Inverses

Let $F: V\longrightarrow W$ be a mapping. $F$ is said to be invertible if there exists a map $G: W\longrightarrow V$ such that
$$G\circ F:I_V,\ F\circ G=I_W,$$
where $I_V: V\longrightarrow V$ and $I_W: W\longrightarrow W$ the identity maps on $V$ and $W$, respectively. The map $G$ is called the inverse of $F$ and is denoted by $F^{-1}$.

Theorem. A map $F: V\longrightarrow W$ has an inverse if and only if it is one-to-one (injective) and onto (surjective).

Remark. If a map $F: V\longrightarrow W$ has an inverse, it is unique.

Example. The inverse map of $T_u$ is $T_{-u}$.

Example. Let $A$ be a square matrix. Then $L_A: \mathbb{R}^n\longrightarrow\mathbb{R}^n$ is invertible if and only if $A$ is invertible.

Proof. If $A$ is invertible, then one can immediately see that $L_A\circ L_{A^{-1}}=L_{A^{-1}}\circ L_A=I$. On the other hand, any linear map from $\mathbb{R}^n$ to $\mathbb{R}^n$ may be written as $L_B:\mathbb{R}^n\longrightarrow\mathbb{R}^n$ for some $n\times n$ square matrix $B$ as seen here. Suppose that $L_B:\mathbb{R}^n\longrightarrow\mathbb{R}^n$ is an inverse of $L_A: \mathbb{R}^n\longrightarrow\mathbb{R}^n$. Then $L_A\circ L_B=L_B\circ L_A=I$ i.e. for any $X\in\mathbb{R}^n$,
$$(AB)X=(BA)X=IX=X.$$
This implies that $AB=BA=I$. (Why?) Therefore, $A$ is invertibale and $B=A^{-1}$.

Theorem. Let $F: U\longrightarrow V$ be a linear map, and assume that this map has an inverse map  $G:V\longrightarrow U$. Then $G$ is a linear map.

Proof. The proof is straightforward and is left for an exercise.

Recall that a linear map $F: V\longrightarrow W$ is injective if and only if $\ker F=\{O\}$ as seen here.

Theorem. Let $V$ be a vector space of $\dim n$. If $W\subset V$ is a subspace of $\dim n$, then $W=V$.

Proof. It follows from the fact that a basis of a vector space is the maximal set of linearly independent vectors.

Theorem. Let $F: V\longrightarrow W$ be a linear map. Assume that $\dim V=\dim W$. Then

(i) If $\ker F=\{O\}$ then $F$ is invertible.

(ii) If $F$ is surjective then $F$ is invertible.

Proof. It follows from the formula
$$\dim V=\dim\ker F+\dim\mathrm{Im}F.$$

Example. Let $F:\mathbb{R}^2\longrightarrow\mathbb{R}^2$ be the linear map such that
$$F(x,y)=(3x-y,4x+2y).$$
Show that $F$ is invertible.

Solution. Suppose that $(x,y)\in\ker F$. Then
\left\{\begin{aligned} 3x-y&=0,\\ 4x+2y&=0. \end{aligned}\right.
This system of linear equations has only the trvial solution $(x,y)=(0,0)$, so $\ker F=\{O\}$. Therefore, $F$ is invertible.

A linear map $F: V\longrightarrow W$ which is invertible is called an isomorphism. If there is an isomorphism from $V$ onto $W$, we say $V$ is isomorphic to $W$ and write $V\cong W$ or $V\stackrel{F}{\cong}W$ in case we want to specify the isomorphism $F$. If $V\cong W$, as vector spaces they are identical and we do not distinguish them.

The following theorem says that there is essentially only one vector space of dimension $n$, $\mathbb{R}^n$.

Theorem. Let $V$ be a vector space of dimension $n$. Then $\mathbb{R}^n\cong V$.

Proof. Let $\{v_1,\cdots,v_n\}$ be a basis of $V$. Define a map $L:\mathbb{R}^n\longrightarrow V$ by
$$L(x_1,\cdots,x_n)=x_1v_1+\cdots+x_nv_n$$
for each $(x_1,\cdots,x_n)\in\mathbb{R}^n$. Then $L$ is an isomorphism.

