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 $W$.

Proof.

To prove that $W$ is a subspace of $V$, the $3\times 3$ zero matrix $\mathbf{0}$ is the zero vector of $V$, and since $\mathbf{0}$ is clearly skew-symmetric, $\mathbf{0}\in W$. Next, let $A,B\in W$. Then $A^{T}=-A$ and $B^{T}=-B$. Therefore,
\[
(A+B)^{T}
=A^{T}+B^{T}
=-A+(-B)
=-(A+B).
\] Thus $A+B$ is skew-symmetric, and therefore $A+B\in W$. Next, take any $A\in W$ and $r$ in the scalar field. Then
\[
(rA)^{T}
=rA^{T}
=r(-A)
=-rA.
\] Thus $rA$ is skew symmetric, which implies $rA\in W$. Therefore, $W$ is a subspace of $V$.

To exhibit a spanning set for $W$, first note that
\[
W=
\left\{
A\left|
A=
\begin{bmatrix}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{bmatrix}
,\;a,b,c\in\R
\right.\right\}.
\] Let
\[
A_{1}=
\begin{bmatrix}
0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
,\;
A_{2}=
\begin{bmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
-1 & 0 & 0
\end{bmatrix}
,\;
A_{3}=
\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & -1 & 0
\end{bmatrix}
.
\] Then any $A\in W$ can be written as
\[
A=
\begin{bmatrix}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{bmatrix}
=aA_{1}+bA_{2}+cA_{3},
\] which is a linear combination of $A_{1},A_{2},A_{3}$. Thus $\{A_{1},A_{2},A_{3}\}$ is a spanning set for $W$.

Find a basis for $\Span(S)$ where $S=
\left\{
\begin{bmatrix}
1 \\ 2 \\ 1
\end{bmatrix}
,
\begin{bmatrix}
-1 \\ -2 \\ -1
\end{bmatrix}
,
\begin{bmatrix}
2 \\ 6 \\ -2
\end{bmatrix}
,
\begin{bmatrix}
1 \\ 1 \\ 3
\end{bmatrix}
\right\}$.

Solution.

We will first use the leading $1$ method. Consider the system
\[
x_{1}
\begin{bmatrix}
1 \\ 2 \\ 1
\end{bmatrix}
+x_{2}
\begin{bmatrix}
-1 \\ -2 \\ -1
\end{bmatrix}
+x_{3}
\begin{bmatrix}
2 \\ 6 \\ -2
\end{bmatrix}
+x_{4}
\begin{bmatrix}
1 \\ 1 \\ 3
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}
.\] The augmented matrix for this system is
\[
\left[\begin{array}{cccc|c}
1 & -1 & 2 & 1 & 0 \\
2 & -2 & 6 & 1 & 0 \\
1 & -1 & -2 & 3 & 0
\end{array}\right] \xrightarrow{\substack{R_{2}-2R_{1} \\ R_{3}-R_{1}}}
\left[\begin{array}{cccc|c}
1 & -1 & 2 & 1 & 0 \\
0 & 0 & 2 & -1 & 0 \\
0 & 0 & -4 & 2 & 0
\end{array}\right] \] \[
\to
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 0 \\
0 & 0 & 1 & -1/2 & 0 \\
0 & 0 & 0 & 0
\end{array}\right].
\] Since the above matrix has leading $1$’s in the first and third columns, we can conclude that the first and third vectors of $S$ form a basis of $\Span(S)$. Thus
\[
\left\{
\begin{bmatrix}
1 \\ 2 \\ 1
\end{bmatrix}
,
\begin{bmatrix}
2 \\ 6 \\ -2
\end{bmatrix}
\right\}
\] is a basis for $\Span(S)$.

Let $S=\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4},\mathbf{v}_{5}\}$ where
\[
\mathbf{v}_{1}=
\begin{bmatrix}
1 \\ 2 \\ 2 \\ -1
\end{bmatrix}
,\;\mathbf{v}_{2}=
\begin{bmatrix}
1 \\ 3 \\ 1 \\ 1
\end{bmatrix}
,\;\mathbf{v}_{3}=
\begin{bmatrix}
1 \\ 5 \\ -1 \\ 5
\end{bmatrix}
,\;\mathbf{v}_{4}=
\begin{bmatrix}
1 \\ 1 \\ 4 \\ -1
\end{bmatrix}
,\;\mathbf{v}_{5}=
\begin{bmatrix}
2 \\ 7 \\ 0 \\ 2
\end{bmatrix}
.\] Find a basis for the span $\Span(S)$.

 

We will give two solutions.

Solution 1.

