02. 생성과 일차독립Span and Linear Independence

Solution. (a) We seek scalars $a, b$ with $(-2,0,3) = a(1,3,0) + b(2,4,-1)$, equivalently

$$a + 2b = -2, \quad 3a + 4b = 0, \quad -b = 3.$$

The third equation gives $b=-3$; substituting into the first gives $a = -2-2(-3) = 4$. Checking the second equation, $3(4)+4(-3) = 12-12=0$, so it is satisfied. Hence $a=4, b=-3$ solve the system, and

$$(-2,0,3) = 4(1,3,0) - 3(2,4,-1),$$

so $(-2,0,3)$ can be written as a linear combination of the other two vectors.

(b) We seek scalars $a, b$ with $x^3-3x+5 = a(x^3+2x^2-x+1) + b(x^3+3x^2-1)$. Comparing coefficients of each power gives

$$a+b=1, \quad 2a+3b=0, \quad -a=-3, \quad a-b=5.$$

From the third equation $a=3$, and the first gives $b=1-3=-2$. Substituting into the second and fourth equations, $2(3)+3(-2)=0$ and $3-(-2)=5$, both hold. Hence

$$x^3-3x+5 = 3(x^3+2x^2-x+1) - 2(x^3+3x^2-1),$$

so $x^3-3x+5$ is also a linear combination of the other two polynomials.

Solution. (a) We seek scalars $a,b,c$ with $\begin{pmatrix}1&2\\-3&4\end{pmatrix} = a\begin{pmatrix}1&0\\-1&0\end{pmatrix} + b\begin{pmatrix}0&1\\0&1\end{pmatrix} + c\begin{pmatrix}1&1\\0&0\end{pmatrix}$. Comparing entries gives

$$a+c=1,\quad b+c=2,\quad -a=-3,\quad b=4.$$

The third equation gives $a=3$, the fourth gives $b=4$, and the first gives $c=1-3=-2$. Substituting into the second equation, $b+c=4-2=2$, which holds. Hence

$$\begin{pmatrix}1&2\\-3&4\end{pmatrix} = 3\begin{pmatrix}1&0\\-1&0\end{pmatrix} + 4\begin{pmatrix}0&1\\0&1\end{pmatrix} - 2\begin{pmatrix}1&1\\0&0\end{pmatrix},$$

so the first matrix is a linear combination of the other three.

(b) Since $S=\{0,1\}$, every function in $\mathcal{F}(S,\mathbb{R})$ is completely determined by the pair $(f(0), f(1)) \in \mathbb{R}^2$. Computing,

$$f = (f(0), f(1)) = (0, 2), \quad g = (g(0), g(1)) = (1, 2), \quad h = (h(0), h(1)) = (-1, 0).$$

Seeking $a, b$ with $f = ag + bh$ gives

$$a - b = 0, \quad 2a = 2,$$

so $a=1$ and $b=1$. Indeed $g+h = (1,2)+(-1,0) = (0,2) = f$, so

$$f = g + h,$$

and $f$ is a linear combination of $g$ and $h$.

Proof. (a) For any $(a_1, \cdots, a_n) \in F^n$, the definition of $e_i$ gives

$$a_1 e_1 + \cdots + a_n e_n = (a_1, \cdots, a_n).$$

Thus every vector in $F^n$ is a linear combination of $e_1, \cdots, e_n$, so $F^n \subset \operatorname{span}(\{e_1, \cdots, e_n\})$. The reverse inclusion $\operatorname{span}(\{e_1, \cdots, e_n\}) \subset F^n$ is immediate, so $\operatorname{span}(\{e_1, \cdots, e_n\}) = F^n$; that is, $\{e_1, \cdots, e_n\}$ generates $F^n$.

(b) Any $p(x) \in P_n(F)$ has $\deg p(x) \leq n$, so it can be written uniquely as

$$p(x) = a_0 + a_1 x + \cdots + a_n x^n \quad (a_0, \cdots, a_n \in F).$$

This shows $p(x)$ is a linear combination of $\{1, x, \cdots, x^n\}$, so $P_n(F) \subset \operatorname{span}(\{1, x, \cdots, x^n\})$. The reverse inclusion is immediate, so $P_n(F) = \operatorname{span}(\{1, x, \cdots, x^n\})$; that is, $P_n(F)$ is generated by $\{1, x, \cdots, x^n\}$.