Theorem. A square matrix $A$ is invertible if and only if its columns $A^1,\cdots,A^n$ are linearly indepdendent.

Proof. Let $L_A:\mathbb{R}^n\longrightarrow\mathbb{R}^n$ be the map such that $L_AX=AX$. For $X=\begin{pmatrix} x_1\\ \vdots\\ x_n \end{pmatrix}\in\mathbb{R}^n$,
$$L_AX=x_1A^1+\cdots+x_nA^n.$$
Hence, $\ker L_A=\{O\}$ if and only if $A^1,\cdots,A^n$ are linearly independent.

Composition of Linear Maps

Let $F: U\longrightarrow V$ and $G: V\longrightarrow W$ be two maps. The composite map $G\circ F: U\longrightarrow W$ is defined by
$$G\circ F(v)=G(F(v))$$
for each $v\in U$.

Example. Let $A$ be an $m\times n$ matrix and let $B$ be a $q\times m$ matrix. Let $L_A: \mathbb{R}^n\longrightarrow\mathbb{R}^n$ be the linear map such that $L_AX=AX$ for each $X\in\mathbb{R}^n$, and let $L_B:\mathbb{R}^m\longrightarrow\mathbb{R}^m$ such that $L_BY=Y$ for each $Y\in\mathbb{R}^m$. Then
\begin{align*}
L_B\circ L_A(X)&=L_B(L_A(X))\\
&=L_B(L_A(X))\\
&=B(AX)\\
&=BA(X)\\
&=L_{BA}(X)
\end{align*}
for each $X\in\mathbb{R}^n$. Therefore, $L_B\circ L_A=L_{BA}$.

Example. Let $A$ be an $m\times n$ matrix, and let $L_A:\mathbb{R}^n\longrightarrow\mathbb{R}^m$ be the linear map such that $L_A(X)=AX$ for each $X\in\mathbb{R}^n$. Let $C$ be a vector in $\mathbb{R}^m$ and let $T_C:\mathbb{R}^m\longrightarrow\mathbb{R}^m$ be the translation by $C$
$$T_C(Y)=Y+C$$
for each $Y\in\mathbb{R}^m$. Then for each $X\in\mathbb{R}^n$,
\begin{align*}
T_C\circ L_A(X)&=T_C(AX)\\
&=AX+C.
\end{align*}

Example. Let $V$ be a vector space, and let $w$ be an element of $V$. Let $T_w: V\longrightarrow V$ be the translation by $w$ i.e.
$$T_w(v)=v+w$$
for each $v\in V$. Then
\begin{align*}
T_{w_1}\circ T_{w_2}(v)&=T_{w_1}(T_{w_2}(v))\\
&=T_{w_1}(v+w_2)\\
&=v+w_2+w_1
\end{align*}
for each $v\in V$. Therefore, $T_{w_1}\circ T_{w_2}=T_{w_1+w_2}$ i.e. the composite of translations is again a translation.

Remark. Note that translations are not linear. One easy way to see this is that $T_w(O)=O+w=w$. So $T_w$ is not linear unless $w=O$ in which case $T_w$ is the identity map.

Example. [Rotations] Let $\theta$ be a number, and $A(\theta)$ the matrix
$$A(\theta)=\begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}.$$
Let $R_\theta:\mathbb{R}^2\longrightarrow\mathbb{R}^2$ be the rotation by angle $\theta$ i.e.
$$R_\theta(X)=A(\theta)X$$
for each $X\in\mathbb{R}^2$. Clearly, rotations are linear.
Now,
\begin{align*}
R_{\theta_1}\circ R_{\theta_2}(X)&=R_{\theta_1}(R_{\theta_2}(X))\\
&=R_{\theta_1}(A(\theta_2)X)\\
&=A(\theta_1)A(\theta_2)X\\
&=A(\theta_1+\theta_2)X\\
&=R_{\theta_1+\theta_2}.
\end{align*}

Theorem. Let $U,V,W$ be vector spaces. Let $F:U\longrightarrow V$ and $G:V\longrightarrow W$ be linear maps. Then the composite map $G\circ F:U\longrightarrow W$ is also linear.