We apply the leading 1 method.
Let $A$ be the matrix whose column vectors are vectors in the set $S$:
\[
A=
\begin{bmatrix}
1 & 1 & 1 & 1 & 2 \\
2 & 3 & 5 & 1 & 7 \\
2 & 1 & -1 & 4 & 0 \\
-1 & 1 & 5 & -1 & 2
\end{bmatrix}
.\] Applying the elementary row operations to $A$, we obtain
\begin{align*}
A=\begin{bmatrix}
1 & 1 & 1 & 1 & 2 \\
2 & 3 & 5 & 1 & 7 \\
2 & 1 & -1 & 4 & 0 \\
-1 & 1 & 5 & -1 & 2
\end{bmatrix}
\xrightarrow[R_4+R_1]{\substack{R_2-2R_1 \\ R_3-2R_1}}
\begin{bmatrix}
1 & 1 & 1 & 1 & 2 \\
0 & 1 & 3 & -1 & 3 \\
0 & -1 & -3 & 2 & -4 \\
0 & 2 & 6 & 0 & 4
\end{bmatrix}\\[6pt] \xrightarrow[R_4-2R_2]{\substack{R_1-R_2 \\ R_3+R_2}}
\begin{bmatrix}
1 & 0 & -2 & 2 & -1 \\
0 & 1 & 3 & -1 & 3 \\
0 & 0 & 0 & 1 & -1 \\
0 & 0 & 0 & 2 & -2
\end{bmatrix}
\xrightarrow[R_4-2R_3]{\substack{R_1-2R_3 \\ R_2+R_3}}
\begin{bmatrix}
1 & 0 & -2 & 0 & 1 \\
0 & 1 & 3 & 0 & 2 \\
0 & 0 & 0 & 1 & -1 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}=\rref(A).
\end{align*}
Observe that the first, second, and fourth column vectors of $\rref(A)$ contain the leading 1 entries.
Hence, the first, second, and fourth column vectors of $A$ form a basis of $\Span(S)$.
Namely,
\[\left\{ \begin{bmatrix}
1 \\
2 \\
2 \\
-1
\end{bmatrix}, \begin{bmatrix}
1 \\
3 \\
1 \\
1
\end{bmatrix}, \begin{bmatrix}
1 \\
1 \\
4 \\
-1
\end{bmatrix}\right \}\] is a basis for $\Span(S)$.

Solution 2.

Let
\[
A=
\begin{bmatrix}
1 & 1 & 1 & 1 & 2 \\
2 & 3 & 5 & 1 & 7 \\
2 & 1 & -1 & 4 & 0 \\
-1 & 1 & 5 & -1 & 2
\end{bmatrix}
.\] Then $\Span(S)$ is the column space of $A$, which is the row space of $A^{T}$. Using row operations, we have
\[
A^{T}=
\begin{bmatrix}
1 & 2 & 2 & -1 \\
1 & 3 & 1 & 1 \\
1 & 5 & -1 & 5 \\
1 & 1 & 4 & -1 \\
2 & 7 & 0 & 2
\end{bmatrix}
\to
\begin{bmatrix}
1 & 2 & 2 & -1 \\
0 & 1 & -1 & 2 \\
0 & 3 & -3 & 6 \\
0 & -1 & 2 & 0 \\
0 & 3 & -4 & 4
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 4 & 5 \\
0 & 1 & -1 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 2 \\
0 & 0 & -1 & -2
\end{bmatrix}
\] \[
\to
\begin{bmatrix}
1 & 0 & 0 & -13 \\
0 & 1 & 0 & 4 \\
0 & 0 & 1 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
.
\] Therefore, the set of nonzero rows
\[
\left\{
\begin{bmatrix}
1 \\ 0 \\ 0 \\ -13
\end{bmatrix}
,
\begin{bmatrix}
0 \\ 1 \\ 0 \\ 4
\end{bmatrix}
,
\begin{bmatrix}
0 \\ 0 \\ 1 \\ 2
\end{bmatrix}
\right\}
\] is a basis for the row space of $A^{T}$, which equals $\Span(S)$.

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 $V$?

Solution.

Yes, in general, $S_2$ can be a spanning set.

As an example, consider the vectors
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}
2 \\
0 \\
0
\end{bmatrix}, \mathbf{v}_3=\begin{bmatrix}
3 \\
0 \\
0
\end{bmatrix}.\] Then, as we have $\mathbf{v}_2=2\mathbf{v}_1$ and $\mathbf{v}_3=\mathbf{3}\mathbf{v}_1$, we see that
the set $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of the subspace $V=\Span(\mathbf{v}_1)$, and clearly $S_2=\{\mathbf{v}_1\}$ is a spanning set for $V$.

Geometrically, $V=\Span(\mathbf{v}_1)$ is a line in $\R^3$ passing through the origin and $\mathbf{v}_1$. The vectors $\mathbf{v}_2, \mathbf{v}_3$ are also on the same line. Thus, we can omit them from the spanning set $S_1$.

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.

Proof.

We prove that $S_2$ is also a spanning set for $V$, that is, we prove that
\[\Span(S_2)=V.\]

Prove $\Span(S_2) \subset V$

We first show that $\Span(S_2)$ is contained in $V$. Let $\mathbf{x}$ be an element in $\Span(S_2)$. Then there exist scalars $c_1, c_2, c_3, c_4$ such that
\[\mathbf{x}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3 \mathbf{v}_3 + c_4 \mathbf{v}_4.\] Since $\Span(S_1)=V$, we know that $c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3 \mathbf{v}_3$ is a vector in $V$. As $\mathbf{v}_4\in V$, we have $c_4\mathbf{v}_4 \in V$.
Since $V$ is a vector space, the sum of two elements in $V$ is in $V$.
So, \[\mathbf{x}=(c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3 \mathbf{v}_3) + (c_4 \mathbf{v}_4) \in V.\] This proves that $\Span(S_2) \subset V$.

