Linear Algebra Done Right (4th Edition) Solutions
3E
11. Suppose $ U = \{ (x1,x2,\dots) \in \mathbf{F}^\infty : x_k \ne 0 \ \text{for only finitely many k} \} $.
(b) Prove that $ \mathbf{F}^\infty/U $ is infinite-dimensional.
Goal and strategy
We show $\mathbf{F}^\infty/U$ is infinite-dimensional, i.e. not finite-dimensional. We use the standard criterion:
A vector space is finite-dimensional of dimension $n$ only if every linearly independent list has length $\le n$. So a space is infinite-dimensional if, for every positive integer $m$, it contains a linearly independent list of length $m$.
Accordingly, we build one infinite list of vectors $v_1, v_2, \dots$ in $\mathbf{F}^\infty$ such that for every $m$, the cosets $v_1 + U, \dots, v_m + U$ are linearly independent in the quotient. That gives independent lists of every length, so the quotient cannot be finite-dimensional.
Translating independence in the quotient
Before constructing anything, let’s record exactly what we must verify. For scalars $a_1, \dots, a_m$,
\[a_1(v_1 + U) + \cdots + a_m(v_m + U) = (a_1 v_1 + \cdots + a_m v_m) + U,\]and a coset $x + U$ is the zero element of $\mathbf{F}^\infty/U$ precisely when $x \in U$. Therefore
\[v_1 + U, \dots, v_m + U \text{ are linearly independent} \iff \Big(\, a_1 v_1 + \cdots + a_m v_m \in U \ \Rightarrow\ a_1 = \cdots = a_m = 0 \,\Big). \tag{$\ast$}\]Since $U$ is the set of sequences with only finitely many nonzero entries, the right-hand side says: no nontrivial linear combination of the $v_j$ is finitely supported. That is the property our vectors must have.
Constructing the vectors
The idea is to give each $v_j$ an infinite “footprint” of slots, with the footprints of different vectors disjoint, so that combinations cannot cancel down to finitely many nonzero terms.
Partition the index set ${1, 2, 3, \dots}$ into infinitely many pairwise disjoint infinite subsets $S_1, S_2, S_3, \dots$. One explicit choice: write each positive integer uniquely as $n = 2^{\,j-1}(2k-1)$ with $j, k \ge 1$, and let
\[S_j = \{\, 2^{\,j-1}(2k-1) : k \ge 1 \,\}.\]Each $S_j$ is infinite, the $S_j$ are pairwise disjoint, and they cover every index. (Any partition into infinitely many infinite blocks works; only these three properties matter.)
Define $v_j \in \mathbf{F}^\infty$ to be the indicator sequence of $S_j$: its entry in slot $n$, written $(v_j)_n$, is
\[(v_j)_n = \begin{cases} 1 & n \in S_j, \\ 0 & n \notin S_j. \end{cases}\]Each $v_j$ has infinitely many nonzero entries (one for each element of the infinite set $S_j$), so $v_j \notin U$.
Verifying independence of the cosets
Fix any $m$, and suppose some combination lies in $U$:
\[w := a_1 v_1 + a_2 v_2 + \cdots + a_m v_m \in U.\]Here $w$ is itself a sequence in $\mathbf{F}^\infty$; write $w_n$ for its entry in slot $n$, so $w_n = a_1 (v_1)_n + \cdots + a_m (v_m)_n$.
Each block carries a single coefficient. Fix $j$ with $1 \le j \le m$ and let $n \in S_j$. Because the supports are disjoint, $n$ belongs to $S_j$ and to no other $S_i$; hence $(v_j)_n = 1$ while $(v_i)_n = 0$ for all $i \ne j$. The sum defining $w_n$ collapses to its single surviving term:
\[w_n = a_j \qquad \text{for every } n \in S_j. \tag{$\dagger$}\]Membership in $U$ forces each coefficient to vanish. Suppose, toward a contradiction, that $a_j \ne 0$ for some $j \le m$. By $(\dagger)$, $w_n = a_j \ne 0$ at every index $n \in S_j$. Since $S_j$ is infinite, $w$ then has infinitely many nonzero entries — contradicting $w \in U$. Hence $a_j = 0$, and as $j \le m$ was arbitrary,
\[a_1 = a_2 = \cdots = a_m = 0.\]By the criterion $(\ast)$, the cosets $v_1 + U, \dots, v_m + U$ are linearly independent.
