03. 기저와 차원Bases and Dimension

Solution. In each case, compare the number of given vectors with $\dim V$, and check linear independence when needed.

(a) Since $\dim \mathbb{R}^3 = 3$ and there are $3$ given vectors, being a basis is equivalent to being linearly independent. The determinant of $$\begin{pmatrix} 1 & 0 & -1 \\ 2 & 5 & 1 \\ 0 & -4 & 3 \end{pmatrix}$$ is $1(15+4) - 0 + (-1)(-8-0) = 19+8 = 27 \neq 0$, so the three vectors are linearly independent. Hence this set is a basis of $\mathbb{R}^3$.

(b) There are $4$ vectors while $\dim \mathbb{R}^3 = 3 \lt 4$, so the set must be linearly dependent (any set of more vectors than the dimension is automatically dependent). Hence it cannot be a basis.

(c) $\dim P_2(\mathbb{R}) = 3$ and there are $3$ vectors, so it suffices to check linear independence. Using coordinate vectors $(1,2,1), (3,0,1), (0,1,1)$, the determinant is $1(0-1) - 2(3-0) + 1(3-0) = -1-6+3 = -4 \neq 0$, so the set is linearly independent. Hence it is a basis of $P_2(\mathbb{R})$.

(d) $\dim P_3(\mathbb{R}) = 4$ but there are only $3$ vectors, so this set cannot generate $P_3(\mathbb{R})$ (a generating set must have at least as many vectors as the dimension). Hence it is not a basis.

Solution. Since $\dim M_{2\times 2}(F) = 4$, any linearly independent set of $4$ matrices is a basis.

1. The standard basis $\beta_1 = \{E^{11}, E^{12}, E^{21}, E^{22}\}$.

2. $\beta_2 = \{E^{11},\ E^{11}+E^{12},\ E^{11}+E^{12}+E^{21},\ E^{11}+E^{12}+E^{21}+E^{22}\}$. If $c_1E^{11}+c_2(E^{11}+E^{12})+c_3(E^{11}+E^{12}+E^{21})+c_4(E^{11}+E^{12}+E^{21}+E^{22}) = O$, comparing entries gives $c_1+c_2+c_3+c_4=0$ (entry $(1,1)$), $c_2+c_3+c_4=0$ (entry $(1,2)$), $c_3+c_4=0$ (entry $(2,1)$), and $c_4=0$ (entry $(2,2)$). Substituting from the bottom up yields $c_4=c_3=c_2=c_1=0$, so $\beta_2$ is linearly independent.

3. $\beta_3 = \{I,\ E^{12},\ E^{21},\ E^{11}-E^{22}\}$. If $aI+bE^{12}+cE^{21}+d(E^{11}-E^{22}) = \begin{pmatrix} a+d & b \\ c & a-d \end{pmatrix} = O$, then $b=c=0$ and $a+d=0=a-d$, so $a=d=0$; thus $\beta_3$ is linearly independent.

Each set consists of $4$ linearly independent matrices, so all three are bases of $M_{2\times 2}(F)$.

Solution. First compute the relations among the five vectors.

$u_3 = -4(2,-3,1) = -4u_1$, so $u_3$ is dependent on $u_1$.

Solving $au_1+bu_2=u_4$ (i.e. $2a+b=1$, $-3a+4b=37$, $a-2b=-17$) gives $a=-3$, $b=7$, which satisfies all three equations, so $u_4 = -3u_1+7u_2$.

On the other hand, $au_1+bu_2=u_5$ has no solution (the first two equations give $a=-7/11$ while the first and third give $a=2/5$, a contradiction). Hence $u_5 \notin \operatorname{span}(\{u_1,u_2\})$.