Prove $\Span(S_2) \supset V$

Note that since $S_1$ is a spanning set for $V$, every element of $S_1$ can be written as a linear combination of the vectors $\mathbf{v}_1, \mathbf{v}_2$, and $\mathbf{v}_3$.
That is, for any $\mathbf{v}\in V$, there exist scalars $c_1, c_2, c_3$ such that
\[\mathbf{v}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3.\] Observe that this can be written as follows.
\[\mathbf{v}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3+0\mathbf{v}_4.\] This tells us that $\mathbf{v}$ is a linear combination of $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$, and $\mathbf{v}_4$.
Hence, any vector in $V$ can be written as a linear combination of the vectors in $S_2$.
Thus, $V\subset \Span(S_2)$.

Putting these inclusion together yields that $V=\Span(S_2)$, and hence $S_2$ is a spanning set for $V$.

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

Solution.

By definition, the range $\calR(A)$ of the matrix $A$ is given by
\[\calR(A)=\left \{ \mathbf{b} \in \R^3 \quad \middle | \quad A\mathbf{x}=\mathbf{b} \text{ for some } \mathbf{x} \in \R^4 \right \}.\]

Thus, a vector $\mathbf{b}=\begin{bmatrix}
b_1 \\
b_2 \\
b_3
\end{bmatrix}$ in $\R^3$ is in the range $\calR(A)$ if and only if the system $A\mathbf{x}=\mathbf{b}$ is consistent.
So, let us find the conditions on $\mathbf{b}$ so that the system is consistent.

To do this, we consider the augmented matrix of the system and reduce it as follows.
\begin{align*}
\left[\begin{array}{rrrr|r}
2 & 4 & 1 & -5 & b_1\\
1 &2 & 1 & -2 & b_2 \\
1 & 2 & 0 & -3 &b_3
\end{array}\right] \xrightarrow{R_1 \leftrightarrow R_2}
\left[\begin{array}{rrrr|r}
1 &2 & 1 & -2 & b_2 \\
2 & 4 & 1 & -5 & b_1\\
1 & 2 & 0 & -3 &b_3
\end{array}\right] \xrightarrow[R_3-R_1]{R_2-2R_1}\\[6pt] \left[\begin{array}{rrrr|r}
1 &2 & 1 & -2 & b_2 \\
0 & 0 & -1 & -1 & b_1-2b_2\\
0 & 0 & 0 & -1 & b_3-b_2
\end{array}\right] \xrightarrow{-R_2}
\left[\begin{array}{rrrr|r}
1 &2 & 1 & -2 & b_2 \\
0 & 0 & 1 & 1 & -b_1+2b_2\\
0 & 0 & -1 & -1 & b_3-b_2
\end{array}\right]\\[6pt] \xrightarrow[R_3+R_2]{R_1-R_2}
\left[\begin{array}{rrrr|r}
1 &2 & 0 & -3 & b_1-b_2 \\
0 & 0 & 1 & 1 & -b_1+2b_2\\
0 & 0 & 0 & 0 & -b_1 +b_2 +b_3
\end{array}\right].
\end{align*}

Note that if the $(3, 5)$-entry $-b_1+b_2+b_3$ is not zero, then the system $A\mathbf{x}=\mathbf{0}$ is inconsistent because this implies $0=1$.
On the other hand, if $-b_1+b_2+b_3=0$, then we see that the system is consistent.
Hence, the vector $\mathbf{b}$ is in the range $\calR(A)$ if and only if $-b_1+b_2+b_3=0$.

In summary, we have
\[\calR(A)=\left\{ \begin{bmatrix}
b_1 \\
b_2 \\
b_3
\end{bmatrix}\in \R^3 \quad \middle | \quad -b_1+b_2+b_3=0 \right \}.\]

Spanning set for the range

With a little bit additional computation, we can find the spanning set for the range as follows.

Thus, $\mathbf{b} \in \calR(A)$ if and only if
\begin{align*}
\mathbf{b}=\begin{bmatrix}
b_1 \\
b_2 \\
b_3
\end{bmatrix}=\begin{bmatrix}
b_2+b_3 \\
b_2 \\
b_3
\end{bmatrix}=b_2\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}+b_3\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}.
\end{align*}

In summary, we have
\begin{align*}
\calR(A)&=\left\{ \mathbf{b} \in \R^3 \quad \middle | \quad \mathbf{b}=b_2\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}+b_3\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix} \right \}\\[6pt] &=\Span\left\{ \begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix} \right \}.
\end{align*}
Hence, the spanning set is
\[ \left\{ \begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix} \right \}.\]

This spanning set is linearly independent, hence it’s a basis for the range.

Thus, the dimension of the range is $2$.