Proof. The proof is straightforward and is left for an exercise.

The Matrix Associated with a Linear Map

Given an $m\times n$ matrix $A$, there is an associated linear map $L_A: \mathbb{R}^n\longrightarrow\mathbb{R}^m$ as seen here. Conversely, given a linear map $L: \mathbb{R}^n\longrightarrow\mathbb{R}^m$ there exists an $m\times n$ matrix $A$ such that $L=L_A$. To see this, consider the unit column vectors $E^1,\cdots,E^n$ of $\mathbb{R}^n$. For each $j=1,\cdots,n$, let $L(E^j)=A^j$, where $A^j$ is a column vector in $\mathbb{R}^m$. For each $X\in\mathbb{R}^n$,
$$X=x_1E^1+\cdots+x_nE^n=\begin{pmatrix} x_1\\ \vdots\\ x_n \end{pmatrix}$$
and hence
\begin{align*}
LX&=x_1L(E^1)+\cdots+x_nL(E^n)\\
&=x_1A^1+\cdots+x_nA^n\\
&=AX
\end{align*}
where $A$ is the matrix whose column vectors are $A^1,\cdots,A^n$. The matrix $A$ is called the matrix associated with the linear map $L$.

Example. Let $L:\mathbb{R}^3\longrightarrow\mathbb{R}^2$ be the projection given by $$L\begin{pmatrix} x\\ y\\ z \end{pmatrix}=\begin{pmatrix} x\\ y \end{pmatrix}.$$
Find the matrix associated with $L$.

Solution. $$L(E^1)=\begin{pmatrix} 1\\ 0\end{pmatrix},\ L(E^2)=\begin{pmatrix} 0\\ 1 \end{pmatrix},\ L(E^3)=\begin{pmatrix} 0\\ 0 \end{pmatrix}.$$
Thus, the matrix associated with $L$ is
$$A=\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \end{pmatrix}.$$

Let us now consider a more general case. Let $V$ be a vector space and $\{v_1,\cdots,v_n\}$ a given basis of $V$. Let $L:V\longrightarrow V$ be a linear map. Then there exist numbers $c_{ij}$, $i,j=1,\cdots,n$ such that
\begin{align*}
Lv_1&=c_{11}v_1+\cdots+c_{1n}v_n,\\
&\vdots\\
Lv_n&=c_{n1}v_1+\cdots+c_{nn}v_n.
\end{align*}
Let $v=x_1v_1+\cdots+x_nv_n$. Then
\begin{align*}
Lv&=\sum_{i=1}^nx_iLv_i\\
&=\sum_{i=1}^nx_i\sum_{j=1}^nc_{ij}v_j\\
&=\sum_{j=1}^n(\sum_{i=1}^nx_ic_{ij})v_j.
\end{align*}
Hence, we have the following theorem.

Theorem. If $C=(c_{ij})$ is the matrix such that $Lv_i=\sum_{j=1}^nc_{ij}v_j$ and $X=\begin{pmatrix} x_1\\ \vdots\\ x_n \end{pmatrix}$ is the coordinate vector of $v$, then the coordinate vector of $Lv$ is ${}^tCX$ i.e. the matrix associated with $L$ is ${}^tC$ with respect to the basis $\{v_1,\cdots,v_n\}$.

Example. Let $L: V\longrightarrow V$ be a linear map. Let $\{v_1,v_2,v_3\}$ be a basis of $V$ such that
\begin{align*}
L(v_1)&=2v_1-v_2,\\
L(v_2)&=v_1+v_2-4v_3,\\
L(v_3)&=5v_1+4v_2+2v_3.
\end{align*}
The matrix associated with $L$ is
$$\begin{pmatrix} 2 & 1 & 5\\ -1 & 1 & 4\\ 0 & -4 & 2 \end{pmatrix}.$$

The Kernel and Image of a Linear Map

Let $F:V\longrightarrow W$ be a linear map. The image of $F$ is the set
$$\mathrm{Im}F=\{w\in W: F(v)=w\ \mbox{for some}\ v\in V\}.$$

Proposition. The image of $F$ is a subspace of $W$.

