Fall 2026: Math 590 Quiz 3 Solutions

Problem 1

Let $V = \mathrm{Span}\{v_1, \ldots, v_r\}$ and $w_1, \ldots, w_s\in V$ be linearly independent. What conclusions can you draw from the Exchange Theorem? (4 points)

Solution. Since $w_1, \ldots, w_s$ is a linearly independent subset of $V = \mathrm{Span}\{v_1, \ldots, v_r\}$, the Exchange Theorem allows us to conclude:

(a) $s\le r$, and
(b) after a suitable reordering of $v_1, \ldots, v_r$, the set $\{w_1, \ldots, w_s,\ v_{s+1}, \ldots, v_r\}$ spans $V$.

Problem 2

Extend $\begin{pmatrix} 1\\0\\1\end{pmatrix}, \begin{pmatrix} -1\\1\\-1\end{pmatrix}$ to a basis for $\mathbb{R}^3$. You must justify your answer. (3 points)

Solution. Let $v_1 = \begin{pmatrix} 1\\0\\1\end{pmatrix}$ and $v_2 = \begin{pmatrix} -1\\1\\-1\end{pmatrix}$. Since $\mathbb{R}^3$ has dimension $3$, we need one more vector, not in $\mathrm{Span}\{v_1, v_2\}$, to extend $\{v_1, v_2\}$ to a basis.

Try $v_3 = \begin{pmatrix} 1\\0\\0\end{pmatrix}$. Suppose $v_3 = av_1+bv_2$ for some $a, b\in\mathbb{R}$. Comparing entries gives

a-b = 1\ (\text{1st entry}), \qquad b = 0\ (\text{2nd entry}), \qquad a-b = 0\ (\text{3rd entry}).

The 1st and 3rd equations require $a-b$ to equal both $1$ and $0$ simultaneously, a contradiction. So $v_3\notin\mathrm{Span}\{v_1, v_2\}$, and $\{v_1, v_2, v_3\}$ is linearly independent.

As a check, we verify directly that $\{v_1, v_2, v_3\}$ is linearly independent by computing the determinant of the matrix with these vectors as columns:

\det\begin{pmatrix} 1 & -1 & 1\\0 & 1 & 0\\1 & -1 & 0\end{pmatrix}.

Expanding along the second row (which has only one nonzero entry):

\det = (-1)^{2+2}(1)\det\begin{pmatrix} 1 & 1\\1 & 0\end{pmatrix} = 1\cdot(0-1) = -1\ne 0.

Since the determinant is nonzero, $\{v_1, v_2, v_3\}$ is linearly independent. As $3$ linearly independent vectors in the $3$-dimensional space $\mathbb{R}^3$, they automatically form a basis. Hence

\left\{\begin{pmatrix} 1\\0\\1\end{pmatrix},\ \begin{pmatrix} -1\\1\\-1\end{pmatrix},\ \begin{pmatrix} 1\\0\\0\end{pmatrix}\right\}

is a basis for $\mathbb{R}^3$ extending the given vectors.

Problem 3

Let $T: V\to W$ be a linear transformation between vector spaces over $F$. Let $K$ be the set of vectors $w\in V$ such that $T(w) = \vec{0}$. Show that $K$ is closed under addition and scalar multiplication. (3 points)

Solution. Closed under addition: Let $v_1, v_2\in K$, so $T(v_1) = \vec{0}$ and $T(v_2) = \vec{0}$. Since $T$ is linear,

T(v_1+v_2) = T(v_1)+T(v_2) = \vec{0}+\vec{0} = \vec{0}.

Hence $v_1+v_2\in K$.

Closed under scalar multiplication: Let $v\in K$ and let $c\in F$ be a scalar. Since $T$ is linear,

T(cv) = c\,T(v) = c\cdot\vec{0} = \vec{0}.

Hence $cv\in K$.

Therefore $K$ is closed under both addition and scalar multiplication.