In this problem, we use the following vectors in $\R^2$.
\[\mathbf{a}=\begin{bmatrix}
1 \\
0
\end{bmatrix}, \mathbf{b}=\begin{bmatrix}
1 \\
1
\end{bmatrix}, \mathbf{c}=\begin{bmatrix}
2 \\
3
\end{bmatrix}, \mathbf{d}=\begin{bmatrix}
3 \\
2
\end{bmatrix}, \mathbf{e}=\begin{bmatrix}
0 \\
0
\end{bmatrix}, \mathbf{f}=\begin{bmatrix}
5 \\
6
\end{bmatrix}.\] For each set $S$, determine whether $\Span(S)=\R^2$. If $\Span(S)\neq \R^2$, then give algebraic description for $\Span(S)$ and explain the geometric shape of $\Span(S)$.

(a) $S=\{\mathbf{a}, \mathbf{b}\}$
(b) $S=\{\mathbf{a}, \mathbf{c}\}$
(c) $S=\{\mathbf{c}, \mathbf{d}\}$
(d) $S=\{\mathbf{a}, \mathbf{f}\}$
(e) $S=\{\mathbf{e}, \mathbf{f}\}$
(f) $S=\{\mathbf{a}, \mathbf{b}, \mathbf{c}\}$
(g) $S=\{\mathbf{e}\}$

Solution.

By definition, the subspace $\Span(S)$ spanned by $S$ is the set of all linear combinations of vectors in $S$. Thus, $\Span(S)$ is a subset in $\R^2$. The question is whether all of the vectors in $\R^2$ are linear combinations of vectors in $S$ or not.

(a) $S=\{\mathbf{a}, \mathbf{b}\}$

Let $\begin{bmatrix}
a \\
b
\end{bmatrix}$ be an arbitrary vector in $\R^2$. We determine whether it is a linear combination of $\mathbf{a}$ and $\mathbf{b}$:
\[\begin{bmatrix}
a \\
b
\end{bmatrix}=c_1\mathbf{a}+c_2\mathbf{b}.\] The augmented matrix is
\begin{align*}
\left[\begin{array}{rr|r}
1 & 1 & a \\
0 & 1 & b
\end{array} \right] \xrightarrow{R_1-R_2}
\left[\begin{array}{rr|r}
1 & 0 & a-b \\
0 & 1 & b
\end{array} \right].
\end{align*}
Hence, the solution is $c_1=a-b$, $c_2=b$.
Thus, we have
\[\begin{bmatrix}
a \\
b
\end{bmatrix}=(a-b)\mathbf{a}+b\mathbf{b},\] and this implies that every vector in $\R^2$ is a linear combination of $\mathbf{a}$ and $\mathbf{b}$.
Hence, $\Span(S)=\R^2$.

(b) $S=\{\mathbf{a}, \mathbf{c}\}$

Just like part (a), let $\begin{bmatrix}
a \\
b
\end{bmatrix}$ be any vector in $\R^2$.
Are there $c_1, c_2$ satisfying
\[\begin{bmatrix}
a \\
b
\end{bmatrix}=c_1\mathbf{a}+c_2\mathbf{c}?\] The augmented matrix is
\begin{align*}
\left[\begin{array}{rr|r}
1 & 2 & a \\
0 & 3 & b
\end{array} \right] \xrightarrow{\frac{1}{3} R_2}
\left[\begin{array}{rr|r}
1 & 2 & a \\
0 & 1 & b/3
\end{array} \right] \xrightarrow{R_1-2R_2}
\left[\begin{array}{rr|r}
1 & 0 & a-\frac{2}{3}b \\
0 & 1 & b/3
\end{array} \right].
\end{align*}
Hence, we have $c_1=a-\frac{2}{3}b$, $c_2=b/3$.
Thus,
\[\begin{bmatrix}
a \\
b
\end{bmatrix}=\frac{2b}{3} \mathbf{a}+\frac{b}{3}\mathbf{c}.\] This yields that every vector in $\R^2$ is in $\Span(S)$. Hence $\Span(S)=\R^2$.

(c) $S=\{\mathbf{c}, \mathbf{d}\}$

The strategy for this problem is the exactly same as before. So let us consider
\begin{align*}
\left[\begin{array}{rr|r}
2 & 3 & a \\
3 & 2 & b
\end{array} \right] \xrightarrow{R_2-R_1}
\left[\begin{array}{rr|r}
2 & 3 & a \\
1 & -1 & b-a
\end{array} \right] \xrightarrow{R_1\leftrightarrow R_2}
\left[\begin{array}{rr|r}
1 & -1 & b-a \\
2 & 3 & a
\end{array} \right]\\[6pt] \xrightarrow{R_2- 2 R_1}
\left[\begin{array}{rr|r}
1 & -1 & b-a \\
0 & 5 & 3a-2b
\end{array} \right] \xrightarrow{\frac{1}{5}R_2}
\left[\begin{array}{rr|r}
1 & -1 & b-a \\
0 & 1 & \frac{3}{5}a-\frac{2}{5}b
\end{array} \right] \\[6pt] \xrightarrow{R_1+R_2}
\left[\begin{array}{rr|r}
1 & 0 & -\frac{2}{5}a+\frac{3}{5}b \\
0 & 1 & \frac{3}{5}a-\frac{2}{5}b
\end{array} \right].
\end{align*}
It follows that any vector $\begin{bmatrix}
a \\
b
\end{bmatrix}\in \R^2$ can be written as a linear combination
\[\begin{bmatrix}
a \\
b
\end{bmatrix}=\left(-\frac{2}{5}a+\frac{3}{5}b \right)\mathbf{c}+\left( \frac{3}{5}a-\frac{2}{5}b \right) \mathbf{d}.\] Hence, $\Span(S)=\R^2$.

