**Definition**. A subset $U$ of a vector space $\mathbb{V}$ is called a **subspace** of $\mathbb{V}$ if it has the following three properties

(1) $U$ contains the zero vector in $\mathbb{V}$,

(1) $U$ is closed under addition: if ${u}$ and ${v}$ are in $U$ then so is ${u}+{v}$, and

(3) $U$ is closed under scalar multiplication: if ${v}$ is in $U$ and $a$ is any scalar, then $a{v}$ is in $U.$

**Example**. Let $U$ be the subset of $\mathbb{R}^5$ defined by $$ U=\{(x_1,x_2,x_3,x_4,x_5)\in\mathbb{R}^5 \, \mid \,x_1=3x_2 \text{ and } x_3=7x_4\}. $$ Show $U$ is a subspace of $\mathbb{R}^5.$

The zero vector ${0}=(0,0,0,0,0,0)$ is in $U$ since $0=3(0)$ and $0=7(0).$ Let ${u}=(u_1, u_2, u_3, u_4, u_5)$ and ${v}=(v_1, v_2, v_3, v_4, v_5)$ be vectors in $U$ and let $a$ be a scalar. Then ${u}+{v}$ is in $U$ since $$ u_1=3 u_2 \text{ and } v_1=3 v_2 \text{ imply } u_1+v_1=3 (u_2+v_2)$$ and $$ u_3=7 u_4 \text{ and } v_3=7 v_4 \text{ imply } u_3+v_3=7 (u_4+v_4).$$ Also, $a {u}$ is in $U$ since $u_1=3u_2$ implies $a u_1=3 (a u_2).$ Hence we have, $U$ is a subspace of $\mathbb{R}^5.$

**Example**. Give an example of a nonempty subset $U$ of $\mathbb{R}^2$ such that $U$ is closed under addition and under taking additive inverses, but $U$ is not a subspace of $\mathbb{R}^2.$ The subset $\mathbb{Z}^2$ of $\mathbb{R}^2$ is closed under additive inverses and addition, however $\mathbb{Z}^2$ is not a subspace of $\mathbb{R}^2$ since $\sqrt{2} \in \mathbb{R}, (1,1)\in \mathbb{Z}^2$ however $(\sqrt{2} , \sqrt{2}) \not \in \mathbb{Z}^2.$

**Example**. Give an example of a nonempty subset $U$ of $\mathbb{R}^2$ such that $U$ is closed under scalar multiplication, but $U$ is not a subspace of $\mathbb{R}^2.$ The set $\{(x_1,x_2)\in \mathbb{R}^2 \mid x_1 x_2=0\}=M$ is closed under scalar multiplication because, if $\lambda\in \mathbb{R}$ and $(x_1,x_2)\in M$, then $(\lambda x_1, \lambda x_2)\in M$ holds since $\lambda x_1 \lambda x_2=0.$ However, $M$ is not a subspace because $(0,1)+(1,0)=(1,1)\not \in M$ even though $(0,1),(1,0)\in M.$

**Example**. Show that the set of all solutions of an $m\times n$ homogenous linear system of equations is a subspace of $\mathbb{V}$ (called the **null space**).

Let $A{x}={0}$ be an $m\times n$ homogenous system of linear equations and let $U$ be the set of solutions to this system. Of course $A{0}={0}$ and so the zero vector is in $U.$ Let ${u}$ and ${v}$ be in $U$ and let $a$ be a scalar. Then $$A({u}+{v})=A{u}+A{v} ={0}+{0}={0}$$ and $$ A(a {u})=a(A{u})=a{0}={0} $$ shows ${u}+{v}$ and $a{u}$ are in $U.$ Hence, $U$ is a subspace of $\mathbb{V}.$

**Definition**. Let ${v}_1, {v}_2, \ldots, {v}_m$ be vectors in the vector space $\mathbb{V}.$ The set of all linear combinations $$ \mathop{span}(v_1, v_2, \ldots, v_m ) =\left\{ c_1 v_1 + c_2 v_2 + \cdots + c_m v_m \mid c_1, c_2, \ldots, c_m\in k \right\}$$ is called the **spanning set** of the vectors $v_1, v_2, \ldots, v_m .$

**Example**. Show that the spanning set of the vectors $v_1, v_2, \ldots, v_m $ in $\mathbb{V}$ is a subspace of $\mathbb{V}.$ Let $U=\mathop{span}(v_1, v_2, \ldots, v_m ).$ Notice ${0}\in U$ since ${0}=0 v_1 + 0 v_2 + \cdots + 0 v_m $ where $0\in k.$ Let ${u}$ and ${v}$ be vectors in $U$ and let $a$ be a scalar. From above, there exists scalars $c_1, c_2, \ldots, c_m$ and scalars $d_1, d_2, \ldots, d_m$ such that $$ {u} =c_1 v_1 + c_2 v_2 + \cdots + c_m v_m \quad \text{ and } \quad {v} =d_1 v_1 + d_2 v_2 + \cdots + d_m v_m $$ Then $$ {u}+{v}=\sum_{i=1}^m c_i {v}_i + \sum{i=1}^m d_i {v}_i =\sum{i=1}^m (c_i+d_i) {v}_i $$ and $$ a{u}=a\left(\sum{i=1}^m c_i {v}_i\right)= \sum{i=1}^m (a c_i) {v}_i

$$ show ${u}+{v}$ and $a{u}$ are in $U$; and thus $U$ is a subspace of $\mathbb{V}.$

We say that a nonzero ${v}_i$ is **redundant** in the list ${v}_1 , \ldots, {v}_i, \ldots, {v}_m$ if $v_i$ can be written as a linear combination of the other nonzero vectors in the list. An equation of the form $a_1 {v}_1 + \cdots +a_m {v}_m = {0}$ is called a **linear relation** among the vectors ${v}_1 , \ldots, {v}_m$; and is called a **nontrivial relation** if at least one of the $a_i$’s is nonzero.