Since $u_1$ and $u_2$ are not scalar multiples of each other, they are linearly independent, and since $u_5 \notin \operatorname{span}(\{u_1,u_2\})$, the set $\{u_1,u_2,u_5\}$ is linearly independent. Indeed, $$\det\begin{pmatrix} 2 & -3 & 1 \\ 1 & 4 & -2 \\ -3 & -5 & 8\end{pmatrix} = 2(32-10)+3(8-6)+1(-5+12) = 44+6+7=57 \neq 0,$$ confirming independence. Since $\dim \mathbb{R}^3 = 3$, three linearly independent vectors form a basis of $\mathbb{R}^3$.

Hence $\{u_1, u_2, u_5\}$ is the desired basis ($u_3 = -4u_1$ and $u_4=-3u_1+7u_2$ are linear combinations of the other vectors, so they are excluded).

Proof.

(a) Since $\{u,v\}$ is a basis of $V$, $\dim V = 2$ and $\{u,v\}$ is linearly independent.

Suppose $c_1(u+v)+c_2(au)=0$. This gives $(c_1+c_2a)u+c_1v=0$, so by the linear independence of $\{u,v\}$, $c_1=0$ and $c_1+c_2a=0$. Since $c_1=0$, $c_2a=0$, and since $a\neq0$, $c_2=0$. Hence $\{u+v,au\}$ is linearly independent, and since $\dim V=2$, a linearly independent set of $2$ vectors is a basis of $V$.

Next, suppose $c_1(au)+c_2(bv)=0$. This gives $(c_1a)u+(c_2b)v=0$, so $c_1a=0$ and $c_2b=0$. Since $a,b\neq0$, $c_1=c_2=0$. Hence $\{au,bv\}$ is also linearly independent, and by the same reasoning is a basis of $V$.

(b) Since $\{u,v,w\}$ is a basis of $V$, $\dim V=3$ and $\{u,v,w\}$ is linearly independent. Suppose $c_1(u+v+w)+c_2(v+w)+c_3w=0$. Then $$c_1u+(c_1+c_2)v+(c_1+c_2+c_3)w=0,$$ so $c_1=0$, $c_1+c_2=0$, and $c_1+c_2+c_3=0$. Substituting successively gives $c_1=c_2=c_3=0$, so $\{u+v+w,v+w,w\}$ is linearly independent. Since $\dim V=3$, this set is a basis of $V$.

Solution.

(1) $V$ is the solution set of two homogeneous linear equations. The zero vector $(0,0,0)$ satisfies both equations, so $(0,0,0)\in V$. Let $x=(x_1,x_2,x_3), y=(y_1,y_2,y_3)\in V$ and $c\in\mathbb{R}$. Since the left-hand side of each equation is linear in the coordinates, the coordinates of $x+y$ and $cx$ also satisfy both equations. For instance, $$(x_1+y_1)-2(x_2+y_2)+(x_3+y_3) = (x_1-2x_2+x_3)+(y_1-2y_2+y_3)=0+0=0,$$ and similarly for the other equation and for $cx$. Hence $V$ is a subspace of $\mathbb{R}^3$.

(2)-(3) Solving the system, the first equation gives $x_1=2x_2-x_3$; substituting into the second equation, $$2(2x_2-x_3)-3x_2+x_3 = x_2-x_3 = 0,$$ so $x_2=x_3$ and $x_1=2x_3-x_3=x_3$. That is, $x_1=x_2=x_3=t$ for some $t\in\mathbb{R}$, so $$V=\{t(1,1,1):t\in\mathbb{R}\}=\operatorname{span}(\{(1,1,1)\}).$$ Since $(1,1,1)\neq0$, $\{(1,1,1)\}$ is linearly independent and generates $V$, so it is a basis of $V$. Hence $\dim V=1$.

Solution. Let $E^{ij}$ denote the $n\times n$ matrix whose $(i,j)$-entry is $1$ and all other entries are $0$. Then $\{E^{ij} : 1\le i,j\le n\}$ is the standard basis of $M_{n\times n}(F)$, and $\dim M_{n\times n}(F)=n^2$.

