% This is lecture no. 8 of MH1200/: Linear Algebra I.

\chapter{Theory of Matrices}%
\label{cha:theory_of_matrices}

\section{Definitions, Vocabulary, and Conventional Notation}%
\label{sec:definitions_vocabulary_and_conventional_notation}



\begin{definition}
	Matrix.
	A matrix (plural matrices) is a rectangular array of numbers.
	The numbers in the array are called entries in the matrix.
	The size of a matrix is given by $m\times n$ where $m$ is the number of rows and $n$ is the number of columns.
	The $(i,j)$-entry of a matrix is the number which is in the $i$-the row and $j$-the column of a matrix.
\end{definition}

\begin{exmp}
	The $3\times 2$ matrix:
	\[
	\begin{bmatrix}
		1 & 2\\
		3 & 4\\
		0 & -1\\
	\end{bmatrix}
	.\]
	Is the same as
	\[
	\begin{pmatrix}
		1 & 2\\
		3 & 4\\
		0 & -1\\
	\end{pmatrix}
	.\]
\end{exmp}

The notation for the set of all $p\times q$ matrices whose entries are real number is as such:
\[
	\mathbb{M}_{p\times q}(\mathbb{R})
.\]
where $p$ denotes the number of rows, and $q$ denotes the number of columns.

\begin{definition}
	Column Matrix/Column Vector.
	A column matrix/column vector is a matrix with only one column.
\end{definition}

\begin{definition}
	Row Matrix/Row Vector.
	A row matrix/row vector is a matrix with only one column.
\end{definition}

\subsection{Correspondence to $\mathbb{R}^n$}%
\label{sub:correspondence_to_r_n}

The element of $\mathbb{M}_{n\times 1}$ and $\mathbb{M}_{1\times n}$ seem very similar to the elements of $\mathbb{R}^n$.
They are all just lists of $n$ real numbers, but presented differently.
For example, consider $n=3$.
We have three closely related but different things:

\bigbreak
\noindent
A vector:
\[
	\vec{x} = (a,b,c)\in \mathbb{R}^3
.\]
\noindent
A row vector:
\[
	X = 
	\begin{bmatrix}
		a & b & c
	\end{bmatrix}
	\in \mathbb{M}_{1\times 3}
.\]
A column vector:
\[
	X^T =
	\begin{bmatrix}
		a\\
		b\\
		c\\
	\end{bmatrix}
	\in \mathbb{M}_{1\times 3}
.\]

These three sets are different but isomorphic.
Consider $\mathbb{R}^n$ and $\mathbb{M}_{1\times n}$.
There are functions going from each of the two sets to the other one.
These functions set up a correspondence (called a bijection) between the sets.
They tell which element in the second set an element in the first set corresponds to:
\[
	\mathbb{R}^n \leftrightharpoons \mathbb{M}_{1\times n}
.\]

The specific entries of an $m\times n$ matrix are often written using the following notation $a_{ij}$ :
\[
A =
\begin{bmatrix}
	a_{11} & a_{12} & \cdots & a_{1n}\\
	a_{21} & a_{22} & \cdots & a_{2n}\\
	\vdots & \vdots & \ddots & \vdots\\
	a_{m1} & a_{m2} & \cdots & a_{mn}\\
\end{bmatrix}
.\]
If we have defined the collection of numbers $a_{ij}$ first then we can define the matrix $A$ by writing:
\[
	A = [a_{ij}]
.\]
Sometimes we write:
\[
	A = [a_{ij}]_{m\times n}
.\]
when we want to emphasize the shape of the matrix.

\begin{exmp}
	$A = (a_{ij})_{2\times 3}$ where $a_{ij}=i+j$:
	\[
	A =
	\begin{bmatrix}
		2 & 3 & 4\\
		3 & 4 & 5\\
	\end{bmatrix}
	.\]
	$B=(b_{ij})_{3\times 2}$ where:
	\[
		b_{ij} =
		\begin{cases}
			1 & \text{if}\ i+j\ \text{is even.}\\
			-1 & \text{if}\ i+j\ \text{is odd.}\\
		\end{cases}
	\]
	\[
	B =
	\begin{bmatrix}
		1 & -1\\
		-1 & 1\\
		1 & -1\\
	\end{bmatrix}
	.\]
	