(d) $S=\{\mathbf{a}, \mathbf{f}\}$

This is also similar to previous problems. For any vector $\begin{bmatrix}
a \\
b
\end{bmatrix}\in \R^2$, we have
\begin{align*}
\left[\begin{array}{rr|r}
1 & 5 & a \\
0 & 6 & b
\end{array} \right] \xrightarrow{\frac{1}{6}R_2}
\left[\begin{array}{rr|r}
1 & 5 & a \\
0 & 1 & b/6
\end{array} \right] \xrightarrow{R_1-5R_2}
\left[\begin{array}{rr|r}
1 & 0 & a-\frac{5}{6}b \\
0 & 1 & b/6
\end{array} \right].
\end{align*}
This yields that
\[\begin{bmatrix}
a \\
b
\end{bmatrix}=\left(a-\frac{5b}{6} \right)\mathbf{a}+\left(\frac{b}{6}\right)\mathbf{f}.\] Hence, $\Span(S)=\R^2$.

(e) $S=\{\mathbf{e}, \mathbf{f}\}$

Note that as the vector $\mathbf{e}$ is the zero vector, any linear combination of $\mathbf{e}$ and $\mathbf{f}$ is just a scalar multiple of $\mathbf{f}$:
\[c_1\mathbf{e}+c_2\mathbf{f}=c_2\mathbf{f}.\] Thus, the algebraic description is
\begin{align*}
\Span(S)=\Span(\mathbf{e}, \mathbf{f})=\Span(\mathbf{f})=\{\mathbf{x}\in \R^2 \mid \mathbf{x}=c\mathbf{f} \text{ for some } c\in \R\}.
\end{align*}
Geometrically, the span is the line parametrized by $c\mathbf{f}=c\begin{bmatrix}
5 \\
6
\end{bmatrix}$ for any $c\in \R$.

(f) $S=\{\mathbf{a}, \mathbf{b}, \mathbf{c}\}$

Note that $\Span(\mathbf{a}, \mathbf{b})$ is contained in $\Span(\mathbf{a}, \mathbf{b}, \mathbf{c})$ because any linear combination of $\mathbf{a}$ and $\mathbf{b}$ is a linear combination of $\mathbf{a}, \mathbf{b}$, and $\mathbf{c}$ by taking the coefficient of $\mathbf{c}$ to be $0$.
We already saw in part (a) that $\Span(\mathbf{a}, \mathbf{b})=\R^2$. Hence, we must have $\Span(\mathbf{a}, \mathbf{b}, \mathbf{c})=\R^2$ as well.

(g) $S=\{\mathbf{e}\}$

Note that $\Span(\mathbf{e})$ is the set of all linear combination of $\mathbf{e}$, which is just the zero vector. So,
\[\Span(\mathbf{e})=\left\{\begin{bmatrix}
0 \\
0
\end{bmatrix}\right\}.\]Geometrically, this is just one point: the origin in the $x$-$y$ plane.

Let $V$ be a vector space over a scalar field $K$.
Let $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$ be vectors in $V$ and consider the subset
\[W=\{a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k \mid a_1, a_2, \dots, a_k \in K \text{ and } a_1+a_2+\cdots+a_k=0\}.\] So each element of $W$ is a linear combination of vectors $\mathbf{v}_1, \dots, \mathbf{v}_k$ such that the sum of the coefficients is zero.

Prove that $W$ is a subspace of $V$.

 

We give two proofs.

Proof 1. (Subspace Criteria)

We use the following subspace criteria.
The subset $W$ is a subspace of $V$ if the following three conditions are met.

The zero vector $\mathbf{0}$ of $V$ can be written as
\[\mathbf{0}=0\mathbf{v}_1+0\mathbf{v}_2+\cdots+0\mathbf{v}_k.\] Clearly the sum of the coefficient is zero, hence $\mathbf{0} \in W$.
So condition 1 is met.

To verify condition 2, let
\[\mathbf{v}=a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k\] and
\[\mathbf{v}’=b_1\mathbf{v}_1+b_2\mathbf{v}_2+\cdots+ b_k\mathbf{v}_k\] be arbitrary elements in $W$. Thus
\[a_1+a_2+\cdots+a_k=0 \text{ and } b_1+b_2+\cdots+b_k=0. \tag{*}\] The sum $\mathbf{v}+\mathbf{v}’$ is
\begin{align*}
\mathbf{v}+\mathbf{v}’&=(a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k)+(b_1\mathbf{v}_1+b_2\mathbf{v}_2+\cdots+ b_k\mathbf{v}_k)\\
&=(a_1+b_1)\mathbf{v}_1+(a_2+b_2)\mathbf{v}_2+\cdots+(a_k+b_k)\mathbf{v}_k.
\end{align*}
The the sum of the coefficients of the above linear combination is
\begin{align*}
&(a_1+b_1)+(a_2+b_2)+\cdots+(a_k+b_k)\\
&=(a_1+a_2+\cdots+a_k)+(b_1+b_2+\cdots+b_k) \stackrel{(*)}{=} 0+0=0.
\end{align*}
It follows that the sum $\mathbf{v}+\mathbf{v}’$ is in $W$, and hence condition 2 is met.