Proof. The proof is straightforward. It is left for an exercise.

The preimage of the identity element $O$ under the linear map $F$ i.e. the set of elements $v\in V$ such that $F(v)=O$ is called the kernel of $F$ and is denoted by $\ker F$.

Proposition. The kernel of $F$ is a subspace of $V$.

Proof. It is straightforward and left for an exercise.

Example. Let $L: \mathbb{R}^3\longrightarrow\mathbb{R}$ be the map defined by
$$L(x,y,z)=3x-2y+z.$$
If we write $A=(3,-2,1)$, then $L(X)$ may be written as
$$L(X)=X\cdot A.$$
So, the kernel of $L$ is the set of all $X$ that are perpendicular to $A$.

Example. Let $A$ be an $m\times n$ matrix and let $L_A:\mathbb{R}^n\longrightarrow\mathbb{R}^m$ be the linear map defined by
$$L_A(X)=AX.$$
The kernel of $L_A$ is the subspace of solutions $X$ of the system of linear equations
$$AX=O.$$

Example. Let $\mathcal{F}$ be the vector space of smooth functions. Let $a_1,\cdots,a_m$ be numbers and let
$$L=a_m\frac{d^m}{dx^m}+a_{m-1}\frac{d^{m-1}}{dx^{m-1}}+\cdots+a_1.$$
Then $L:\mathcal{F}\longrightarrow\mathcal{F}$ is a linear map. $\ker L$ is the space of solutions of the homogeneous linear differential equation
$$a_m\frac{d^mf}{dx^m}+a_{m-1}\frac{d^{m-1}f}{dx^{m-1}}+\cdots+a_1f=0.$$
If there exists one solution $h_0$ for the non-homogeneous linear differential equation $L(h)=g$, then any solution $h$ may be written as $h=f+h_0$ where $f$ is a solution of the homogeneous equation $L(f)=0$. The proof is left as an exercise.

Theorem. Let $F: V\longrightarrow W$ be a linear map such that $\ker F=\{O\}$. If $v_1,\cdots,v_n$ are linearly independent elements of $V$, then $F(v_1),\cdots,F(v_n)$ are linearly independent elements of $W$.

Proof. The proof is straightforward and is left for an exercise.

Theorem. Let $F: V\longrightarrow W$ be a linear map. $F$ is one-to-one if and only if $\ker F=\{O\}$.

Proof. Suppose that $F$ is one-to-one. Let $v\in\ker F$. Then $F(v)=O=F(O)$. Since $F$ is one-to-one, $v=O$. So, $\ker F=\{O\}$. Suppose that $\ker F=\{O\}$. Let $F(v_1)=F(v_2)$. Then $F(v_1-v_2)=O$ and so $v_1-v_2\in\ker F=\{O\}$ i.e. $v_1=v_2$. Hence, $F$ is one-to-one.

Given a linear map $L: V\longrightarrow W$, there is a relationship between the dimensions of $V$, $\ker L$, and $\mathrm{Im}L$, namely
$$\dim V=\dim\ker L+\dim\mathrm{Im}L.$$
We will not prove it here but those interested may find proof in [1].

Example. Consider the linear map $L:\mathbb{R}^3\longrightarrow\mathbb{R}$ given by
$$L(x,y,z)=3x-2y+z.$$
The image is not $\{O\}$, so it is $\mathbb{R}$. Therefore the dimension of $\ker L$, the space of all solutions of $3x-2y+z=0$ is 2. $3x-2y+z=0$ is indeed equation of a plane through the origin and we know that the dimension of a plane as a vector space is 2.

References:

[1] Serge Lang, Introduction to Linear Algebra, Second Edition, Undergraduate Texts in Mathematics, Springer, 1986

Linear Maps

Let $V, W$ be two vector spaces. A map $L:V\longrightarrow W$ is called a linear map if it satisfies the following properties: for any elements $u,v\in V$ and any scalar $c$,

LM 1. $L(u+v)=L(u)+L(v)$.