Proof. Let $W$ denote the set of all symmetric $2\times 2$ matrices. Given any $\begin{pmatrix} a & b \\ b & d \end{pmatrix} \in W$,

$$a M_1 + d M_2 + b M_3 = a\begin{pmatrix}1&0\\0&0\end{pmatrix} + d\begin{pmatrix}0&0\\0&1\end{pmatrix} + b\begin{pmatrix}0&1\\1&0\end{pmatrix} = \begin{pmatrix} a & b \\ b & d \end{pmatrix},$$

so $W \subset \operatorname{span}(\{M_1, M_2, M_3\})$. Conversely, any element of $\operatorname{span}(\{M_1,M_2,M_3\})$ has the form $xM_1+yM_2+zM_3 = \begin{pmatrix} x & z \\ z & y \end{pmatrix}$ for some $x,y,z \in \mathbb{R}$, which is always symmetric, so $\operatorname{span}(\{M_1,M_2,M_3\}) \subset W$. Combining both inclusions, $\operatorname{span}(\{M_1,M_2,M_3\}) = W$.

Solution. (a) $\operatorname{span}(\{x\}) = Fx = \{ax : a \in F\}$. Indeed $Fx$ contains $0 = 0x$, is closed under addition ($ax+bx=(a+b)x$) and scalar multiplication ($c(ax)=(ca)x$), so it is a subspace of $V$, and it contains $x = 1x$. Moreover, any subspace $W$ containing $x$ is closed under scalar multiplication, so $ax \in W$ for every $a \in F$, i.e., $Fx \subset W$. Hence $Fx$ is the smallest subspace containing $x$, so $\operatorname{span}(\{x\}) = Fx$.

(b) $\operatorname{span}(\{x_1, \cdots, x_n\}) = \{a_1x_1 + \cdots + a_nx_n : a_1, \cdots, a_n \in F\}$. Denote this set by $U$. Then $U$ contains $0 = 0x_1+\cdots+0x_n$, is closed under addition and scalar multiplication (sums and scalar multiples of elements of $U$ again have this form), so $U$ is a subspace of $V$, and each $x_i = 0x_1+\cdots+1x_i+\cdots+0x_n \in U$. On the other hand, any subspace containing $x_1, \cdots, x_n$ is closed under addition and scalar multiplication, so it must contain every linear combination of them, i.e., it contains $U$. Hence $U$ is the smallest subspace containing $x_1, \cdots, x_n$, so $\operatorname{span}(\{x_1, \cdots, x_n\}) = U$.

Proof. Let $W = \bigcap \{U : U \text{ is a subspace of } V \text{ with } S \subset U\}$, the intersection of all subspaces of $V$ containing $S$. We show $\operatorname{span}(S) = W$ by proving both inclusions.

($\operatorname{span}(S) \subset W$) Let $U$ be any subspace of $V$ containing $S$. Since $U$ is closed under addition and scalar multiplication, every linear combination of elements of $S$ also lies in $U$, so $\operatorname{span}(S) \subset U$. As this holds for every subspace $U$ in the intersection, $\operatorname{span}(S) \subset W$.

($W \subset \operatorname{span}(S)$) The set $\operatorname{span}(S)$ is itself a subspace of $V$ containing $S$ (it is closed under addition and scalar multiplication, contains $0 \in \operatorname{span}(S)$, and $S \subset \operatorname{span}(S)$), so it is one of the subspaces $U$ over which the intersection is taken. Hence $W$, being the intersection of $\operatorname{span}(S)$ together with the other such subspaces, satisfies $W \subset \operatorname{span}(S)$.

Combining both inclusions, $\operatorname{span}(S) = W$.

Proof. ($\Rightarrow$) Suppose $W$ is a subspace of $V$. Since each $w \in W$ is the linear combination $w = 1w$ of itself, $W \subset \operatorname{span}(W)$. Conversely, since $W$ is closed under addition and scalar multiplication, every linear combination of elements of $W$ again lies in $W$, so $\operatorname{span}(W) \subset W$. Combining both inclusions, $\operatorname{span}(W) = W$.