**Definition**. The vectors $v_1, v_2, \ldots, v_m $ in $\mathbb{V}$ are called **linear independent** if the only choice for $ a_1 v_1 + a_2 v_2 + \cdots + a_m v_m = {0} $ is $a_1 = a_2=\cdots =a_m = 0.$ Otherwise the vectors $v_1, v_2, \ldots, v_m $ are called **linear dependent**.

**Lemma**. Show that any set $S={v_1, v_2, \ldots, v_m }$ of vectors in $\mathbb{V}$ is a linearly dependent set of vectors if and only if at least one of the vectors in the set can be written as a linear combination of the others.

**Proof**. Assume the vectors in the set $S$ are linearly dependent. There exists scalars $c_1, c_2, \ldots, c_m$ (not all zero) such that $c_1 v_1 + c_2 v_2 + \cdots + c_m v_m ={0}.$ Let $i$ be the least index such that $c_i$ is nonzero. Thus $c_1=c_2=\cdots =c_{i-1}=0.$ So $c_i {v}_i=-c_{i+1}{v}_{i+1}-\cdots -c_m {v}_m$ for some $i.$ Since $c_i\neq 0$ and $c_i\in k$, $c_i^{-1}$ exists and thus $${v}_i =\left(\frac{-c{i+1}}{c_i}\right){v}_{i+1} \cdots + \left(\frac{-c_m}{c_i}\right){v}_m $$ which shows ${v}_i$ is a linear combination of the others, via $$ {v}_i= 0{v}_1+\cdots+0{v}{i-1}+\left(\frac{-c_{i+1}}{c_i}\right){v}_{i+1} \cdots + \left(\frac{-c_m}{c_i}\right){v}_m. $$ Now assume one of the vectors in the set $S$ can be written as a linear combination of the others, say $$ {v}_k =c_1 {v}_1+\cdots +c_{k-1}{v}_{k-1} +c_{k+1} {v}_{k+1}

+\cdots c_m {v}_m $$ where $c_1, c_2, \ldots, c_m$ are scalars. Thus, $$

{0}=c_1 {v}_1+\cdots c_{k-1} {v}_{k-1} +(-1) {v}_k +c_{k+1} {v}_{k+1} +\cdots + c_m {v}_m $$ and so, ${v}_1, \ldots, {v}_m$ are linearly dependent.

For a list of vectors $v_1, v_2, \ldots, v_m $ in $\mathbb{V}$ the following equivalent statements follow from the appropriate definitions:

- vectors $v_1, v_2, \ldots, v_m $ are linearly independent,
- none of the vectors $v_1, v_2, \ldots, v_m $ are redundant,
- none of the vectors $v_1, v_2, \ldots, v_m $ can be written as a linear combination of the other vectors in the list,
- there is only the trivial relation among the vectors $v_1, v_2, \ldots, v_m $,
- the only solution to the equation $a_1 v_1 + a_2 v_2 + \cdots + a_m v_m =0$ is $a_1 = a_2=\cdots =a_m= 0$, and
- $\mathop{rank}(A)=n$

where $A$ is the $n\times m$ matrix whose columns are the vectors $v_1, v_2, \ldots, v_m .$

**Example**. Determine whether the following vectors ${u}$, ${v}$, and ${w}$ are linearly independent. $$ {u}=\begin{bmatrix}1 \\ 1 \\ 1 \\ 1\end{bmatrix} \qquad {v}=\begin{bmatrix}1 \\ 2 \\ 3 \\ 4\end{bmatrix} \qquad {w}=\begin{bmatrix}1 \\ 4 \\ 7 \\ 10\end{bmatrix} $$ Without interchanging rows, we use elementary row operations to find $$ \mathop{rref} \begin{bmatrix} {u} & {v} & {w} \end{bmatrix} =\begin{bmatrix}1 & 0 & -2 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. $$ From this we infer the nontrivial relation $0=(-2){u}+(3){v}+(-1){w}.$ Therefore the given vectors are linearly dependent.

**Theorem**. Let $$ {v}_1=\begin{bmatrix}a_{11} \\ a_{21} \\ \vdots \\ a_{n1}\end{bmatrix} \quad {v}_2=\begin{bmatrix}a_{12} \\ a_{22} \\ \vdots \\ a_{n2}\end{bmatrix} \quad \cdots \quad {v}_m =\begin{bmatrix}a_{1s} \\ a_{2m} \\ \vdots \\ a_{nm}\end{bmatrix} $$ be $s$ vectors in $\mathbb{V}.$ These vectors are linearly dependent if and only if there exists a solution to the system of linear equations \begin{equation*} \label{lincomsys} \begin{cases} a_{11}x_1+a_{12}x_2+\cdots+a_{1m}x_m=0 \\ a_{21}x_1+a_{22}x_2+\cdots+a_{2m}x_m=0 \\ \qquad \qquad \vdots \\ a_{n1}x_1+a_{n2}x_2+\cdots+a_{nm}x_m=0 \\ \end{cases} \end{equation*} different from $x_1=x_2=\cdots=x_m=0.$