Finally, let us check condition 3. Let
\[\mathbf{v}=a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k\] be an arbitrary vector in $W$ and let $c\in K$.
Since $\mathbf{v}\in W$, we have
\[a_1+a_2+\cdots+a_k=0.\] Then the scalar product is
\begin{align*}
c\mathbf{v}&=c(a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k)\\
&=ca_1\mathbf{v}_1+ca_2\mathbf{v}_2+\cdots+ ca_k\mathbf{v}_k.
\end{align*}
The sum of the coefficients of the linear combination is
\begin{align*}
ca_1+ca_2+\cdots+ ca_k=c(a_1+a_2+\cdots+a_k)=a0=0.
\end{align*}
Hence $c\mathbf{v}\in V$, and condition 3 is met.

Therefore by the subspace criteria, we conclude that $W$ is a subspace of $V$.

Proof 2. (Span)

Consider an arbitrary vector in $W$:
\[a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k \text{ with } a_1+a_2+\cdots+a_k=0.\] Substituting the relation $a_k=-(a_1+a_2+\cdots+a_{k-1})$, we obtain
\begin{align*}
&a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+a_{k-1}\mathbf{v}_{k-1}+ a_k\mathbf{v}_k\\
&=a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots +a_{k-1}\mathbf{v}_{k-1}-(a_1+a_2+\cdots+a_{k-1})\mathbf{v}_k\\
&=a_1(\mathbf{v}_1-\mathbf{v}_k)+a_2(\mathbf{v}_2-\mathbf{v}_k)+\cdots+a_{k-1}(\mathbf{v}_{k-1}-\mathbf{v}_{k-1}).
\end{align*}
This computation yields that every vector in $W$ is a linear combination of vectors in
\[S:=\{\mathbf{v}_1-\mathbf{v}_k, \mathbf{v}_2-\mathbf{v}_k,\dots, \mathbf{v}_{k-1}-\mathbf{v}_{k-1}\}.\] That is, we have $W\subset \Span(S)$.

On the other hand, let
\[\mathbf{v}=c_1(\mathbf{v}_1-\mathbf{v}_k)+c_2(\mathbf{v}_2-\mathbf{v}_k)+\cdots+c_{k-1}(\mathbf{v}_{k-1}-\mathbf{v}_{k-1})\] be an arbitrary vector in $\Span(S)$.
Then we have
\begin{align*}
\mathbf{v}&=c_1\mathbf{v}_1+c_2\mathbf{v}_2+\cdots +c_{k-1}\mathbf{v}_{k-1}-(c_1+c_2+\cdots+c_{k-1})\mathbf{v}_k.
\end{align*}
This is a linear combination of $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$, and the sum of the coefficients is
\[c_1+c_2+\cdots +c_{k-1}-(c_1+c_2+\cdots+c_{k-1})=0.\] Therefore $\mathbf{v}\in W$. Thus we also have $\Span(S)\subset W$.

Putting together these inclusion yields that $W=\Span(S)$.
As the span is always a subspace, we conclude that $W$ is a subspace of $V$.

Determine whether each of the following sets is a basis for $\R^3$.

(a) $S=\left\{\, \begin{bmatrix}
1 \\
0 \\
-1
\end{bmatrix}, \begin{bmatrix}
2 \\
1 \\
-1
\end{bmatrix}, \begin{bmatrix}
-2 \\
1 \\
4
\end{bmatrix} \,\right\}$

(b) $S=\left\{\, \begin{bmatrix}
1 \\
4 \\
7
\end{bmatrix}, \begin{bmatrix}
2 \\
5 \\
8
\end{bmatrix}, \begin{bmatrix}
3 \\
6 \\
9
\end{bmatrix} \,\right\}$

(c) $S=\left\{\, \begin{bmatrix}
1 \\
1 \\
2
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
7
\end{bmatrix} \,\right\}$

(d) $S=\left\{\, \begin{bmatrix}
1 \\
2 \\
5
\end{bmatrix}, \begin{bmatrix}
7 \\
4 \\
0
\end{bmatrix}, \begin{bmatrix}
3 \\
8 \\
6
\end{bmatrix}, \begin{bmatrix}
-1 \\
9 \\
10
\end{bmatrix} \,\right\}$

Definition (A Basis of a Subspace).

A subset $S$ of a vector space $V$ is called a basis if

1. $S$ is linearly independent, and
2. $S$ is a spanning set.

Solution.

Recall that any three linearly independent vectors form a basis of $\R^3$.
(See the post “Three Linearly Independent Vectors in $\R^3$ Form a Basis. Three Vectors Spanning $\R^3$ Form a Basis.” for the proof of this fact.)