LM 2. $L(cu)=cL(u)$.

That is, linear maps are maps that preserve addition and scalar multiplication.

Proposition. A map $L: V\longrightarrow W$ is linear if and only if for any elements $u,v\in V$ and scalars $a,b$,
$$L(au+bv)=aL(u)+bL(v).$$

Proof. It is straightforward and left as an exercise.

Example. Let $A$ be an $m\times n$ matrix. Define
$$L_A:\mathbb{R}^n\longrightarrow\mathbb{R}^m$$
by
$$L_A(X)=A\cdot X.$$ Then $L_A$ is linear.

Example. Let $A=(a_1,\cdots,a_n)$ be a fixed vector in $\mathbb{R}^n$. Define $L_A:\mathbb{R}^n\longrightarrow\mathbb{R}$ by
$$L_A(X)=A\cdot X.$$
Then $L_A$ is a linear map. The dot product $A\cdot X$ can be viewed as a matrix multiplication if we view $A$ as a row vector and $X$ as a column vector. So this example is a spacial case of the previous example.

Example. Let $\mathcal{F}$ be the set of all smooth functions. Then the derivative $D:\mathcal{F}\longrightarrow\mathcal{F}$ is a linear map.

Example. Define $\wp: \mathbb{R}^3\longrightarrow\mathbb{R}^2$ by $\wp(x,y,z)=(x,y)$, i.e. $\wp$ is a projection. It is a linear map.

Proposition. Let $L: V\longrightarrow W$ be a linear map. Then $L(O)=O$.

Proof. Let $v\in V$. Then
$$L(O)=L(v-v)=L(v-v)=L(v)-L(v)=O.$$

Example. Let $L:\mathbb{R}^2\longrightarrow\mathbb{R}^2$ be a linear map. Suppose that
$$L(1,1)=(1,4)\ \rm{and}\ L(2,-1)=(-2,3).$$
Find $L(3,-1)$.

Solution. $(3,-1)$ is written as a linear combination of $(1,1)$ and $(2,-1)$ as
$$(3,-1)=\frac{1}{3}(1,1)+\frac{4}{3}(-2,3).$$
Hence,
$$L(3,1)=\frac{1}{3}L(1,1)+\frac{4}{3}L(-2,3)=\frac{1}{3}(1,4)+\frac{4}{3}(-2,3)=\left(-\frac{7}{3},\frac{16}{3}\right).$$

The coordinates of a linear map

Consider a map $F: V\longrightarrow\mathbb{R}^n$. For any $v\in V$, $F(v)\in\mathbb{R}^n$ so $F(v)$ may be written as
$$F(v)=(F_1(v),F_2(v),\cdots,F_n(v))$$
where each $F_i$ is a function $F_i:V\longrightarrow\mathbb{R}$ called the $i$-th coordinate function.

Proposition. A map $F_i: V\longrightarrow\mathbb{R}^n$ is linear if and only if each coordinate function $F_i$ is linear.

Proof. Straightforward. Left as an exercise.

Example. Let $F:\mathbb{R}^2\longrightarrow\mathbb{R}^3$ be the map
$$F(x,y)=(2x-y,3x+4y,x-5y).$$
Then
$$F_1(x,y)=2x-y,\ F_2(x,y)=3x+4y,\\ F_3(x,y)=x-5y.$$
These coordinate functions can be written as
$$F_1(x,y)=\begin{pmatrix} 2 & -1 \end{pmatrix}\begin{pmatrix} x\\y \end{pmatrix},\ F_2(x,y)=\begin{pmatrix} 3 & 4 \end{pmatrix}\begin{pmatrix} x\\y \end{pmatrix},\ F_3(x,y)=\begin{pmatrix} 1 & -5 \end{pmatrix}\begin{pmatrix} x\\y \end{pmatrix}.$$
Hence, each $F_i$ is linear, $i=1,2,3$ and therefore $F$ is linear by the Proposition. In fact, $F$ may be written as $L_A:\mathbb{R}^2\longrightarrow\mathbb{R}^3$ where
$$A=\begin{pmatrix} 2 & -1\\ 3 & 4\\ 1 & -5 \end{pmatrix}.$$