**Proof**. Assume $v_1, v_2, \ldots, v_m $ are linear dependent. There exists scalars $c_1, c_2, \ldots, c_m$ such that \begin{equation} \label{lincomeq} c_1 v_1 + c_2 v_2 + \cdots + c_m v_m ={0} \end{equation} and not all $c_i$’s are zero. Thus we have a system $$ \begin{cases} a_{11}c_1+a_{12}c_2+\cdots+a_{1m}c_m=0 \\ a_{21}c_1+a_{22}c_2+\cdots+a_{2m}c_m=0 \\ \qquad \qquad \vdots \\ a_{n1}c_1+a_{n2}c_2+\cdots+a_{nm}c_m=0 \\ \end{cases} $$ with solution $x_1=c_1$, $x_2=c_2, \ldots, x_m=c_m.$ Since not all $c_i$’s are zero we have a solution different from $x_1=x_2=\cdots=x_m=0.$ Assume the system has a solution ${x}$ with ${x} \neq {0}$, say $x_i.$ We can write \begin{equation*} {0}=A{x}= x_1 v_1 + x_2 v_2 + \cdots + x_m v_m \end{equation*} Isolating the term $x_i {v}_i$ yields \begin{equation*} x_i {v}_i=-x_i {v}_1-\cdots -x_{i-1}{v}_{i-1}-x_{i+1}{v}_{i+1} -\cdots -x_m {v}_m \end{equation*} and since $x_i \neq 0$, $(x_i)^{-1}$ exists. Therefore, \begin{equation*} {v}_1=\left(-\frac{x_1}{x_i}\right){v}_1-\cdots -\left(-\frac{x_{i-1}}{x_i}\right){v}_{i+1}-\cdots – \left(-\frac{x_{m}}{x_i}\right){v}_{m} \end{equation*} shows the vectors $v_1, v_2, \ldots, v_m $ are linearly dependent.

**Theorem**. The $n\times m$ linear system of equations $A{x}={b}$ has a solution if and only if the vector ${b}$ is contained in the subspace of $\mathbb{V}$ generated by the column vectors of $A.$

**Proof**. Let ${x}$ be a solution to $A{x}={b}$ with $A=\begin{bmatrix}{v}_1 & {v}_2 & \cdots & {v}_m \end{bmatrix}.$ Thus ${b}=A{x}=x_1 v_1 + x_2 v_2 + \cdots + x_m v_m $ and thus ${b}\in \mathop{span}(v_1, v_2, \ldots, v_m )$ as needed. Conversely, assume ${b}$ is in the subspace generated by the column vectors of $A$; that is assume ${b}\in \mathop{span}(v_1, v_2, \ldots, v_m ).$ There exists scalars $c_1, c_2, \ldots, c_m$ such that ${b}=c_1 v_1 + c_2 v_2 + \cdots + c_m v_m .$ Thus, ${b}=c_1 v_1 + c_2 v_2 + \cdots + c_m v_m =A{c}$ where the components of ${c}$ are the $c_i$’s. Thus the system $A{x}={b}$ has a solution, namely ${c}.$

**Example**. Let $U$ and $\mathbb{V}$ be finite subsets of a vector space $\mathbb{V}$ with $U\subseteq V.$

- If $U$ is linear dependent, then so is $\mathbb{V}.$
- If $\mathbb{V}$ is linear independent, then so is $U.$

Let $U=\{u_1, u_2, \ldots, u_s\}$ and $V=\{v_1, v_2, \ldots, v_t\}.$

- If $U$ is linear dependent, then thee exists a vector, say ${u}_k$ such that ${u}_k$ is a linear combination of the other ${u}_i$’s. Since $U\subseteq V$ all ${u}_i$’s are in $\mathbb{V}.$ Thus we have a vector ${u}_k$ in $\mathbb{V}$ that is a linear combination of other vectors in $\mathbb{V}.$ Therefore, $\mathbb{V}$ is linear dependent.
- Let $c_1, c_2, \ldots, c_s$ be scalars such that \begin{equation} \label{lincombcus} c_1 u_1 + c_2 u_2 + \cdots + c_s u_s ={0}. \end{equation} Since $U\subseteq V$, we know $u_i\in V$ for $1\leq i \leq s.$ Since $\mathbb{V}$ is linear independent, $c_1=c_2=\cdots =c_m=0.$ Thus $U$ is linear independent as well.

**Corollary**. Any vector in $\mathbb{V}$, written as a column matrix, can be expressed (uniquely) as a linear combination of ${v}_1, {v}_2, \ldots, {v}_m$ if and only if $A {x}={v}$ has unique solution, where $A=\begin{bmatrix}{v}_1 & {v}_2 & \cdots & {v}_m\end{bmatrix}.$ When there is a solution, the components $x_1, x_2, \ldots, x_m$ of ${x}$ give the coefficients for the linear combination.

**Theorem**. The vectors $v_1, v_2, \ldots, v_n $ in $\mathbb{V}$ form a linearly independent set of vectors if and only if $\begin{bmatrix}{v}_1& {v}_2 & \cdots & {v}_n \end{bmatrix}$ is row equivalent to $I_n.$

**Theorem**. Let $\mathbb{V}$ be a vector space and assume that the vectors $v_1, v_2, \ldots, v_n $ are linearly independent and $\mathop{span}(s_1, s_2, \ldots, s_m )=V.$ Then $n\leq m.$