(a) Let $W$ be the set of matrices with trace zero. We claim $$\beta=\{E^{ij}: i\neq j\}\cup\{E^{ii}-E^{nn} : 1\le i \le n-1\}$$ is a basis of $W$. For $A=(a_{ij})\in W$, since $a_{nn}=-\sum_{i=1}^{n-1}a_{ii}$, $$A=\sum_{i\neq j}a_{ij}E^{ij}+\sum_{i=1}^{n-1}a_{ii}(E^{ii}-E^{nn}),$$ so $\beta$ generates $W$. If $\sum_{i\neq j}c_{ij}E^{ij}+\sum_{i=1}^{n-1}d_i(E^{ii}-E^{nn})=O$, comparing the $(i,j)$-entry for $i\neq j$ gives $c_{ij}=0$, and comparing the $(i,i)$-entry for $1\le i\le n-1$ gives $d_i=0$; hence $\beta$ is linearly independent. Since $|\beta| = (n^2-n)+(n-1)=n^2-1$, we get $\dim W = n^2-1$.

(b) The set $\{E^{ij} : 1\le i\le j\le n\}$ is a subset of the standard basis, hence linearly independent, and it generates the space of upper triangular matrices since any upper triangular $A=(a_{ij})$ equals $\sum_{i\le j}a_{ij}E^{ij}$. The number of pairs $(i,j)$ with $1\le i\le j\le n$ is $n+(n-1)+\cdots+1=n(n+1)/2$, so the dimension is $n(n+1)/2$.

(c) Assume $\operatorname{char} F\neq 2$ (otherwise the diagonal entries need not vanish). A skew-symmetric matrix $A=(a_{ij})$ satisfies $a_{ii}=0$ (from $a_{ii}=-a_{ii}$) and $a_{ji}=-a_{ij}$ for $i \lt j$, so $A=\sum_{i \lt j}a_{ij}(E^{ij}-E^{ji})$; hence $\{E^{ij}-E^{ji} : 1\le i \lt j\le n\}$ generates the space of skew-symmetric matrices. If $\sum_{i \lt j}c_{ij}(E^{ij}-E^{ji})=O$, the $(i,j)$-entry for each pair $i \lt j$ gives $c_{ij}=0$, since distinct pairs contribute to disjoint entries; hence the set is linearly independent. Its size is $\binom{n}{2}=n(n-1)/2$, so the dimension is $n(n-1)/2$.

Solution.

(a) We show $W$ is a subspace of $V$. The zero sequence has (vacuously) only finitely many nonzero terms, so it lies in $W$. Let $x=(x_n), y=(y_n)\in W$ and $c\in F$. Let $S_x, S_y$ be the (finite) sets of indices where $x,y$ are nonzero, respectively. The set of indices where $x+y$ is nonzero is contained in $S_x\cup S_y$, which is finite, and the set of indices where $cx$ is nonzero is contained in $S_x$, also finite. Hence $x+y, cx\in W$, so $W$ is a subspace of $V$.

(b) For each $i=1,2,3,\ldots$, let $e_i$ be the sequence whose $i$-th term is $1$ and all other terms are $0$. We show $\beta=\{e_1,e_2,e_3,\ldots\}$ is a basis of $W$. Any $w=(w_1,w_2,\ldots)\in W$ has only finitely many nonzero terms $w_{i_1},\ldots,w_{i_m}$, so $w=w_{i_1}e_{i_1}+\cdots+w_{i_m}e_{i_m}$ is a (finite) linear combination of elements of $\beta$; hence $\beta$ generates $W$. Also, for any finite subset $\{e_{i_1},\ldots,e_{i_m}\}\subseteq\beta$, if $c_1e_{i_1}+\cdots+c_me_{i_m}=0$, comparing the $i_k$-th term gives $c_k=0$ for each $k$, so $\beta$ is linearly independent. Hence $\beta$ is a basis of $W$. (Since linear independence and generation are always tested against finite subsets by definition, the fact that $\beta$ is infinite causes no difficulty.)