Rank of a Matrix

Consider an $m\times n$ matrix

$$A=\begin{pmatrix}a_{11} & \cdots & a_{1n}\\\vdots & & \vdots\\a_{m1} & \cdots & a_{mn}\end{pmatrix}.$$

The columns of $A$ generate a vector space, which is a subspace of $\mathbb{R}^m$, called the column space of $A$. The dimension of the subspace is called the column rank of $A$. Similarly the rows of $A$ generate a subspace of $\mathbb{R}^n$, called the row space of $A$ and the dimension of this subspace is called the row rank of $A$. It turns out that the column rank and the row rank must be equal. So, we simply call the column rank or the row rank of $A$, the rank of $A$.

There are a couple important theorems regarding the rank of a matrix. They are introduced without proofs.

Theorem. Row and column operations do not change the row rank of a matrix, nor do they change the column rank.

Remark. Row and column operations only change basis of row space or column space.

Theorem. Let $A$ be a matrix of rank $r$. By a succession of row and column operations, the matrix can be transformed to the matrix having components equal to $1$ on the diagonal of the first $r$ rows and columns, and $0$ everywhere else.

$$\begin{pmatrix}1 & 0 & \cdots & 0 & 0 &\cdots &0\\0 & 1 & \cdots & 0 & 0 & \cdots &0\\\vdots & &\ddots &\vdots&\vdots& &\vdots\\0 & 0 &\cdots & 1& 0 &\cdots &0\\0 & 0 &\cdots & 0& 0 &\cdots &0\\\vdots & & &\vdots&\vdots&\ddots&\vdots\\0 & 0 &\cdots & 0& 0 &\cdots &0\end{pmatrix}$$

Example. Find the rank of the matrix $\begin{pmatrix}2 & 1 & 1\\0 & 1 & -1\end{pmatrix}$.

Solution. There are only two rows, so the rank will be at most 2. On the other hand, the column vectors $\begin{pmatrix}2\\0\end{pmatrix}$ and $\begin{pmatrix}1\\1\end{pmatrix}$ are linear independent. Therefore, the rank is 2.

Example. Find the rank of the matrix

$$\begin{pmatrix}1 & 2 & -3\\2 & 1 & 0\\-2 & -1 & 3\\-1 & 4 & -2\end{pmatrix}.$$

Solution. Since there are three columns, the rank will be at most 3. Subtract 2 times column 1 from column 2; add 3 times column 1 to column 3. The resulting matrix is

$$\begin{pmatrix}1 & 0 & 0\\2 & -3 & 6\\-2 & 3 & -3\\-1 & 6 & -5\end{pmatrix}.$$

Add 2 times column 2 to column 3. The resulting matrix is

$$\begin{pmatrix}1 & 0 & 0\\2 & -3 & 0\\-2 & 3 & 3\\-1 & 6 & 7\end{pmatrix}.$$

This matrix is in column echelon form and one can easily see that the first three rwo vectors are linearly independent. Therefore, the rank is 3.

Linear Independence

Let $V$ be a vector space. $v_1,\cdots,v_n\in V$ are said to be linearly dependent if there exist numbers $a_1,\cdots,a_n$ not all equal to $0$ such that
$$a_1v_1+\cdots+a_n=O.$$
If there do not exist such numbers, then we say $v_1,\cdots,v_n$ are linearly independent. That is, $v_1,\cdots,v_n$ are linearly independent if whenever $a_1v_1+\cdots+a_nv_n=O$, $a_1=\cdots=a_n=0$.

Example. In $\mathbb{R}^n$, the standard unit vectors $E_1,\cdots,E_n$ are linearly independent.

Example. The vectors $(1,1)$ and $(-3,2)$ are linearly indepdent in $\mathbb{R}^2$.

A set of vectors $\{v_1,\cdots,v_n\}\subset V$ is said to be a basis of $V$ if $v_1,\cdots,v_n$ generate $V$ and that they are linearly independent.