**Proof**. We are given $$ \mathop{span}(s_1, s_2, \ldots, s_m )=V

\quad \text{and} \quad v_1, v_2, \ldots, v_n \text{ are linearly independent.} $$ Since ${v}_1$ is a linear combination of the vectors ${s}_1$, ${s}_2$, \ldots, ${s}_m$ we obtain $$ \mathop{span}({v}_1,{s}_2,\ldots,{s}_m)=V \quad \text{and} \quad {v}_2, \ldots, {v}_n \text{ are linearly independent,} $$ respectively. Since ${v}_2$ is a linear combination of ${v}_1$, ${s}_2, \ldots, {s}_m$ we obtain $$ \mathop{span}({v}_1,{v}_2,{s}_3,\ldots,{s}_m)=V \quad \text{and} \quad {v}_3, \ldots, {v}_n \text{ are linearly independent,} $$ respectively. Now if $$v_{m+1}, \ldots, {v}_n \text{ are linearly independent.} $$ This is a contradiction since ${v}_n$ is not in $\mathop{span}(v_1, v_2, \ldots, v_m)$; and whence $n\leq m.$

**Theorem**. A set $S=\{v_1, v_2, \ldots, v_m \}$ of vectors in $\mathbb{V}$ is linearly independent if and only if for any vector ${u}$, if ${u} = u_1 v_1 + u_2 v_2 + \cdots + u_m v_m $, then this representation is unique.

**Proof**. Assume the vectors in $S$ are linearly independent and assume ${u}$ is an arbitrary vector with \begin{equation*} {u}=a_1 v_1 + a_2 v_2 + \cdots + a_m v_m \qquad \text{and} \qquad {u}=b_1 v_1 + b_2 v_2 + \cdots + b_m v_m \end{equation*} as both representations of ${u}$ as linear combinations of the vectors in $S.$ Then $$ {0}={u}-{u} =(a_1-b_1){v}_1+(a_2-b_2){v}_2+\cdots +(a_m-b_m){v}_m. $$ Since $S$ is linearly independent $a_1-b_1=a_2-b_2=\cdots =a_m-b_m=0$ and thus $a_1=b_1$, $a_2=b_2$, \ldots, $a_m=b_m.$ Therefore, the representation of ${u}$ as a linear combination of the vectors in $S$ is unique. Conversely, assume for nay vector ${u}$ which can be written as a linear combination of the vectors in $S$, the representation is unique. If $c_1, c_2, \ldots, c_m$ are scalars such that $c_1 v_1 + c_2 v_2 + \cdots + c_m v_m ={0}$ then $c_1=c_2=\cdots =c_m=0$ much hold since $0{v}_1+0{v}_2+\cdots +0{v}_m={0}$ and this representation is unique. Therefore, the vectors in $S$ are linearly independent.

**Definition**. The vectors $v_1, v_2, \ldots, v_m $ in $\mathbb{V}$ are called a **basis** of a linear subspace $\mathbb{V}$ if they span $\mathbb{V}$ and are linearly independent.

**Example**. Find a basis for $\mathbb{V}$ for $n=1,2,3,\ldots,m.$ For $n=2$, the vectors $\begin{bmatrix}1 \\ 0\end{bmatrix}$, $\begin{bmatrix}0 \\ 1\end{bmatrix}$ form a basis for $k^2.$ For $n=3$, the vectors $\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$, $\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}$, $\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$ form a basis for $k^3.$ In general, for a positive integer $n$, the following $n$ vectors of $\mathbb{V}$ form a basis (called the **standard basis**) of $\mathbb{V}.$ \begin{equation} \label{stba} {e}_1=\begin{bmatrix}1 \\ 0 \\ \vdots \\ 0\end{bmatrix} \qquad {e}_2=\begin{bmatrix}0 \\ 1 \\ \vdots \\ 0\end{bmatrix} \qquad \cdots \qquad {e}_n=\begin{bmatrix}0 \\ 0 \\ \vdots \\ 1\end{bmatrix} \end{equation} The vectors in a standard basis are linearly independent. Given any vector ${v}$ in $\mathbb{V}$ with components $v_i$, we can write $$ {v}=v_1 e_1 + v_2 e_2 + \cdots + v_n e_n , $$ and thus $k^n=\mathop{span}(e_1, e_2, \ldots, e_n )$ which shows that any standard basis is in fact a basis.

**Example**. Show the following vectors ${v}_1$, ${v}_2$, ${v}_3$, and ${v}_4$ form a basis for $\mathbb{R}^4.$ $$ v_1=\begin{bmatrix} 1 \\ 1\\ 1 \\ 1 \end{bmatrix} \qquad v_2=\begin{bmatrix} 1 \\ -1\\ 1 \\ -1 \end{bmatrix} \qquad v_3=\begin{bmatrix} 1 \\ 2 \\ 4 \\ 8 \end{bmatrix}\qquad v_4=\begin{bmatrix} 1 \\ -2 \\ 4 \\ -8 \end{bmatrix} $$ We determine $\mathop{rref}(A)=I_4$ where $A$ is the matrix with column vectors $v_1, v_2, v_3, v_4.$ Thus we have, $v_1, v_2, v_3, v_4$ are linearly independent. Since $v_1, v_2, v_3, v_4$ also span $\mathbb{R}^4$, they form a basis of $\mathbb{R}^4.$

**Example**. Let $U$ be the subspace of $\mathbb{R}^5$ defined by $$

U=\{(x_1,x_2,x_3,x_4,x_5)\in\mathbb{R}^5 \mid x_1=3x_2 \text{ and } x_3=7x_4\}. $$ Find a basis of $U.$ The following vectors belong to $U$ and are linearly independent in $\mathbb{R}^5.$ $$ v_1=\begin{bmatrix}3 \\ 1 \\ 0 \\ 0 \\ 0\end{bmatrix} \qquad v_2=\begin{bmatrix} 0 \\ 0 \\ 7 \\ 1 \\ 0\end{bmatrix} \qquad v_3=\begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 1\end{bmatrix} $$ If $u\in U$, then the representation $$ u =\begin{bmatrix}u_1 \\ u_2 \\ u_3 \\ u_4 \\ u_5\end{bmatrix} =\begin{bmatrix}3u_2 \\ u_2 \\ 7u_4 \\ u_4 \\ u_5\end{bmatrix} =u_2\begin{bmatrix}3 \\ 1 \\ 0 \\ 0 \\ 0\end{bmatrix} + u_4\begin{bmatrix}0 \\ 0 \\ 7 \\ 1 \\ 0\end{bmatrix} + u_5\begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 1\end{bmatrix} $$ shows that they also span $U$, and thus form a basis of $U$.