(c) $\beta$ is infinite, and $W$ cannot be generated by finitely many vectors: the span of any finite subset $\{e_{i_1},\ldots,e_{i_m}\}\subset\beta$ consists of sequences that vanish outside the indices $i_1,\ldots,i_m$, so it cannot produce an element of $W$ that is nonzero at some index outside $\{i_1,\ldots,i_m\}$ (e.g., $e_j$ for $j\notin\{i_1,\ldots,i_m\}$). Hence $W$ is infinite-dimensional, i.e. $\dim W=\infty$.

Solution. First we show $\gamma=\{1,(x-a),(x-a)^2,\ldots,(x-a)^n\}$ is a basis of $P_n(\mathbb{R})$. It has $n+1$ elements and $\dim P_n(\mathbb{R})=n+1$, so it suffices to show linear independence. Suppose $\sum_{k=0}^n c_k(x-a)^k=0$ as a polynomial. The only term of degree $n$ comes from $k=n$, forcing $c_n=0$; proceeding inductively by decreasing degree gives $c_{n-1}=\cdots=c_0=0$. Hence $\gamma$ is linearly independent and is a basis of $P_n(\mathbb{R})$.

Let $W=\{f\in P_n(\mathbb{R}):f(a)=0\}$. Writing any $f\in P_n(\mathbb{R})$ uniquely as $f=\sum_{k=0}^n c_k(x-a)^k$ with respect to $\gamma$, every term with $k\ge1$ vanishes at $x=a$, so $f(a)=c_0$. Hence $$f\in W \iff c_0=0 \iff f\in\operatorname{span}(\{(x-a),(x-a)^2,\ldots,(x-a)^n\}).$$ So $\beta=\{(x-a),(x-a)^2,\ldots,(x-a)^n\}$ generates $W$. Since $\beta\subset\gamma$ and $\gamma$ is linearly independent, $\beta$ is also linearly independent. Hence $\beta$ is a basis of $W$ consisting of $n$ elements, so $\dim W = n$.

Proof. The necessary and sufficient condition is $W_2\subseteq W_1$ (assuming $\dim W_2 \lt \infty$).

($\Leftarrow$) If $W_2\subseteq W_1$, then $W_1\cap W_2=W_2$, so $\dim(W_1\cap W_2)=\dim W_2$ holds trivially.

($\Rightarrow$) Suppose $\dim(W_1\cap W_2)=\dim W_2$. Since $W_1\cap W_2$ is a subspace of $W_2$, $W_2$ is finite-dimensional, and the two have equal dimension, the theorem that a subspace of a finite-dimensional space with the same dimension equals the whole space gives $W_1\cap W_2=W_2$. Hence $W_2\subseteq W_1\cap W_2\subseteq W_1$, so $W_2\subseteq W_1$.

Proof. The necessary and sufficient condition is $v\in\operatorname{span}(\{v_1,v_2,\cdots,v_k\})=W_1$.

($\Leftarrow$) If $v\in W_1$, then $v$ is a linear combination of $v_1,\ldots,v_k$, so $\operatorname{span}(\{v_1,\ldots,v_k,v\})=\operatorname{span}(\{v_1,\ldots,v_k\})$, i.e. $W_2=W_1$. Hence $\dim W_1=\dim W_2$ holds trivially.

($\Rightarrow$) Suppose $\dim W_1=\dim W_2$. Since $W_1$ is generated by $k$ vectors it is finite-dimensional, and $W_1\subseteq W_2$ with equal dimension, so $W_2$ is finite-dimensional as well. By the theorem that a subspace of a finite-dimensional space with the same dimension equals the whole space, $W_1=W_2$. Hence $v\in W_2=W_1=\operatorname{span}(\{v_1,\ldots,v_k\})$.