Chapter 2 — Vector Spaces

Solution

If \( 0 \) and \( 0' \) are both null vectors, \( 0 = 0 + 0' = 0' \). If \(w, w'\) are both inverses of \(v\), \( w = w + 0 = w + (v + w') = (w + v) + w' = 0 + w' = w' \).

Solution

\( \alpha\,0_V = \alpha(0_V + 0_V) = \alpha\,0_V + \alpha\,0_V \), so \( \alpha\,0_V = 0_V \). Similarly \( 0_\F\,v = (0_\F + 0_\F)v = 0_\F v + 0_\F v \Rightarrow 0_\F v = 0_V \). If \( \alpha v = 0_V \) with \( \alpha \neq 0 \), then \( v = 1\cdot v = (\alpha^{-1}\alpha)v = \alpha^{-1}(\alpha v) = \alpha^{-1}0_V = 0_V \).

Solution

Function space. Each of the eight axioms reduces pointwise to the field axioms; e.g. \( (f + g)(x) = f(x) + g(x) = g(x) + f(x) = (g + f)(x) \), the zero is \( 0_{\mathcal F}(x) = 0_\F \), and the inverse is \( (-f)(x) = -f(x) \).

\( \ell^1 \). For \( \ell_1 = (x_n) \), \( \ell_2 = (y_n) \in \ell^1 \), the triangle inequality gives \( \sum_n |x_n + y_n| \le \sum_n |x_n| + \sum_n |y_n| < \infty \), so \( \ell_1 + \ell_2 \in \ell^1 \); the zero sequence and \( -\ell = (-x_n) \) lie in \( \ell^1 \). So \( \ell^1 \) is a vector space.

\( \ell^2 \). Using \( (a - b)^2 \ge 0 \Rightarrow 2ab \le a^2 + b^2 \), \( \sum_n |x_n + y_n|^2 = \sum_n(|x_n|^2 + |y_n|^2 + 2|x_n||y_n|) \le 2\sum_n|x_n|^2 + 2\sum_n|y_n|^2 < \infty \), so \( \ell^2 \) is a vector space.

Solution

No. The zero matrix \(O\) is not in \( \mathrm{GL}_n(\C) \) (not invertible). Also \( I_n, -I_n \in \mathrm{GL}_n(\C) \) but \( I_n + (-I_n) = O \notin \mathrm{GL}_n(\C) \), so it is not closed under addition.

Solution

Sums and scalar multiples of continuous functions are continuous, so the set is closed under both operations; the axioms hold pointwise, the zero is the zero function, and \( -f \) is the inverse.

Solution

\( 0 \in H \) since \( \sum_i a_i\cdot 0 = 0 \). For \( u = (x_i), v = (y_i) \in H \), \( \sum_i a_i(x_i + y_i) = \sum_i a_i x_i + \sum_i a_i y_i = 0 \), so \( u + v \in H \); and \( \sum_i a_i(\alpha x_i) = \alpha\sum_i a_i x_i = 0 \), so \( \alpha u \in H \). Hence \(H\) is a subspace.

Solution

Take \( U = X - x_0 \) (any \( x_0 \in X \)). Then \( 0 = x_0 - x_0 \in U \). For \( u = u_x - x_0, v = v_x - x_0 \in U \), \( u + v = (u_x + v_x) - 2x_0 = 2\big[\tfrac12(u_x + v_x) - x_0\big] \in U \) (using the given closure with \( t = \tfrac12 \)), and \( \alpha u = \alpha(u_x - x_0) \in U \). So \(U\) is a subspace, and \( X = U + x_0 \) follows since \( U + x_0 = \{ u_x : u_x \in X \} = X \).

Solution

It contains \(0\) and is closed under \(+\) and scalar multiplication (componentwise in each \(U_i\)), so it is a subspace. If \(W\) is a subspace containing each \(U_i\), then any \( u_1 + \dots + u_m \) (with \( u_i \in U_i \subseteq W \)) lies in \(W\) by closure; hence \( U_1 + \dots + U_m \subseteq W \).

Solution

(9.1) If the sum is direct, \(0\) has the unique representation \( 0 = 0 + \dots + 0 \). Conversely, if \( v = \sum u_i = \sum w_i \) are two representations, subtracting gives \( 0 = \sum(u_i - w_i) \); by uniqueness for \(0\), \( u_i = w_i \), so the representation of every \(v\) is unique.

(9.2) If \( U_1 + U_2 \) is direct and \( v \in U_1 \cap U_2 \), then \( v \in U_1 \), \( -v \in U_2 \), and \( 0 = v + (-v) \); uniqueness forces \( v = 0 \), so \( U_1 \cap U_2 = \{0\} \). Conversely, if \( U_1 \cap U_2 = \{0\} \) and \( u_1 + u_2 = 0 \) (\( u_i \in U_i \)), then \( u_1 = -u_2 \in U_1 \cap U_2 = \{0\} \), so \( u_1 = u_2 = 0 \).