**Theorem**. Let $S=\{v_1, v_2, \ldots, v_n \}$ be a set of vectors in a vector space $\mathbb{V}$ and let $W=\mathop{span}(S).$ Then some subset of $S$ is a basis for $W.$

**Proof**. Assume $W=\mathop{span}(S)$ and suppose $S$ is a linearly independent set of vectors. Thus, in this case, $S$ is a basis of $W$. So we can assume $S$ is a linearly dependent set of vectors. There exists $i$ such that $1\leq i \leq m$ and ${v}_i$ is a linear combination of the other vectors in $S.$ It is left for an exercise to show that $$ W=\mathop{span}(S)=\mathop{span}(S_1) $$ where $S_1=S/{{v}_i}.$ If $S_1$ is linearly independent set of vectors, then $S_1$ is a basis of $W.$ Otherwise, $S_1$ is a linear dependent set and we can delete a vector from $S_1$ that is a linear combination of the other vectors in $S_1.$ We obtain another subset $S_2$ of $S$ with $$ W=\mathop{span}(S)=\mathop{span}(S_1)=\mathop{span}(S_2). $$ Since $S$ is finite, if we continue, we find a linearly independent subset of $S$ and thus a basis of $W.$

**Corollary**. All bases of a subspace $U$ of a vector space $\mathbb{V}$ consists of the same number of vectors.

**Proof**. Let $S=\{v_1, v_2, \ldots, v_n\}$ and $T=\{w_1, w_2, \ldots, w_m\}$ be bases of a subspace $U.$ Then $\mathop{span}(S)=V$ and $T$ is a lineal independent set of vectors. Hence, $m\leq n.$ Similarly, since $\mathop{span}(T)=V$ and $S$ is a linearly independent set of vectors, $n\leq m.$ Therefore, $m=n$ as desired.

**Corollary**. The vectors ${v}_1, {v}_2, \ldots, {v}_n$ form a basis of $\mathbb{V}$ if and only if the reduced row echelon form of the $n\times n$ matrix $\begin{bmatrix}{v}_1& {v}_2 & \cdots & {v}_n \end{bmatrix}$ is $I_n.$

**Proof**. Suppose the vectors $v_1, v_2, \ldots, v_n $ form a basis of $\mathbb{V}$ and consider the $n\times n$ linear system $$ \begin{cases} v_{11} x_1+v_{12} x_2+\cdots +v_{1n} x_n=0 \\ v_{21} x_1+v_{22} x_2+\cdots +v_{2n} x_n=0 \\ \qquad \qquad \vdots \\ v_{n1} x_1+v_{n2} x_2+\cdots +v_{nn} x_n=0 \end{cases} $$ where the $v_{ij}$’s are the components of the ${v}_j$’s. Since ${v_1, v_2, \ldots, v_n }$ is a basis, ${v}_1={v}_2=\cdots ={v}_n={0}$ and this linear system can not have another solution. Therefore we have, $\mathop{rref}(A)=I_n$ where $A=\begin{bmatrix}{v}_1& {v}_2 & \cdots & {v}_n \end{bmatrix}.$

**Definition**. The number of vectors in a basis of a subspace $U$ of $\mathbb{V}$ is called the **dimension** of $U$, and is denoted by $\mathop{dim} U.$

**Example**. Find a basis of the subspace of $\mathbb{R}^4$ that consists of all vectors perpendicular to both of the following vectors ${v}_1$ and $v_2.$ $$ v_1=\begin{bmatrix}1 \\ 0 \\ -1 \\ 1\end{bmatrix} \qquad v_2=\begin{bmatrix}0 \\ 1 \\ 2 \\ 3\end{bmatrix} $$ We need to find all vectors $x$ in $\mathbb{R}^4$ such that $x \cdot v_1=0$ and $x \cdot v_2=0.$ We solve both $$ \begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{bmatrix} \cdot \begin{bmatrix}1 \\ 0 \\ -1 \\ 1\end{bmatrix}=0 \qquad \text{and} \qquad \begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{bmatrix} \cdot \begin{bmatrix}0 \\ 1 \\ 2 \\ 3\end{bmatrix}=0 $$ which leads to the system and matrix $$ \begin{cases} x_1-x_3+x_4 =0 \\ x_2+2x_3+3x_4 =0 \end{cases} \qquad \text{and} \qquad A= \begin{bmatrix} 1 & 0 & -1 & 1 \\ 0 & 1 & 2 & 3 \end{bmatrix}. $$ All solutions are given by $$ \begin{bmatrix}x_1 \\ x_2 \\ x_3 \ x_4\end{bmatrix} =t \begin{bmatrix}-1 \\ -3 \\ 0 \\ 1\end{bmatrix} + u\begin{bmatrix}1 \\ -2 \\ 1 \\ 0\end{bmatrix} \quad \text{ where $t, u\in\mathbb{R}$}. $$ It follows the vectors $\begin{bmatrix}-1 \\ -3 \\ 0 \\ 1\end{bmatrix}$, $\begin{bmatrix}1 \\ -2 \\ 1 \\ 0\end{bmatrix}$ form a basis of the desired subspace.