Conclusion
For every positive integer $m$, the quotient $\mathbf{F}^\infty/U$ contains a linearly independent list of length $m$, namely $v_1 + U, \dots, v_m + U$. No finite-dimensional space has independent lists of arbitrary length, so
\[\mathbf{F}^\infty / U \text{ is infinite-dimensional.} \qquad \blacksquare\]Remark
Conceptually, $U$ is the subspace of sequences that eventually vanish, so passing to $\mathbf{F}^\infty/U$ discards everything except a sequence’s long-run “tail.” The vectors $v_j$ were designed to have persistent, non-overlapping tails: each lives forever on its own block $S_j$, and disjointness — distilled in equation $(\dagger)$ — means a linear combination reads off coefficient $a_j$ across the whole of $S_j$, with no way for different vectors to interfere and cancel. So no nontrivial combination can die out, giving infinitely many independent directions in the quotient.
15. Suppose $U$ is a subspace of $V$ and $ v_1 + U, \dots, v_m + U $ is a basis of $ V/U $ and $ u_1, \dots, u_n $ is a basis of $U$. Prove that $ v_1, \dots, v_m, u_1, \dots, u_n $ is a basis of 𝑉.
Spanning
Let $v \in V$ be arbitrary; $v + U \in V/U$. Since $v_1 + U, \dots, v_m + U$ spans $V/U$, there are scalars $b_1, \dots, b_m$ with
\[v + U = b_1(v_1 + U) + \cdots + b_m(v_m + U) = (b_1 v_1 + \cdots + b_m v_m) + U.\]Equal cosets have a difference lying in $U$, so
\[v - (b_1 v_1 + \cdots + b_m v_m) \in U.\]Since $u_1, \dots, u_n$ spans $U$, this difference equals $c_1 u_1 + \cdots + c_n u_n$ for some scalars $c_1, \dots, c_n$. Rearranging,
\[v = b_1 v_1 + \cdots + b_m v_m + c_1 u_1 + \cdots + c_n u_n.\]Thus $v \in \operatorname{span}(v_1, \dots, v_m, u_1, \dots, u_n)$. As $v$ was arbitrary, the combined list spans $V$.
Linear independence
Suppose scalars satisfy
\[a_1 v_1 + \cdots + a_m v_m + c_1 u_1 + \cdots + c_n u_n = 0. \tag{1}\]Move the $u$-terms to the other side:
\[a_1 v_1 + \cdots + a_m v_m = -(c_1 u_1 + \cdots + c_n u_n).\]The right-hand side lies in $U$, so $a_1 v_1 + \cdots + a_m v_m \in U$. Passing to the quotient, this says
\[a_1(v_1 + U) + \cdots + a_m(v_m + U) = (a_1 v_1 + \cdots + a_m v_m) + U = 0.\]Since $v_1 + U, \dots, v_m + U$ is a basis of $V/U$, it is linearly independent, forcing
\[a_1 = a_2 = \cdots = a_m = 0.\]Substituting back into $(1)$ leaves $c_1 u_1 + \cdots + c_n u_n = 0$. Since $u_1, \dots, u_n$ is a basis of $U$, it is linearly independent, forcing
\[c_1 = c_2 = \cdots = c_n = 0.\]All coefficients vanish, so the combined list is linearly independent.
Conclusion
The list $v_1, \dots, v_m, u_1, \dots, u_n$ spans $V$ and is linearly independent, hence is a basis of $V$. $\blacksquare$