\end{exmp}

\begin{definition}
	Square Matrix.
	A matrix is called a square matrix if it has the same number of rows and columns.
	In particular, an $n\times n$ matrix is called a square matrix of order $n$.
\end{definition}

\begin{definition}
	Diagonal of a Matrix.
	Given a square matrix $A=(a_{ij})$ of order $n>0$, the diagonal of $A$ is the sequence of entries:
	\[
		a_{11},a_{22},\ldots,a_{nn}
	.\]
	The entries $a_{ii}$ are called the diagonal entries while $a_{ij},i\neq j$ are called non-diagonal entries.
	\[
	\begin{bmatrix}
		&&i<j\\
		&i=j&\\
		i>j&&\\
	\end{bmatrix}
	.\]
\end{definition}

\begin{definition}
	Diagonal Matrix.
	A square matrix is called a diagonal matrix if all its non-diagonal entries are 0:
	\[
		A = (a_{ij})_{n\times n}\ \text{is diagonal if}\ a_{ij}=0\ \text{whenever}\ i\neq j
	.\]
	For example:
	\[
	\begin{bmatrix}
		4\\
	\end{bmatrix}
	\quad
	\begin{bmatrix}
		1 & 0\\
		0 & 2\\
	\end{bmatrix}
	\quad
	\begin{bmatrix}
		1 & 0 & 0\\
		0 & 3 & 0\\
		0 & 0 & 2\\
	\end{bmatrix}
	\quad
	\begin{bmatrix}
		2 & 0 & 0 & 0\\
		0 & 3 & 0 & 0\\
		0 & 0 & 0 & 0\\
		0 & 0 & 0 & 1\\
	\end{bmatrix}
	.\]
\end{definition}

\begin{definition}
	Scalar Matrix.
	A diagonal matrix is called a scalar matrix if all its non-diagonal entries are the same:
	$A=(a_{ij})_{n\times n}$ is scalar if for some constant $c\in \mathbb{R}$ :
	\[
		a_{ij} =
		\begin{cases}
			c & \text{if }i=j\\
			0 & \text{if }i\neq j\\
		\end{cases}
	.\]
	
	For example:
	\[
	\begin{bmatrix}
		4\\
	\end{bmatrix}
	\quad
	\begin{bmatrix}
		2 & 0\\
		0 & 2\\
	\end{bmatrix}
	\quad
	\begin{bmatrix}
		3 & 0 & 0\\
		0 & 3 & 0\\
		0 & 0 & 3\\
	\end{bmatrix}
	\quad
	\begin{bmatrix}
		3 & 0 & 0 & 0\\
		0 & 3 & 0 & 0\\
		0 & 0 & 3 & 0\\
		0 & 0 & 0 & 3\\
	\end{bmatrix}
	.\]
\end{definition}

\begin{definition}
	Identity Matrix.
	A diagonal matrix is called the identity matrix if all its non-diagonal entries are 1:
	We use $I_n$ to denote the identity matrix of order $n$.
	For example:
	\[
	I_1 =
	\begin{bmatrix}
		1\\
	\end{bmatrix}
	\quad
	I_2 =
	\begin{bmatrix}
		1 & 0\\
		0 & 1\\
	\end{bmatrix}
	\quad
	I_3 =
	\begin{bmatrix}
		1 & 0 & 0\\
		0 & 1 & 0\\
		0 & 0 & 1\\
	\end{bmatrix}
	\quad
	I_4 =
	\begin{bmatrix}
		1 & 0 & 0 & 0\\
		0 & 1 & 0 & 0\\
		0 & 0 & 1 & 0\\
		0 & 0 & 0 & 1\\
	\end{bmatrix}
	.\]
\end{definition}
\begin{definition}
	Zero Matrix.
	We use $0_{m\times n}$ to denote the identity matrix of size $m\times n$.
	Sometimes we write 0 instead of $0_{m\times n}$ when there is no danger of confusion.
	For example:
	\[
		0_{1\times 1} =
	\begin{bmatrix}
		0\\
	\end{bmatrix}
	\quad
	0_{2\times 4} =
	\begin{bmatrix}
		0 & 0 & 0 & 0\\
		0 & 0 & 0 & 0\\
	\end{bmatrix}
	\quad
	0_{4\times 3} =
	\begin{bmatrix}
		0 & 0 & 0\\
		0 & 0 & 0\\
		0 & 0 & 0\\
		0 & 0 & 0\\
	\end{bmatrix}
	.\]
\end{definition}

