Exercises 2B — Axler, Linear Algebra Done Right, 4th Edition

Exercises

Exercise 1. Find all vector spaces that have exactly one basis.

Solution. The vector spaces with exactly one basis are:

$V_1 = \{0\}$, the vector space containing only the zero vector (its unique basis is the empty list).
$V_2 = $ any one-dimensional vector space over $\mathbf{F}_2 = \{0,1\}$ (its unique basis is $(v)$, where $v$ is the only nonzero vector).

If $\mathbf{F} = \mathbb{R}$ or $\mathbf{F} = \mathbb{C}$, then only $V_1$ is possible. But for arbitrary fields $\mathbf{F}$ it would be any one-dimensional vector space over $\mathbf{F}_2 = \{0,1\}$. $\qquad\square$

Exercise 5. Suppose $V$ is finite-dimensional and $U, W$ are subspaces of $V$ such that $V = U + W$. Prove there exists a basis of $V$ consisting of vectors in $U \cup W$.

Proof. Let $u_1,\dots,u_m$ be a basis in $U$ and let $w_1,\dots,w_n$ be a basis in $W$. Since $V = U + W$, every $v \in V$ can be written $v = u + w$, $u \in U$, $w \in W$. Let

\[ u = a_1 u_1 + \dots + a_m u_m \quad\text{and}\quad w = b_1 w_1 + \dots + b_n w_n, \qquad a_i, b_j \in \mathbf{F},\ i=1,\dots,m,\ j=1,\dots,n. \]

This means every $v \in V$ can be written

\[ v = a_1 u_1 + \dots + a_m u_m + b_1 w_1 + \dots + b_n w_n. \]

Thus $(u_1,\dots,u_m, w_1,\dots,w_n)$ spans $V$, and from that spanning list of vectors in $U \cup W$ we can form a basis of $V$ by removing vectors until the list becomes linearly independent while still spanning $V$ and thus a basis of $V$. $\qquad\square$

Exercise 10. If $V = U \oplus W$ and $u_1,\dots,u_m$ is a basis of $U$ and $w_1,\dots,w_n$ is a basis of $W$, prove that $u_1,\dots,u_m,\,w_1,\dots,w_n$ is a basis of $V$.

Proof. From Exercise 5, we know $u_1,\dots,u_m,\,w_1,\dots,w_n$ spans $V$. Note $U \cap W = \{0\}$ since $V = U \oplus W$. So every nonzero vector in $W$ cannot be created as a linear combination of vectors in $U$ and vice versa. Hence $u_1,\dots,u_m,\,w_1,\dots,w_n$ is a linearly independent set of vectors in $V$. Since this set also spans $V$, it is a basis of $V$. $\qquad\square$