**Theorem**. Let $U$ be a subspace of $k^m$ with $\dim U=n$, then

(1) any list of linearly independent vectors contains $n$ elements,

(2) any list of vectors that spans $U$ contains at least $n$ elements,

(3) if $n$ vectors are linearly independent then they form a basis, and

(4) if $n$ vectors span $U$, then they form a basis of $U.$

**Example**. Determine the values of $a$ for which the following vectors ${u}_1$, ${u}_2$, ${u}_3$, and ${u}_4$ form a basis of $\mathbb{R}^4.$ $$ {u}_1=\begin{bmatrix}1 \\ 0 \\ 0 \\ 4\end{bmatrix} \qquad {u}_2=\begin{bmatrix}0 \\ 1 \\ 0 \\ 6\end{bmatrix} \qquad {u}_3=\begin{bmatrix}0 \\ 0 \\ 1 \\ 8\end{bmatrix} \qquad {u}_4=\begin{bmatrix}4 \\ 5 \\ 6 \\ a\end{bmatrix} $$ Let $A=\begin{bmatrix}{u}_1 & {u}_2 & {u}_3& {u}_4\end{bmatrix}.$ Using row operations we find the row-echelon form of $A$ to be the following matrix. $$ \begin{bmatrix} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 6\\ 0 & 0 & 8 & a-94 \end{bmatrix} $$ Thus, $\mathop{rref}(A)=I_4$ if and only if $a=95.$ Therefore, $B={{u}_1, {u}_2, {u}_3, {u}_4}$ is a basis if and only if $a=95.$

**Theorem**. The dimension of the row space of a matrix $A$ is equal to the dimension of the column space of $A.$

Proof. The proof is left for the reader.

Read about Orthonormal Bases here.

## Exercises on Subspaces

**Exercise**. Determine whether the following collection of vectors in $\mathbb{R}^3$ are linearly independent or linearly dependent.

- $(0,1,1), (1,2,1), (0,4,6), (1,0,-1)$
- $(0,1,0), (1,2,1), (0,-4,6), (-1,1,-1)$

**Exercise**. Determine whether the following collection of vectors in $\mathbb{R}^4$ are linearly independent or linearly dependent.

- $(0,1,1,1), (1,2,1,1), (0,4,6,2), (1,0,-1, 2)$
- $(0,1,0,1), (1,2,1,3), (0,-4,6,-2), (-1,1,-1, 2)$

**Exercise**. Show that the given vectors do not form a basis for the vector space $\mathbb{V}.$

- $(21,-7), (-6, 1)$; $V=\mathbb{R}^2$
- $(21,-7,14), (-6, 1,-4), (1,0,0)$; $V=\mathbb{R}^3$
- $(48,24,108,-72), (-24, -12,-54,36), (1,0,0,0), (1,1,0,0)$; $V=\mathbb{R}^4$

**Exercise**. Reduce the vectors to a basis of the vector space $\mathbb{V}.$

- $(1,0), (1,2), (2,4)$, $V=\mathbb{R}^2$
- $(1,2,3), (-1, -10, 15), (1, 2, -3), (2,0,6), (1, -2, 3)$, $V=\mathbb{R}^3$

**Exercise**. Which of the following collection of vectors in $\mathbb{R}^3$ are linearly dependent? For those that are express one vector as a linear combination of the rest.

- $(1,1,0), (0,2,3), (1,2,3)$
- $(1,1,0), (3,4,2), (0,2,3)$

**Exercise**. Let $S=\{v_1, v_2, \ldots, v_k\}$ be a set of vectors in a a vector space $\mathbb{V}.$ Prove that $S$ is linearly dependent if and only if one of the vectors in $S$ is a linear combination of all other vectors in $S.$

**Exercise**. Suppose that $S=\{v_1, v_2, v_3\}$ is a linearly independent set of vector in a vector space $\mathbb{V}.$ Prove that $T=\{u_1, u_2, u_3\}$ is also linearly independent where $u_1=v_1$, $u_2=v_1+v_2$, and $u_3=v_1+v_2+v_3.$

**Exercise**. Which of the following sets of vectors form a basis for the vector space $\mathbb{V}.$

- $(1,3), (1,-1)$; $V=\mathbb{R}^2$
- $(1,3),(-2,6)$; $V=\mathbb{R}^2$
- $(3,2,2), (-1,2,1), (0,1,0)$; $V=\mathbb{R}^3$
- $(3,2,2), (-1,2,0), (1,1,0)$; $V=\mathbb{R}^3$
- $(2,2,2,2), (3,3,3,2), (1,0,0,0), (0,1,0,0)$; $V=\mathbb{R}^4$
- $(1,1,2,0), (2,2,4,0), (1,2,3,1), (2,1,3,-1), (1,2,3,-1) $; $V=\mathbb{R}^4$

**Exercise**. Find a basis for the subspace of the vector space $\mathbb{V}.$

- All vectors of the form $(a,b,c)$ where $b=a+c$ where $V=\mathbb{R}^3.$
- All vectors of the form $(a,b,c)$ where $b=a-c$ where $V=\mathbb{R}^3.$
- All vectors of the form $\begin{bmatrix}b-a \ a+c \ b+c \ c\end{bmatrix}$ where $V=\mathbb{R}^4.$

**Exercise**. Let ${v}_1=\begin{bmatrix}0 \ 1 \ 1\end{bmatrix}$, ${v}_2=\begin{bmatrix}1 \ 0 \ 0\end{bmatrix}$ and $S=\mathop{span}({v}_1,{v}_2).$

- Is $S$ a subspace of $\mathbb{R}^3$?
- Find a vector ${u}$ in $S$ other than ${v}_1$, ${v}_2.$
- Find scalars which verify that $3{u}$ is in $S.$
- Find scalars which verify that ${0}$ is in $S.$

**Exercise**. Let ${u}_1=\begin{bmatrix}0 \ 2 \ 2\end{bmatrix}$, ${u}_2=\begin{bmatrix}2 \ 0 \ 0\end{bmatrix}$ and $T=\mathop{span}({u}_1,{u}_2).$ Show $S=T$ by showing $S\subseteq T$ and $T\subseteq S$ where $S$ is defined above.