\begin{definition}
	Symmetric Matrix.
	A square matrix $(a_{ij})$ is called a symmetric matrix if:
	\[
		a_{ij} = a_{ji} \quad \forall i,j
	.\]
	\noindent
	For example:
	\[
	\begin{bmatrix}
		4\\
	\end{bmatrix}
	\quad
	\begin{bmatrix}
		1 & -1 & 0\\
		-1 & 3 & 2\\
		0 & 2 & 2\\
	\end{bmatrix}
	\quad
	\begin{bmatrix}
		2 & 1 & 6 & -2\\
		1 & 3 & 0 & -1\\
		6 & 0 & 0 & 0\\
		-2 & -1 & 0 & 1\\
	\end{bmatrix}
	.\]
\end{definition}

\begin{definition}
	Triangular Matrix.
	A square matrix $(a_{ij})$ is called a upper triangular matrix if:
	\[
		a_{ij} = 0 \quad \forall i>j
	.\]
	A square matrix $(a_{ij})$ is called a lower triangular matrix if:
	\[
		a_{ij} = 0 \quad \forall i<j
	.\]
	Both upper and lower triangular matrices are called triangular matrices
	\noindent
	For example:
	\[
	\begin{bmatrix}
		4\\
	\end{bmatrix}
	\quad
	\begin{bmatrix}
		1 & 2 & 3\\
		0 & 3 & 2\\
		0 & 0 & 2\\
	\end{bmatrix}
	\quad
	\begin{bmatrix}
		2 & 0 & 0 & 0\\
		0 & 3 & 0 & 0\\
		1 & 0 & 0 & 0\\
		-2 & 3 & 0 & 1\\
	\end{bmatrix}
	.\]
\end{definition}

\begin{definition}
	Equal Matrix.
	Two matrices are said to be equal if they have the same size and their corresponding entries are equal.
	Let $A=(a_{ij})_{m\times n}$ and $B=(b_{ij})_{p\times q}$.
	Then $A$ is equal to $B$ if:
	\[
		m=p,\ n=q\ \text{and}\ a_{ij}=b_{ij}\quad \forall i,j
	.\]
\end{definition}

\begin{definition}
	Trace of a Matrix.
	The trace is an important function defined on square matrices:
	\[
		tr:\{square\ matrices\} \to \mathbb{R}
	.\]
	It is defined to be the sum of the diagonal elements.
	In symbols:
	Given a matrix $(a_{ij})_{n\times n}$ then:
	\[
	tr(A) = \sum_{i=0}^n a_{ii}
	.\]
\end{definition}

\begin{exmp}
	\[
		tr\left(
			\begin{bmatrix}
				1 & 2 & 3 & 4\\
				-1 & 0 & 1 & 2\\
				-3 & -2 & -1 & 0\\
				-5 & 0 & 5 & 10\\
			\end{bmatrix}
		\right)=1 + 0 + (-1) + 10 = 10\in \mathbb{R}
	.\]
	This function is important because it has nice properties, such as cyclic invariance: $tr(AB)=tr(BA)$.
\end{exmp}

\begin{definition}
	Transpose.
	The transpose is defined on matrices of any shape.
	The transpose of a matrix $M$ is denoted $M^T$ :
	\[
		transpose: \mathbb{M}_{p\times q}(\mathbb{R}) \to \mathbb{M}_{q\times p}(\mathbb{R})
	.\]
	If $A=(a_{ij})_{p\times q}$ then $A^T$ is defined to be the matrix $B=(b_{ij})_{q\times p}$ where:
	\[
		b_{ij} = a_{ji}
	.\]
\end{definition}

\begin{exmp}
	\[
	\begin{bmatrix}
		3 & 1 & -1 & 5\\
		4 & 2 & 0 & 10\\
	\end{bmatrix}^T
	=
	\begin{bmatrix}
		3 & 4\\
		1 & 2\\
		-1 & 0\\
		5 & 10\\
	\end{bmatrix}
	.\]
	