($\Leftarrow$) Suppose $\operatorname{span}(W) = W$. Since $\operatorname{span}(W)$ is always a subspace of $V$ (as shown in EX 2.2 and EX 3), $W = \operatorname{span}(W)$ is also a subspace of $V$.

Proof. (a) Suppose $S_1 \subset S_2$. Any linear combination of elements of $S_1$ is also a linear combination of elements of $S_2$, since $S_1 \subset S_2$. Hence $\operatorname{span}(S_1) \subset \operatorname{span}(S_2)$.

(b) Suppose $S_1 \subset S_2$ and $\operatorname{span}(S_1) = V$. By part (a), $\operatorname{span}(S_1) \subset \operatorname{span}(S_2)$, so $V = \operatorname{span}(S_1) \subset \operatorname{span}(S_2)$. Also $\operatorname{span}(S_2)$ is a subspace of $V$, so $\operatorname{span}(S_2) \subset V$. Combining both inclusions, $\operatorname{span}(S_2) = V$.

Solution. (a) We seek scalars $c_1,c_2,c_3$ with $c_1(1,-1,2) + c_2(1,-2,1) + c_3(1,1,4) = 0$. Comparing components gives

$$c_1+c_2+c_3=0, \quad -c_1-2c_2+c_3=0, \quad 2c_1+c_2+4c_3=0.$$

The first equation gives $c_1 = -c_2-c_3$. Substituting into the second, $(c_2+c_3) - 2c_2 + c_3 = -c_2+2c_3=0$, so $c_2=2c_3$, and hence $c_1 = -2c_3-c_3=-3c_3$. Substituting into the third, $2(-3c_3)+2c_3+4c_3 = 0$ holds for every value of $c_3$. Taking $c_3=1$ gives the nontrivial solution $c_1=-3, c_2=2, c_3=1$, and

$$-3(1,-1,2) + 2(1,-2,1) + (1,1,4) = (0,0,0).$$

Hence this set is linearly dependent.

(b) We seek scalars with $c_1(x^3+2x^2) + c_2(-x^2+3x+1) + c_3(x^3-x^2+2x-1) = 0$. Comparing coefficients of each power gives

$$c_1+c_3=0, \quad 2c_1-c_2-c_3=0, \quad 3c_2+2c_3=0, \quad c_2-c_3=0.$$

The fourth equation gives $c_2=c_3$; substituting into the third, $3c_3+2c_3=5c_3=0$, so $c_3=0$, hence $c_2=0$. The first equation gives $c_1=-c_3=0$, and the second, $2(0)-0-0=0$, also holds. Thus the only solution is $c_1=c_2=c_3=0$, so this set is linearly independent.

Solution. (a) Since $\begin{pmatrix}-2&6\\4&-8\end{pmatrix} = -2\begin{pmatrix}1&-3\\-2&4\end{pmatrix}$, the two matrices are scalar multiples of each other. Indeed

$$2\begin{pmatrix}1&-3\\-2&4\end{pmatrix} + \begin{pmatrix}-2&6\\4&-8\end{pmatrix} = O$$

is a nontrivial linear combination (the coefficients $(2,1)$ are not both zero), so this set is linearly dependent.

(b) Let $A_1 = \begin{pmatrix}1&1\\0&0\\0&0\end{pmatrix}, A_2=\begin{pmatrix}0&0\\1&1\\0&0\end{pmatrix}, A_3=\begin{pmatrix}0&0\\0&0\\1&1\end{pmatrix}, A_4=\begin{pmatrix}1&0\\1&0\\1&0\end{pmatrix}, A_5=\begin{pmatrix}0&1\\0&1\\0&1\end{pmatrix}$. Setting $c_1A_1+\cdots+c_5A_5=O$ and comparing the two columns of each row gives

$$c_1+c_4=0,\ \ c_1+c_5=0,\ \ c_2+c_4=0,\ \ c_2+c_5=0,\ \ c_3+c_4=0,\ \ c_3+c_5=0.$$

The first three equations (column 1) give $c_1=c_2=c_3=-c_4$, and the last three (column 2) give $c_1=c_2=c_3=-c_5$, so $c_4=c_5$. Writing $c_4=c_5=t$, the values $c_1=c_2=c_3=-t$ solve the system for every $t$. Taking $t=1$,

$$-A_1-A_2-A_3+A_4+A_5 = O$$

is a nontrivial linear combination (the coefficients $(-1,-1,-1,1,1)$ are not all zero), so this set is linearly dependent.

