$

so $f+g\in W_e$ and $cf\in W_e$. Hence $W_e$ is a subspace of $\mathcal{F}(F_1,F_2)$.

Let $f,g\in W_o$ and $c\in F_2$. For all $x\in F_1$,

$$(f+g)(-x) = f(-x)+g(-x) = -f(x)-g(x) = -(f+g)(x), \quad (cf)(-x)=cf(-x)=-cf(x)=-(cf)(x),$$

so $f+g\in W_o$ and $cf\in W_o$. Hence $W_o$ is also a subspace of $\mathcal{F}(F_1,F_2)$.