\end{exmp}

\section{Matrix Algebra}%
\label{sec:matrix_algebra}

\subsection{Basic Algebraic Operations on Matrices}%
\label{sub:basic_algebraic_operations_on_matrices}

\begin{definition}
	Matrix Addition.
	Take two matrices of the same shape:
	\[
		M_1,M_2 \in \mathbb{M}_{m\times n}(\mathbb{R})
	.\]
	The sum of these two matrices:
	\[
		M_1+M_2 \in \mathbb{M}_{m\times n}(\mathbb{R})
	.\]
	is defined to be the matrix with the smae shape whose $(i,j)$-entry is the sim of the $(i,j)$-entry of $M_1$ plus the $(i,j)$-entry of $M_2$.
\end{definition}

\begin{exmp}
	\[
	\begin{bmatrix}
		1 & 2\\
		3 & 4\\
		5 & 6\\
	\end{bmatrix}
	+
	\begin{bmatrix}
		-2 & 4\\
		-8 & 16\\
		-32 & 64\\
	\end{bmatrix}
	=
	\begin{bmatrix}
		-1 & 6\\
		-5 & 20\\
		-27 & 70\\
	\end{bmatrix}
	.\]
\end{exmp}

\begin{definition}
	Scalar Multiplication of a Matrix.
	Take a scalar $r\in \mathbb{R}$.
	Take a matrix $M\in \mathbb{M}_{m\times n}(\mathbb{R})$.
	The matrix that results when you multiply this matrix $M$ by the scalar $r$, denoted:
	\[
		rM\in \mathbb{M}_{m\times n}(\mathbb{R})
	.\]
	is defined to be the matrix with the same shape as $M$ whose $(i,j)$-entry is the result of multiplying the $(i,j)$-entry of $M$ by the scalar $r$.
\end{definition}

\begin{thrm}
	Let $M,N,P\in \mathbb{M}_{m\times n}(\mathbb{R})$ be matrices of the same shape.
	Recall that $0 \in \mathbb{M}_{m\times n}(\mathbb{R})$ denotes the zero matrix.
	Let $r,s\in \mathbb{R}$.
	Then:
	\begin{align*}
		(M+N)+P & = M+(N+P)\\
		M+N & = N+M\\
		0+M & = M+0\\
		    & = M
	\end{align*}
	And:
	\begin{align*}
		r(sM) & = (rs)M\\
		1M & = M\\
		0M & = 0
	\end{align*}
	And:
	\begin{align*}
		(r+s)M & = rM + sM\\
		r(M+N) & = rM + rN
	\end{align*}
\end{thrm}

\begin{definition}
	Multiplication of two Matrices.
	Take two matrices $A \in \mathbb{M}_{m\times p}(\mathbb{R})$ and $B \in \mathbb{M}_{p\times n}(\mathbb{R})$.
	Assume:
	\[
		A = [a_{ij}]_{m\times p}
		\quad
		\text{and}
		\quad
		B = [b_{ij}]_{p\times n}
	.\]
	The product of these two matrices is the $m\times n$ matrix:
	\[
		AB \in \mathbb{M}_{m\times n}(\mathbb{R})
	.\]
	whose $(i,j)$-entry is given by the formula:
	\[
		a_{i1}b_{1j} + \ldots + a_{ip}b_{pj} = \sum_{k=1}^p a_{ik}b_{kj}
	.\]
	The above is called summation notation.
\end{definition}

\subsection{Properties of Matrix Multiplication}%
\label{sub:properties_of_matrix_multiplication}

\begin{thrm}
	Let $M \in \mathbb{M}_{m\times n}(\mathbb{R})$.

	\noindent
	Then:
	\[
	I_mM = MI_n = M
	.\]
\end{thrm}

This theorem says that the identity matrix behaves very much like the usual number $1 \in \mathbb{R}$.

\begin{thrm}
	Let $M \in \mathbb{M}_{m\times n}(\mathbb{R})$.

	\noindent
	Then:
	\[
		0_{p\times n}M = 0_{p\times n}
	.\]
	and
	\[
		M0_{n\times p} = 0_{m\times p}
	.\]
\end{thrm}

\begin{note}
	The theorem above states that the shape of the zero matrix may change in the equation.
\end{note}