**Exercise**. Prove that the non-empty intersection of two subspaces of $\mathbb{R}^3$ is a subspace of $\mathbb{R}^3.$

**Exercise**. Let $S$ and $T$ be subspaces of $\mathbb{R}^3$ defined by $$ S=\mathop{span}\left(\begin{bmatrix}1 \\ 0 \\ 2\end{bmatrix}, \begin{bmatrix}0 \\ 2 \\ 1\end{bmatrix}\right) \qquad \text{and} \qquad

T=\mathop{span}\left(\begin{bmatrix}2 \\ -2 \\ 3\end{bmatrix}, \begin{bmatrix}3 \\ -4 \\ 4\end{bmatrix}\right). $$ Show they are the same subspace of $\mathbb{R}^3.$

**Exercise**. Let $\{{v}_1,{v}_2, {v}_3\}$ be a linearly independent set of vectors. Show that if ${v}_4$ is not a linear combination of ${v}_1, {v}_2, {v}_3$, then ${{v}_1,{v}_2, {v}_3},{v}_4$ is a linearly independent set of vectors.

**Exercise**. If $\{{v}_1,{v}_2, {v}_3\}$ is a linearly independent set of vectors in $\mathbb{V}$, show that $\{{v}_1,{v}_1+{v}_2, {v}_1+{v}_2+{v}_3\}$ is also a linearly independent set of vectors in $\mathbb{V}.$

**Exercise**. If $\{{v}_1,{v}_2, {v}_3\}$ is a linearly independent set of vectors in $\mathbb{V}$, show that $\{{v}_1 + {v}_2, {v}_2+{v}_3, {v}_3+{v}_1\}$ is also a linearly independent set of vectors in $\mathbb{V}.$

**Exercise**. Let ${{v}_1,{v}_2, {v}_3}$ be a linearly dependent set. Show that at least one of the ${v}_i$ is a linear combination of the others.

**Exercise**. Prove or provide a counterexample to the following statement. If a set of vectors $T$ spans the vector space $\mathbb{V}$, then $T$ is linearly independent.

**Exercise**. Which of the following are not a basis for $\mathbb{R}^3$?

$$ {v_1}=\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}, {v_2} = \begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}, {v_3}=\begin{bmatrix}1 \\ -1 \\ -1\end{bmatrix} $$

$$ {u_1}=\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}, {u_2}=\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}, {u_3}=\begin{bmatrix}2 \\ 3 \\ 4\end{bmatrix} $$

**Exercise**. Let $S$ be the space spanned by the vectors $$ {v}_1=\begin{bmatrix}1 \\ 0 \\ 1 \\ 1\end{bmatrix} \quad {v}_1=\begin{bmatrix}-1 \\ -3 \\ 1 \\ 0\end{bmatrix} \quad {v}_1=\begin{bmatrix}2 \\ 3 \\ 0 \\ 1\end{bmatrix} \quad {v}_1=\begin{bmatrix}2 \\ 0 \\ 2 \\ 2\end{bmatrix} $$ Find the dimension of $S$ and a subset of $T$ which could serve as a basis for $S.$

**Exercise**. Let $\{{v}_1, {v}_2, \ldots, {v}_n\}$ be a basis for $\mathbb{V}$, and suppose that ${u} =a_1 {v_1}+a_2 {v_2}+\cdots + a_n {v_n}$ with $a_1\neq 0.$ Prove that $\{{u}, {v}_2, \ldots, {v}_n\}$ is also a basis for $\mathbb{V}.$

**Exercise**. Let $S=\mathop{span}({v}_1,{v}_2,{v}_3)$ and $T=\mathop{span}({u}_1,{u}_2,{u}_3)$ where ${v}_i$ and ${u}_i$ are defined as follows.

$$ {v}_1=\begin{bmatrix}1 \\ -1 \\ 2 \\ 0\end{bmatrix} \quad {v}_2=\begin{bmatrix}2 \\ 1 \\ 1 \\ 1\end{bmatrix} \quad {v}_3=\begin{bmatrix}3 \\ -1 \\ 2 \\ -1\end{bmatrix}$$

$$ {u}_1=\begin{bmatrix}3 \\ 0 \\ 3 \\ 1\end{bmatrix} \quad {u}_2=\begin{bmatrix}1 \\ 2 \\ -1 \\ 1\end{bmatrix} \quad {u}_3=\begin{bmatrix}4 \\ -1 \\ 5 \\ 1\end{bmatrix} $$

Is one of these two subspaces strictly contained in the other or are they equal?