Solution

For \( v = (x, y) \), solving \( v = a v_1 + b v_2 \) gives \( x = a + b \), \( y = -a + b \), so \( b = \tfrac{x + y}{2} \), \( a = \tfrac{x - y}{2} \). Hence any \( (x, y) = \tfrac{x-y}{2}v_1 + \tfrac{x+y}{2}v_2 \in \operatorname{span}(v_1, v_2) \), so \( \operatorname{span}(v_1, v_2) = \R^2 \).

Solution

\( 0 = \sum_i 0\cdot v_i \in U \); sums and scalar multiples of linear combinations are linear combinations, so \(U\) is a subspace. Any subspace \( W \) containing all \( v_i \) must (by closure) contain every \( \sum a_i v_i \), so \( U \subseteq W \); thus \(U\) is the smallest such subspace.

Solution

For a direct sum we need \( u_1 + \dots + u_m = 0 \Rightarrow u_i = 0 \) with \( u_i \in \operatorname{span}(v_i) \), i.e. \( u_i = a_i v_i \). Then \( \sum a_i v_i = 0 \) and LI of the \( v_i \) give \( a_i = 0 \), so each \( u_i = 0 \); the sum is direct. Both inclusions \( \operatorname{span}(v_1, \dots, v_m) = \sum_i \operatorname{span}(v_i) \) are immediate, so \( \operatorname{span}(v_1, \dots, v_m) = \bigoplus_i \operatorname{span}(v_i) \).

Solution

If it is a basis and \( v = \sum a_i v_i = \sum b_i v_i \), subtracting gives \( \sum(a_i - b_i)v_i = 0 \); LI forces \( a_i = b_i \), so the representation is unique. Conversely, unique representations make the \( v_i \) span (existence) and LI (the representation of \(0\) is unique, namely all-zero), so they form a basis.

Solution

(1) If \(n\) LI vectors did not span, some \( v \in V \setminus \operatorname{span} \) would extend them to \( n+1 \) LI vectors, exceeding the spanning length \(n\) — contradiction. (2) If \(n\) spanning vectors were LD, the linear dependence lemma lets us remove one, leaving \( n-1 \) vectors spanning \(V\), contradicting that any basis has length \(n\).

Solution

A basis of \(U\) is a LI list in \(V\); by Proposition 3 its length is at most that of a spanning list of \(V\), so \( \dim U \le \dim V \).

Solution

If all \( a_i = 0 \), the constraint is vacuous and \( H = \R^n \), so \( \dim H = n \). Otherwise pick \(j\) with \( a_j \neq 0 \); then \( x_j = -\tfrac{1}{a_j}\sum_{i\neq j}a_i x_i \) is determined by the other \( n - 1 \) coordinates. Concretely, the vectors \( f_i = e_i - \tfrac{a_i}{a_j}e_j \) for \( i \neq j \) are \( n - 1 \) linearly independent elements of \(H\) that span it, so \( \dim H = n - 1 \).

Solution

Let \( \dim V = n \), bases \( u_1, \dots, u_m \) of \( U_1 \) and \( w_1, \dots, w_p \) of \( U_2 \) with \( m + p > n \). The combined \( m + p \) vectors lie in \(V\), so (exceeding \(n\)) they are LD. By the linear dependence lemma some \( w_j = a_1 u_1 + \dots + a_m u_m + b_1 w_1 + \dots + b_{j-1}w_{j-1} \), giving a non-zero \( w = w_j - \sum_{i<j} b_i w_i \in U_2 \) that also equals \( \sum_i a_i u_i \in U_1 \). So \( 0 \neq w \in U_1 \cap U_2 \), hence \( \dim(U_1 \cap U_2) \ge 1 \).

Solution

\( \{(1,0)\} \) is a basis of \(U\). For \( W_1 \): any \( v = a(1,0) + b(0,1) \), so \( V = U + W_1 \); and \( a(1,0) + b(0,1) = 0 \Rightarrow a = b = 0 \), so \( U \cap W_1 = \{0\} \) and \( V = U \oplus W_1 \). For \( W_2 \) (basis \( (1,1) \)): \( v = a(1,0) + b(1,1) = (a+b,\ b) \) is solvable (\( b = v_2 \), \( a = v_1 - v_2 \)), so \( V = U + W_2 \); and \( a(1,0) + b(1,1) = 0 \Rightarrow b = 0, a = 0 \), so \( U \cap W_2 = \{0\} \) and \( V = U \oplus W_2 \). Thus \( W_1 \) and \( W_2 \) are both complementary to \(U\).