\begin{proof}
	Associativity.
	Consider three matrices:
	\begin{align*}
		A \in \mathbb{M}_{p\times q}(\mathbb{R})\ \text{with}\ A & = [a_{ij}]_{p\times q}\\
		B \in \mathbb{M}_{q\times r}(\mathbb{R})\ \text{with}\ B & = [b_{ij}]_{q\times r}\\
		C \in \mathbb{M}_{r\times s}(\mathbb{R})\ \text{with}\ C & = [c_{ij}]_{r\times s}
	\end{align*}
	We need to prove that:
	\[
		A(BC) = (AB)C
	.\]
	The two sides of this equation are matrices with the same shape $p\times s$.
	So to prove the equation we need to check that all the corresponding entries are the same.
	Let's focus on some entry, say the $(i,j)$-entry on the left hand side:
	\begin{align*}
		\textit{LHS}_{(i,j)} & = \sum_{k=1}^q a_{ik}(BC)_{(k,j)}\\
				     & = \sum_{k=1}^q a_{ik}\left(\sum_{l=1}^r b_{kl}c_{lj}\right)\\
				     & = \sum_{k=1}^q \sum_{l=1}^ra_{ik} b_{kl}c_{lj}
	\end{align*}
	Now focus on the same $(i,j)$-entry on the right hand side:
	\begin{align*}
		\textit{RHS}_{(i,j)} & = \sum_{l=1}^r (AB)_{(i,l)}c_{lj}\\
				     & = \sum_{l=1}^r \left(\sum_{k=1}^q a_{ik}b_{kl}\right)c_{lj}\\
				     & = \sum_{l=1}^r \sum_{k=1}^q a_{ik}b_{kl}c_{lj}
	\end{align*}
	This is exactly the same as the expression calculated for the $(i,j)$-entry of the left hand side.
\end{proof}

\begin{thrm}
	Let $A,B \in \mathbb{M}_{m\times n}(\mathbb{R})$.
	Let $C,D \in \mathbb{M}_{n\times q}(\mathbb{R})$.

	\noindent
	Then:
	\[
		(A+B)C = AC + BC
	.\]
	and:
	\[
		A(C+D) = AC + AD
	.\]
\end{thrm}

\begin{definition}
	Inverse Matrices.
	Consider a square matrix $M \in \mathbb{M}_{m\times m}(\mathbb{R})$.
	$M$ is said to be invertible if there exists another matrix $M^{-1}\in \mathbb{M}_{m\times m}(\mathbb{R})$ such that:
	\[
		MM^{-1} = M^{-1}M = I_m
	.\]
	If such a matrix exists then it is said to be the inverse of $M$.
\end{definition}

\begin{thrm}
	Let $A \in \mathbb{M}_{m\times n}(\mathbb{R})$.
	Let $B \in \mathbb{M}_{n\times p}(\mathbb{R})$.
	Let $r \in \mathbb{R}$.

	\noindent
	Then:
	\[
		r(AB) = (rA)B = A(rB)
	.\]
\end{thrm}

\section{Writing Linear Systems as Matrix Equations}%
\label{sec:writing_linear_systems_as_matrix_equations}

An $n$-tuple $(x_1,\ldots ,x_n)\in \mathbb{R}^n$ solves the following system of $m$ linear equations in $n$ variables $x_1, \ldots , x_n$ :
\begin{align*}
	a_{11}x_1 + \ldots + a_{1n}x_n & = b_1\\
				       & \;\;\vdots\\
	a_{m1}x_1 + \ldots + a_{mn}x_n & = b_m
\end{align*}
if and only if the following matrix equation is true:
\[
\begin{bmatrix}
	a_{11} & \cdots & a_{1n}\\
	\vdots & \ddots & \vdots\\
	a_{m1} & \cdots & a_{mn}\\
\end{bmatrix}
\begin{bmatrix}
	x_1\\
	\vdots\\
	x_n\\
\end{bmatrix}
=
\begin{bmatrix}
	b_1\\
	\vdots\\
	b_m\\
\end{bmatrix}
.\]

\begin{proof}
	Theorem about the possible number of solutions of a linear system.
	\begin{thrm}