Proof. ($\Rightarrow$, contrapositive) If $r = s$, then $f = g$, so

$$1\cdot f + (-1)\cdot g = f - g = 0$$

is a nontrivial linear combination (the coefficients $(1,-1)$ are not both zero). Hence $\{f,g\}$ is linearly dependent.

($\Leftarrow$) Suppose $r \neq s$, and suppose $af + bg = 0$, i.e., $ae^{rt} + be^{st} = 0$ for all $t \in \mathbb{R}$. Setting $t=0$ gives

$$a+b=0 \quad \Rightarrow \quad b=-a.$$

Differentiating both sides with respect to $t$ gives $ar e^{rt} + bs e^{st} = 0$ for all $t$; setting $t=0$ gives

$$ar+bs=0.$$

Substituting $b=-a$, we get $ar - as = a(r-s) = 0$, and since $r \neq s$, $a=0$. Hence $b=-a=0$ as well, so only the trivial solution exists, and $\{f,g\}$ is linearly independent.

Combining both directions, $\{f,g\}$ is linearly independent if and only if $r \neq s$.

Proof. (a) Suppose $c_1e_1+\cdots+c_ne_n=0$. By the definition of $e_i$, the left side is the vector $(c_1, \cdots, c_n)$ whose $i$th coordinate is $c_i$, so this equation means $(c_1,\cdots,c_n)=(0,\cdots,0)$. Hence $c_i=0$ for every $i$, so $\{e_1,\cdots,e_n\}$ is linearly independent.

(b) Suppose $c_0\cdot 1+c_1x+\cdots+c_nx^n = 0$ (the zero polynomial). Equality of polynomials means the coefficients of every power agree, so $c_0=c_1=\cdots=c_n=0$. Hence $\{1,x,\cdots,x^n\}$ is linearly independent in $P_n(F)$.

(c) Suppose $\sum_{i,j} c_{ij}E^{ij} = O$. By the definition of $E^{ij}$, the $(i,j)$ entry of the left-hand matrix is exactly $c_{ij}$ (every other $E^{i'j'}$ contributes $0$ to the $(i,j)$ entry), so this means $c_{ij}=0$ for all $i,j$. Hence $\{E^{ij} : 1\leq i \leq m, 1\leq j \leq n\}$ is linearly independent.

Moreover, for any $A=(a_{ij}) \in M_{m\times n}(F)$, the $(i,j)$ entry of $\sum_{i,j}a_{ij}E^{ij}$ is likewise $a_{ij}$, so

$$A = \sum_{i,j} a_{ij}E^{ij}.$$

Hence $A$ is a linear combination of $\{E^{ij}\}$, so $\{E^{ij} : 1\leq i \leq m, 1\leq j \leq n\}$ spans $M_{m\times n}(F)$.

Proof. (a) ($\Rightarrow$) Suppose $\{u,v\}$ is linearly dependent, so there exist scalars $a,b$, not both zero, with $au+bv=0$. If $a\neq 0$, then $u = -(b/a)v$, so $u$ is a multiple of $v$. If $a=0$, then $b\neq 0$ and $bv=0$ forces $v=0$, which is $v=0\cdot u$, so $v$ is a multiple of $u$. In either case, $u$ or $v$ is a multiple of the other.

($\Leftarrow$) Suppose $u$ is a multiple of $v$, say $u=cv$ (the other case is symmetric). Then $1\cdot u + (-c)\cdot v = 0$, and since the coefficient of $u$ is $1\neq 0$, this is a nontrivial linear combination. Hence $\{u,v\}$ is linearly dependent.

Combining both directions, $\{u,v\}$ is linearly dependent if and only if $u$ or $v$ is a multiple of the other.

(b) In $\mathbb{R}^3$, consider $u=(1,0,0)$, $v=(0,1,0)$, $w=(1,1,0)$. Since $w=u+v$,

$$u+v-w=0$$

is a nontrivial linear combination, so $\{u,v,w\}$ is linearly dependent. If $v=cu$, the first coordinate gives $0=c$ and the second gives $1=c\cdot 0=0$, a contradiction, so $v$ is not a multiple of $u$; by the same argument $u$ is not a multiple of $v$. If $w=cu$, the first coordinate gives $c=1$ but the second gives $1=c\cdot 0=0$, a contradiction, and if $w=cv$, the first coordinate gives $1=c\cdot 0=0$, also a contradiction, so $w$ is a multiple of neither $u$ nor $v$. Hence $\{u,v,w\}$ is an example of three linearly dependent vectors, none of which is a multiple of another.

