Let
\[
\mathbf{v}_{1}
=
\begin{bmatrix}
1 \\ 1
\end{bmatrix}
,\;
\mathbf{v}_{2}
=
\begin{bmatrix}
1 \\ -1
\end{bmatrix}
.
\]
Let $V=\Span(\mathbf{v}_{1},\mathbf{v}_{2})$. Do $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ form an orthonormal basis for $V$?

Let $W$ be the set of $3\times 3$ skew-symmetric matrices. Show that $W$ is a subspace of the vector space $V$ of all $3\times 3$ matrices. Then, exhibit a spanning set for $W$.

Let $A$ be an $m \times n$ matrix.
Suppose that the nullspace of $A$ is a plane in $\R^3$ and the range is spanned by a nonzero vector $\mathbf{v}$ in $\R^5$. Determine $m$ and $n$. Also, find the rank and nullity of $A$.

(c) Find a basis for the range of $A$ that consists of column vectors of $A$.

(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.

Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^3$. Is it possible that $S_2=\{\mathbf{v}_1\}$ is a spanning set for $V$?

Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^5$. If $\mathbf{v}_4$ is another vector in $V$, then is the set
\[S_2=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}\]
still a spanning set for $V$? If so, prove it. Otherwise, give a counterexample.

For a set $S$ and a vector space $V$ over a scalar field $\K$, define the set of all functions from $S$ to $V$
\[ \Fun ( S , V ) = \{ f : S \rightarrow V \} . \]

For $f, g \in \Fun(S, V)$, $z \in \K$, addition and scalar multiplication can be defined by
\[ (f+g)(s) = f(s) + g(s) \, \mbox{ and } (cf)(s) = c (f(s)) \, \mbox{ for all } s \in S . \]

(a) Prove that $\Fun(S, V)$ is a vector space over $\K$. What is the zero element?

(b) Let $S_1 = \{ s \}$ be a set consisting of one element. Find an isomorphism between $\Fun(S_1 , V)$ and $V$ itself. Prove that the map you find is actually a linear isomorpism.

(c) Suppose that $B = \{ e_1 , e_2 , \cdots , e_n \}$ is a basis of $V$. Use $B$ to construct a basis of $\Fun(S_1 , V)$.

(d) Let $S = \{ s_1 , s_2 , \cdots , s_m \}$. Construct a linear isomorphism between $\Fun(S, V)$ and the vector space of $n$-tuples of $V$, defined as
\[ V^m = \{ (v_1 , v_2 , \cdots , v_m ) \mid v_i \in V \mbox{ for all } 1 \leq i \leq m \} . \]

(e) Use the basis $B$ of $V$ to constract a basis of $\Fun(S, V)$ for an arbitrary finite set $S$. What is the dimension of $\Fun(S, V)$?

(f) Let $W \subseteq V$ be a subspace. Prove that $\Fun(S, W)$ is a subspace of $\Fun(S, V)$.

Let $A=\begin{bmatrix}
2 & 4 & 6 & 8 \\
1 &3 & 0 & 5 \\
1 & 1 & 6 & 3
\end{bmatrix}$. (a) Find a basis for the nullspace of $A$.

(b) Find a basis for the row space of $A$.

(c) Find a basis for the range of $A$ that consists of column vectors of $A$.

(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.

Using the definition of the range of a matrix, describe the range of the matrix
\[A=\begin{bmatrix}
2 & 4 & 1 & -5 \\
1 &2 & 1 & -2 \\
1 & 2 & 0 & -3
\end{bmatrix}.\]

(a) If $AB=B$, then $B$ is the identity matrix. (b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions. (c) If $A$ is invertible, then $ABA^{-1}=B$. (d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix. (e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.

Let $A$ be an $n\times n$ nonsingular matrix. Let $\mathbf{v}, \mathbf{w}$ be linearly independent vectors in $\R^n$. Prove that the vectors $A\mathbf{v}$ and $A\mathbf{w}$ are linearly independent.

Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.
If
\[A\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}=B\begin{bmatrix}
2 \\
6 \\
10
\end{bmatrix},\]
then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not.