		Consider two $n$-tuples:
		\begin{align*}
			(y_1,\ldots,y_n) & \in \mathbb{R}^n\\
			(z_1,\ldots,z_n) & \in \mathbb{R}^n
		\end{align*}
		and when solving the following system of $m$ linear equations in $n$ variables $x_1 , \ldots , x_n$ :
		\begin{align*}
			a_{11}x_1 + \ldots + a_{1n}x_n & = b_1\\
						       & \;\;\vdots\\
			a_{m1}x_1 + \ldots + a_{mn}x_n & = b_m
		\end{align*}
		Then for every scalar $\lambda \in \mathbb{R}$ the following linear combinations:
		\[
			\lambda (y_1,\ldots,y_n) + (1-\lambda )(z_1,\ldots,z_n)
		.\]
		are also solutions.
	\end{thrm}
	First, we must convert everything to equivalent matrix equations and exploiting matrix algebra.
	Let $A$ denote the following matrix:
	\[
	\begin{bmatrix}
		a_{11} & \cdots & a_{1n}\\
		\vdots & \ddots & \vdots\\
		a_{m1} & \cdots & a_{mn}\\
	\end{bmatrix}
	.\]
	Because we know $(y_1,\ldots,y_n)$ and $(z_1,\ldots,z_n)$ are solutions to the system represented by $A$, we know the following matrix equations are true:
	\[
	A
	\begin{bmatrix}
		y_1\\
		\vdots\\
		y_n\\
	\end{bmatrix}
	=
	\begin{bmatrix}
		b_1\\
		\vdots\\
		b_n\\
	\end{bmatrix}
	\quad
	\text{and}
	\quad
	A
	\begin{bmatrix}
		z_1\\
		\vdots\\
		z_n\\
	\end{bmatrix}
	=
	\begin{bmatrix}
		b_1\\
		\vdots\\
		b_n\\
	\end{bmatrix}
	.\]
	We need to show that the linear combination:
	\[
		\lambda(y_1,\ldots ,y_n) + (1-\lambda )(z_1, \ldots ,z_n)
		=
		(\lambda y_1 + (1-\lambda)z_1 , \ldots ,\lambda y_n + (1-\lambda)z_n)
	.\]
	is also a solution.
	We can do this by checking the corresponding matrix equation:
	\begin{align*}
		A
		\begin{bmatrix}
			\lambda y_1 + (1-\lambda)z_1\\
			\vdots\\
			\lambda y_n + (1-\lambda)z_n\\
		\end{bmatrix}
		& = A
		\left(
		\lambda
		\begin{bmatrix}
			y_1\\
			\vdots\\
			y_n\\
		\end{bmatrix}
		+ (1-\lambda)
		\begin{bmatrix}
			z_1\\
			\vdots\\
			z_n\\
		\end{bmatrix}
		\right)\\
		\intertext{Because matrix multiplication is distributive:}
		& = A
		\left(
		\lambda
		\begin{bmatrix}
			y_1\\
			\vdots\\
			y_n\\
		\end{bmatrix}
		\right)
		+ A
		\left(
		(1-\lambda)
		\begin{bmatrix}
			z_1\\
			\vdots\\
			z_n\\
		\end{bmatrix}
		\right)\\
		\intertext{Because scalar multiplication of matrices respects matrix multiplication:}
		& = \lambda
		\left(
		A
		\begin{bmatrix}
			y_1\\
			\vdots\\
			y_n\\
		\end{bmatrix}
		\right)
		+ (1-\lambda)
		\left(
		A
		\begin{bmatrix}
			z_1\\
			\vdots\\
			z_n\\
		\end{bmatrix}
		\right)\\
		\intertext{Because our assumption is that these matrix equations are true:}
		& = \lambda
		\begin{bmatrix}
			b_1\\
			\vdots\\
			b_m\\
		\end{bmatrix}
		+ (1-\lambda)
		\begin{bmatrix}
			b_1\\
			\vdots\\
			b_m\\
		\end{bmatrix}\\
		\intertext{Because scalar multiplication is distributive:}
		& = (\lambda + (1-\lambda))
		\begin{bmatrix}
			b_1\\
			\vdots\\
			b_m\\
		\end{bmatrix}\\
		\intertext{Because $\lambda + (1-\lambda)=1 \in \mathbb{R}$:}
		& =
		\begin{bmatrix}
			b_1\\
			\vdots\\
			b_m\\
		\end{bmatrix}
	\end{align*}
	So, because the corresponding matrix equation is true, it follows that the given linear combination:
	\[
		\lambda(y_1,\ldots ,y_n) + (1-\lambda )(z_1, \ldots ,z_n)
	.\]
	is also a solution.
	Therefore, every linear system has either 0 solutions, 1 solution, or $\infty$ solutions.
	Because as soon as you have two different solutions, combining them as above produces infinitely many different solutions.
\end{proof}