Proof. ($\Leftarrow$) If $u_1=0$, then $1\cdot u_1+0\cdot u_2+\cdots+0\cdot u_n=0$ is a nontrivial linear combination (the coefficient $1\neq 0$), so $S$ is linearly dependent. If $u_{k+1}\in\operatorname{span}(\{u_1,\cdots,u_k\})$ for some $k$ $(1\leq k \lt n)$, then $u_{k+1}=a_1u_1+\cdots+a_ku_k$ for some scalars, and

$$a_1u_1+\cdots+a_ku_k+(-1)u_{k+1}+0\cdot u_{k+2}+\cdots+0\cdot u_n=0$$

is a nontrivial linear combination (the coefficient of $u_{k+1}$ is $-1\neq 0$), so $S$ is again linearly dependent.

($\Rightarrow$) Suppose $S$ is linearly dependent, so there exist scalars $c_1,\cdots,c_n$, not all zero, with $c_1u_1+\cdots+c_nu_n=0$. Let $k$ be the largest index with $c_k\neq 0$ (such $k$ exists since not all coefficients vanish). By maximality of $k$, $c_{k+1}=\cdots=c_n=0$, so

$$c_1u_1+\cdots+c_ku_k=0.$$

If $k=1$, then $c_1u_1=0$ with $c_1\neq 0$, so multiplying by $c_1^{-1}$ gives $u_1=0$, satisfying the first alternative.

If $k>1$, then $c_k\neq 0$, so multiplying by $c_k^{-1}$ and rearranging gives

$$u_k = -\frac{c_1}{c_k}u_1-\cdots-\frac{c_{k-1}}{c_k}u_{k-1} \in \operatorname{span}(\{u_1,\cdots,u_{k-1}\}).$$

Setting $m=k-1$, we have $1\leq m \lt n$ and $u_{m+1}=u_k\in\operatorname{span}(\{u_1,\cdots,u_m\})$, satisfying the second alternative.

In either case, whenever $S$ is linearly dependent, either $u_1=0$ or $u_{k+1}\in\operatorname{span}(\{u_1,\cdots,u_k\})$ for some $k$ $(1\leq k \lt n)$. This proves the equivalence.

Proof. Since linear independence is a condition about finite linear combinations, any linear relation among elements of a finite subset of $S$ is precisely a finite linear relation among elements of $S$ itself. Keeping this in mind, we prove both directions.

($\Rightarrow$) Suppose $S$ is linearly independent, and let $T=\{u_1,\cdots,u_k\}$ be any finite subset of $S$. If $c_1u_1+\cdots+c_ku_k=0$, then since $u_1,\cdots,u_k\in S$, this is a linear relation among elements of $S$, and since $S$ is linearly independent, $c_1=\cdots=c_k=0$. Hence $T$ is linearly independent, and since $T$ was arbitrary, every finite subset of $S$ is linearly independent.

($\Leftarrow$) Suppose every finite subset of $S$ is linearly independent. To show $S$ is linearly independent, we must show that for distinct elements $u_1,\cdots,u_n$ of $S$ and scalars $c_1,\cdots,c_n$, $c_1u_1+\cdots+c_nu_n=0$ implies $c_1=\cdots=c_n=0$. Since $u_1,\cdots,u_n$ are distinct, $T=\{u_1,\cdots,u_n\}$ is a finite subset of $S$ with $n$ elements, and by hypothesis $T$ is linearly independent. Hence $c_1u_1+\cdots+c_nu_n=0$ forces $c_1=\cdots=c_n=0$. Since this holds for every finite linear relation among elements of $S$, $S$ is linearly independent.

(For $S=\emptyset$: $S$ is vacuously linearly independent, and its only finite subset $\emptyset$ is likewise vacuously linearly independent, so this case is consistent with the argument above.)

Combining both directions, $S$ is linearly independent if and only if every finite subset of $S$ is linearly independent.