Example. The vectors $E_1,\cdots,E_n$ form a basis of $\mathbb{R}^n$.

Example. The vectors $(1,1)$ and $(-1,2)$ form a basis of $\mathbb{R}^2$.

In general for $\mathbb{R}$, the following theorem holds.

Theorem. Let $(a,b)$ and $(c,d)$ be two vectors in $\mathbb{R}^2$.

(i) They are linearly depedendent if and only if $ad-bc=0$.

(ii) If they are  indepdendent, they form a basis of $\mathbb{R}^2$.

Proof. Exercise

Let $V$ be a vector space and let $\{v_1,\cdots,v_n\}$ be a basis of $V$. If $v\in V$ is written as a linear combination
$$v=x_1v_1+\cdots+x_nv_n,$$
$(x_1,\cdots,x_n)$ is called the coordinates of $v$ with respect to the basis $\{v_1,\cdots,v_n\}$. For each $i=1,\cdots,n$, $x_i$ is called the i-th coordinate. The following theorem says that there can be only one set of coordinates for a given vector.

Theorem. Let $V$ be a vector space. Let $v_1,\cdots,v_n$ be linearly independent elements of $V$. If $x_1,\cdots,x_n$ and $y_1,\cdots,y_n$ are numbers such that
$$x_1v_1+\cdots+x_nv_n=y_1v_1+\cdots+y_nv_n,$$
then $x_i=y_i$ for all $i=1,\cdots,n$.

Proof. It is strightforward from the definition of linearly independent vectors.

Example. Find the coordinates of $(1,0)$ with respect to the two vectors $(1,1)$ and $(-1,2)$.

Example. The two functions $e^t$ and $e^{2t}$ are linearly independent.

Proof. Exercise.

Theorem. Let $v,w$ be two vectors of a vector space $V$. They are linearly dependent if and only if one of them is a scalr multiple of the other, i.e there is a number $c$ such that $v=cw$ or $w=cv$.

Proof. Exercise.

If one basis of a vector space $V$ has $n$ elements and another basis has $m$ elements, then $n=m$. The number of elements in any basis of a vector space $V$ is called the dimension of $V$ and is denoted by $\dim V$.

Linear Combination

Let $V$ be a vector space and let $v_1,\cdots, v_n\in V$. $V$ is said to be generated by $v_1,\cdots,v_n$ if given an element $v\in V$, there exist numbers $x_1,\cdots, x_n$ such that
$$v=x_1v_1+\cdots+x_nv_n.$$
The expression $x_1v_1+\cdots+x_nv_n$ is called a linear combination of $v_1,\cdots,v_n$. The numbers $x_1,\cdots,x_n$ are called the coefficients of the linear combination.

Example. Let $E_1,\cdots,E_n$ be the standard unit vectors in $\mathbb{R}^n$. Then $E_1,\cdots,E-n$ generate $\mathbb{R}^n$.

Proof. Gievn $X=(x_1,\cdots,x_n)\in\mathbb{R}^n$,
$$X=\sum_{i=1}^nx_iE_i.$$

Proposition. The set of all linear combinations of $v_1,\cdots,v_n$ is a subspace of $V$.

Proof. Straightforward.

Example. Let $v_1$ be a non-zero element of a vector space $V$, and let $w$ be any element of $V$. The set
$$\{w+tv_1: t\in\mathbb{R}\}$$
is the line passing through $w$ in the direction of $v_1$. This line is not a subspace, however if $w=O$, it is a subspace of $V$, generated by a single vector $v_1$.

Example. Let $v_1,v_2$ be two elements of a vector space $V$. The set of all linear combinations of $v_1,v_2$
$$t_1v_1+t_2v_2: t_1,t_2\in\mathbb{R}\}$$
is a plane through the origin and it is a subspace of $V$, generated by $v_1,v_2$. The plane passing through a point $P\in V$, parallel to $v_1,v_2$ is the set
$$\{P+t_1v_1+t_2v_2: t_1,t_2\in\mathbb{R}\}.$$
However, this is not a subspace unless $P=O$.