(a) $S=\left\{\, \begin{bmatrix}
1 \\
0 \\
-1
\end{bmatrix}, \begin{bmatrix}
2 \\
1 \\
-1
\end{bmatrix}, \begin{bmatrix}
-2 \\
1 \\
4
\end{bmatrix} \,\right\}$

Let us check that whether $S$ is a linearly independent set.
Consider the linear combination
\[x_1\begin{bmatrix}
1 \\
0 \\
-1
\end{bmatrix}+x_2 \begin{bmatrix}
2 \\
1 \\
-1
\end{bmatrix}+x_3\begin{bmatrix}
-2 \\
1 \\
4
\end{bmatrix} =\mathbf{0}.\] This is equivalent to the matrix equation
\[\begin{bmatrix}
1 & 2 & -2 \\
0 &1 &1 \\
-1 & -1 & 4
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}=\mathbf{0}.\] To find the solution, consider the augmented matrix.
Applying elementary row operations, we obtain
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 2 & -2 & 0 \\
0 &1 & 1 & 0 \\
-1 & -1 & 4 & 0
\end{array} \right] \xrightarrow{R_3+R_1}
\left[\begin{array}{rrr|r}
1 & 2 & -2 & 0 \\
0 &1 & 1 & 0 \\
0 & 1 & 2 & 0
\end{array} \right]\\[6pt] \xrightarrow[R_3-R_2]{R_1-2R_2}
\left[\begin{array}{rrr|r}
1 & 0 & -4 & 0 \\
0 &1 & 1 & 0 \\
0 & 0 & 1 & 0
\end{array} \right] \xrightarrow[R_2-R_3]{R_1+4R_3}
\left[\begin{array}{rrr|r}
1 & 0 & 0 & 0 \\
0 &1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{array} \right].
\end{align*}
It follows that the solution is $x_1=x_2=x_3=0$.
Hence $S$ is linearly independent.
As $S$ consists of three linearly independent vectors in $\R^3$, it must be a basis of $\R^3$.

(b) $S=\left\{\, \begin{bmatrix}
1 \\
4 \\
7
\end{bmatrix}, \begin{bmatrix}
2 \\
5 \\
8
\end{bmatrix}, \begin{bmatrix}
3 \\
6 \\
9
\end{bmatrix} \,\right\}$

As in part (a), we determine whether the set $S$ is linearly independent or not by considering the following augmented matrix:
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 0 \\
4 &5 & 6 & 0 \\
7 & 8 & 9 & 0
\end{array} \right] \xrightarrow[R_3-7R_1]{R_2-4R_1}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 0 \\
0 &-3 & -6 & 0 \\
0 & -6 & -12 & 0
\end{array} \right]\\[6pt] \xrightarrow{-\frac{1}{3}R_2}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 0 \\
0 & 1 & 2& 0 \\
0 & -6 & -12 & 0
\end{array} \right] \xrightarrow[R_3+6R_2]{R_1-2R_2}
\left[\begin{array}{rrr|r}
1 & 0 & -1 & 0 \\
0 &1 & 2 & 0 \\
0 & 0 & 0 & 0
\end{array} \right].
\end{align*}

Thus, the general solution is $x_1=x_3$, $x_2=-2x_3$, where $x_3$ is a free variable.
Hence, in particular, there is a nonzero solution.
So $S$ is linearly dependent, and hence $S$ cannot be a basis for $\R^3$.

(c) $S=\left\{\, \begin{bmatrix}
1 \\
1 \\
2
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
7
\end{bmatrix} \,\right\}$

A quick solution is to note that any basis of $\R^3$ must consist of three vectors. Thus $S$ cannot be a basis as $S$ contains only two vectors.

Another solution is to describe the span $\Span(S)$.
Note that a vector $\mathbf{v}=\begin{bmatrix}
a \\
b \\
c
\end{bmatrix}$ is in $\Span(S)$ if and only if $\mathbf{v}$ is a linear combination of vectors in $S$.
Equivalently, the vector $\mathbf{v}$ is in $\Span(S)$ if and only if the system
\[\begin{bmatrix}
1 & 0 \\
1 & 1 \\
2 &7
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
=
\begin{bmatrix}
a \\
b \\
c
\end{bmatrix}\] is consistent.

Let us consider the augmented matrix and reduce it by elementary row operations.
\begin{align*}
\left[\begin{array}{rr|r}
1 & 0 & a \\
1 & 1 &b \\
2 & 7 & c
\end{array}\right] \xrightarrow[R_3-2R_1]{R_2-R_1}
\left[\begin{array}{rr|r}
1 & 0 & a \\
0 & 1 &b-a \\
0 & 7 & c-2a
\end{array}\right]\\[6pt] \xrightarrow{R_3-7R_2}
\left[\begin{array}{rr|r}
1 & 0 & a \\
0 & 1 &b-a \\
0 & 0 & 5a-7b+c
\end{array}\right].
\end{align*}
Note that we obtained the $(3,3)$-entry by $c-2a-7(b-a)=5a-7b+c$.
It follows that the system is consistent if and only if
\[5a-7b+c=0.\] Thus, for example, the vector $\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}$ is not in $\Span(S)$ as $5\cdot 1-7\cdot 0+0\neq 0$.
Hence $\Span(S)$ is not $\R^3$, and we conclude that $S$ is not a basis.