**Exercise**. Let $S=\mathop{span}({v}_1,{v}_2,{v}_3)$ where ${v}_i$ are defined as follows. $$ {v}_1=\begin{bmatrix}1 \\ 2 \\ 3 \\ 1\end{bmatrix} \qquad {v}_2=\begin{bmatrix}2 \\ -1 \\ 1 \\ -3\end{bmatrix} \qquad {v}_3=\begin{bmatrix}1 \\ 3 \\ 4 \\ 2\end{bmatrix} \qquad {u}=\begin{bmatrix}1 \\ 2 \\ 3 \\ 1\end{bmatrix} $$ Is the vector ${u}$ in $S$?

**Exercise**. If possible, find a value of $a$ so that the vectors $$ \begin{bmatrix}1 \\ 2 \\ a\end{bmatrix} \qquad \begin{bmatrix}0 \\ 1 \\ a-1\end{bmatrix} \qquad \begin{bmatrix}3 \\ 4 \\ 5\end{bmatrix} \qquad $$ are linearly independent.

**Exercise**. Let $S=\mathop{span}({v}_1,{v}_2,{v}_3)$ where ${v}_i$ are defined as follows. $$ {v}_1=\begin{bmatrix}1 \\ -1 \\ 2 \\ 3\end{bmatrix}\qquad {v}_2=\begin{bmatrix}1 \\ 0 \\ 1 \\ 0\end{bmatrix} \qquad {v}_3=\begin{bmatrix}3 \\ -2 \\ 5 \\ 7\end{bmatrix} \qquad \text{and}\qquad {u}=\begin{bmatrix}1 \\ 1 \\ 0 \\ -1\end{bmatrix} $$ Find a basis of $S$ which includes the vector ${u}.$

**Exercise**. Find a vector ${u}$ in $\mathbb{R}^4$ such that ${u}$ and the vectors $$ {v}_1=\begin{bmatrix}1 \\ -1 \\ -1 \\ 1\end{bmatrix} \qquad {v}_2=\begin{bmatrix}1 \\ 0 \\ 1 \\ 1\end{bmatrix} \qquad {v}_3=\begin{bmatrix}1 \\ 2 \\ 1 \\ 1\end{bmatrix} $$ for a basis of $\mathbb{R}^4.$

**Exercise**. Show that every subspace of $\mathbb{V}$ has no more than $n$ linearly independent vectors.

**Exercise**. Find two bases of $\mathbb{R}^4$ that have only the vectors ${e}_3$ and ${e}_4$ in common.

**Exercise**. Prove that if a list of vectors is linearly independent so is any sublist.

**Exercise**. Suppose ${v}_1,{v}_2, {v}_3$ and ${v}_1, {v}_2, {v}_4$ are two sets of linearly dependent vectors, and suppose that ${v}_1$ and ${v}_2$ are linearly independent. Prove that any set of three vectors chosen from ${v}_1, {v}_2, {v}_3, {v}_4$ is linearly dependent.

**Exercise**. If ${u}$ and ${v}$ are linearly independent vectors in $\mathbb{V}$, prove that the vectors $a{u}+b{v}$ and $c{u}+d{v}$ are also linearly independent if and only if $ad-bc\neq 0.$

**Exercise**. Let $U$ be the collection of vectors that satisfy the equations $x+y+z=0$ and $x+2y-z=0.$ Show $U$ is a subspace of $\mathbb{R}^3$, find a basis for $U$, and find $\dim(U).$

**Exercise**. Let $U$ be the collection of vectors that satisfy the equations $x+y+z=0$, $x+2y-z=0$, and $y-2z=0.$ Show $U$ is a subspace of $\mathbb{R}^3$, find a basis for $U$, and find $\dim(U).$

**Exercise**. Show that the only subspaces of $\mathbb{R}$ are ${{0}}$ and ${\mathbb{R}}.$

**Exercise**. Show that the only subspaces of $\mathbb{R}^2$ are ${{0}}$, ${\mathbb{R}^2}$, and any set consisting of all scalar multiples of a nonzero vector. Describe these subspaces geometrically.

**Exercise**. Determine the various types of subspaces of $\mathbb{R}^3$ and describe them geometrically.

**Exercise**. For ${b}\neq{0}$, show that the set of solutions of the $n\times m$ linear system $A {x}={b}$, is not a subspace of $\mathbb{V}.$

**Exercise**. Suppose that ${v}_1, {v}_2, \ldots, {v}_n$ are linearly independent in $\mathbb{R}^n.$ Show that if $A$ is an $n\times n$ matrix with $\mathop{rref}(A)=I_n$, then $A{v}_1, A{v}_2, \ldots, A{v}_n$ are also linearly independent in $\mathbb{R}^n.$

**Exercise**. Let $S=\{v_1, v_2, \ldots, v_s \}$ and $T=\{u_1, u_2, \ldots, u_t \}$ be two sets of vectors in $\mathbb{V}$ where each ${u}_i$, $(i=1,2,\ldots,t)$ is a linear combination of the vectors in $S.$ Show that ${w}=a_1 u_1 + a_2 u_2 + \cdots + a_t u_t $ is a linear combination of the vectors in $S.$

**Exercise**. Let $S=\{v_1, v_2, \ldots, v_m \}$ be a set of non-zero vectors in a vector space $\mathbb{V}$ such that every vector in $\mathbb{V}$ can be uniquely as a linear combination of the vectors in $S.$ Prove that $S$ is a basis for $\mathbb{V}.$

**Exercise**. Find a basis for the solution space of the homogeneous system $(\lambda I_n-A){x}={0}$ for the given $\lambda$ and $A.$

$$\lambda=1, A= \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & -3 \\ 0 & 1 & 3 \end{bmatrix} $$

$$\lambda=2, A= \begin{bmatrix} -2 & 0 & 0 \\ 0 & -2 & -3 \\ 0 & 4 & 5 \end{bmatrix} $$