Vector Spaces

A vector space $V$ is a set of objects which can be added and multiplied by numbers, in such a way that the sum of two elements of $V$ is again an element of $V$, the product of an element of $V$ by a number is an element of $V$, and the following properties are satisfied:

VS 1. Given $u,v,w\in V$, we have

$$(u+v)+w=u+(v+w).$$

VS 2. There is an element $O\in V$ such that

$$O+u=u+O=u$$

for all $u\in V$.

VS 3. Given $u\in V$, the element $(-1)u$ is such that

$$u+(-1)u=(-1)u+u=O.$$

$(-1)u$ is simply written as $-u$.

VS 4. For all $u,v\in V$, we have

$$u+v=v+u.$$

VS 5. If $c$ is a number, then $c(u+v)=cu+cv$.

VS 6. If $a,b$ are two numbers, then $(a+b)v=av+bv$.

VS 7. If $a,b$ are two numbers, then $(ab)v=a(bv)$.

VS 8. For any $u\in V$, we have $1u=u$.

The axioms VS 1-VS 4 say that $(V,+)$ is an abelian group. The elements of vector space $V$ are called vectors. One can easily verify that vectors in $\mathbb{R}^n$ satisfy the axioms VS 1-VS 8 and hence $\mathbb{R}^n$ is a vector space.

Example. Let $M(m,n)$ denote the set of all $m\times n$ matrices. Then $M(m,n)$ is a vector space. Using the identification

$$\begin{pmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\a_{21} & a_{22} & \cdots & a_{2n}\\\vdots &\vdots& \ddots&\vdots\\a_{m1} & a_{m2} & \cdots & a_{mn}\end{pmatrix}\longleftrightarrow(a_{11},\cdots,a_{1n};a_{21},\cdots,a_{2n};\cdots;a_{m1},\cdots,a_{mn}),$$

we see that $M(m,n)$ may be identified with $\mathbb{R}^{mn}$ as a vector space.

Example. Let $\mathcal{F}$ be the set of all functions from $\mathbb{R}$ to $\mathbb{R}$. For any $f,g\in\mathcal{F}$, define $f+g$ by

$$(f+g)(x)=f(x)+g(x)$$

for all $x\in\mathbb{R}$. For any $f\in\mathbb{R}$ and for any number $c$, define $cf$ by

$$(cf)(x)=cf(x)$$

for all $x\in\mathbb{R}$. Then $\mathcal{F}$ is a vector space called a function space.

Example. Let $V=\{ae^t+be^{2t}: a,b\in\mathbb{R}\}$. Then $V$ is a vector space. Note that $V$ is the set of all solutions of the second order linear differential equation $\frac{d^2x}{dt^2}-3\frac{dx}{dt}+2x=0$.

Subspaces

A subset $U$ of a vector space $V$ is said to be a subspace if $U$ itself is also a vector space. For $U$ to be a vector space, it suffices to satisfy that

(i) For any $v,w\in U$, $v+w\in U$.

(ii) If $v\in U$ and $c$ is a number, $cv\in U$.

(iii) The identity element $O$ of $V$ is also an element of $U$.

Proposition. A nonempty subset $U$ of a vector space $V$ is a subspace if and only if $av+bw\in U$ for any $v,w\in U$ and numbers $a,b$.

Proof. Exercise

Example. Let $U$ be the set of vectors in $\mathbb{R}^n$ whose last coordinate is $0$. Then $U$ is a subspace of $\mathbb{R}^n$. $U$ may be identified with $\mathbb{R}^{n-1}$.

Example. Let $A$ be a vector in $\mathbb{R}^n$. Let $U$ be the set of all vectors $B$ in $\mathbb{R}^n$ such that $B\cdot A=0$ i.e. $B$ is perpendicular to $A$. Then $U$ is a subspace of $V$.

Example. Let $U$ and $W$ be subspaces of a vector space $V$. Then $U\cap W$ is also a subspace of $V$.

Example. Let $U$ and $W$ be subspaces of a vector space $V$. Define the sum of $U$ and $W$

$$U+W=\{u+w: u\in U\ \rm{and}\ w\in W\}.$$

Then $U+W$ is a subspace.