(d) $S=\left\{\, \begin{bmatrix}
1 \\
2 \\
5
\end{bmatrix}, \begin{bmatrix}
7 \\
4 \\
0
\end{bmatrix}, \begin{bmatrix}
3 \\
8 \\
6
\end{bmatrix}, \begin{bmatrix}
-1 \\
9 \\
10
\end{bmatrix} \,\right\}$

The set $S$ contains four $3$-dimensional vectors. Hence $S$ is linearly dependent, and thus $S$ is not a basis.

Let $V$ be a subset of $\R^4$ consisting of vectors that are perpendicular to vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$, where
\[\mathbf{a}=\begin{bmatrix}
1 \\
0 \\
1 \\
0
\end{bmatrix}, \quad \mathbf{b}=\begin{bmatrix}
1 \\
1 \\
0 \\
0
\end{bmatrix}, \quad \mathbf{c}=\begin{bmatrix}
0 \\
1 \\
-1 \\
0
\end{bmatrix}.\]

Namely,
\[V=\{\mathbf{x}\in \R^4 \mid \mathbf{a}^{\trans}\mathbf{x}=0, \mathbf{b}^{\trans}\mathbf{x}=0, \text{ and } \mathbf{c}^{\trans}\mathbf{x}=0\}.\]

(a) Prove that $V$ is a subspace of $\R^4$.

(b) Find a basis of $V$.

(c) Determine the dimension of $V$.

Proof.

(a) Prove that $V$ is a subspace of $\R^4$.

Observe that the conditions
\[\mathbf{a}^{\trans}\mathbf{x}=0, \mathbf{b}^{\trans}\mathbf{x}=0, \text{ and } \mathbf{c}^{\trans}\mathbf{x}=0\] can be combined into the following matrix equation
\[A\mathbf{x}=\mathbf{0},\] where
\[A=\begin{bmatrix}
1 & 0 & 1 & 0 \\
1 &1 & 0 & 0 \\
0 & 1 & -1 & 0
\end{bmatrix}\] and $\mathbf{0}$ is the three dimensional zero vector.
Note that the rows of the matrix $A$ are $\mathbf{a}^{\trans}$, $\mathbf{b}^{\trans}$, and $\mathbf{c}^{\trans}$.
It follows that the subset $V$ is the null space $\calN(A)$ of the matrix $A$.
Being the null space, $V=\calN(A)$ is a subspace of $\R^4$.
(See the post “The Null Space (the Kernel) of a Matrix is a Subspace of $\R^n$“.)

(b) Find a basis of $V$.

In the proof of Part (a), we saw that $V=\calN(A)$.
To find a basis, we determine the solutions of $A\mathbf{x}=\mathbf{0}$.
Applying elementary row operations to the augmented matrix, we see that
\begin{align*}
\left[\begin{array}{rrrr|r}
1 & 0 & 1 & 0 &0 \\
1 & 1 & 0 & 0 & 0 \\
0 & 1 & -1 & 0 & 0
\end{array}\right] \xrightarrow{R_2-R_1}
\left[\begin{array}{rrrr|r}
1 & 0 & 1 & 0 &0 \\
0 & 1 & -1 & 0 & 0 \\
0 & 1 & -1 & 0 & 0
\end{array}\right]\\[6pt] \xrightarrow{R_3-R_2}
\left[\begin{array}{rrrr|r}
1 & 0 & 1 & 0 &0 \\
0 & 1 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{array}\right].
\end{align*}
It follows that the general solution is given by
\[x_1=-x_3, x_2=x_3.\] The vector form solution is
\[\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}=\begin{bmatrix}
-x_3 \\
x_3 \\
x_3 \\
x_4
\end{bmatrix}=x_3\begin{bmatrix}
-1 \\
1 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
0 \\
0 \\
0 \\
1
\end{bmatrix}.\]

Hence we have
\begin{align*}
V&=\calN(A)\\
&=\left\{\, \mathbf{x}\in \R^4 \quad \middle|\quad \mathbf{x}=x_3\begin{bmatrix}
-1 \\
1 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
0 \\
0 \\
0 \\
1
\end{bmatrix}, \text{ where $x_3, x_4\in \R$} \,\right\}\\[6pt] &=\Span \left\{\, \begin{bmatrix}
-1 \\
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
0 \\
0 \\
1
\end{bmatrix} \,\right\}.
\end{align*}
Let $B:=\left\{\, \begin{bmatrix}
-1 \\
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
0 \\
0 \\
1
\end{bmatrix} \,\right\}$.
Then we just showed that $B$ is a spanning set for $V$.
It is straightforward to see that $B$ is linearly independent.
Hence $B$ is a basis for $V$.

(c) Determine the dimension of $V$.

As the basis $B$ for $V$ that we obtained in Part (b) consists of two vectors, the dimension of the subspace $V$ is $\dim(V)=2$.