System of Linear Equations

The fundamental problem of Linear Algebra is solving a system of Linear Equations.

  • A system of linear equations is a system of first order equations with multiple unknowns.

$A \mathbf x = \mathbf b$

Matrix View

Suppose we have a system with $m$ equations and $n$ unknowns:

$\left\{\begin{matrix} a_{11} x_1 + a_{12} x_2 + \ ... \ + a_{1n} x_n = b_1\\ a_{21} x_1 + a_{22} x_2 + \ ... \ + a_{2n} x_n = b_2\\ \vdots \\ a_{m1} x_1 + a_{22} x_2 + \ ... \ + a_{mn} x_n = b_m\\ \end{matrix}\right.$

The coefficients of the unknowns form a Matrix - a rectangular array of numbers:

  • [math]A = \begin{bmatrix} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ \vdots & \vdots & \vdots & \vdots \\ a_{m1} & a_{m2} & ... & a_{mn} \end{bmatrix}[/math]
  • We call this matrix $A$, it's $m \times n$ matrix: with $m$ rows and $n$ columns

The unknowns themselves form a column vector $\mathbf{x}$, and the ... form a column vector $b$

  • [math]\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \\ \end{bmatrix}[/math], [math]\mathbf{b} = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \\ \end{bmatrix}[/math],
  • they're both column vectors, $\mathbf{x}$ has size $n$, and $\mathbf{b}$ has size $m$
  • can write them as rows using the transposition notation: $\mathbf{x} = [x_1, x_2, ..., x_n]^T$ (but they still remain column vectors)

So the system of linear equations can be expressed in a matrix form as $A\mathbf{x} = \mathbf{b}$

Geometry of Linear Equations

Suppose we have the following system of linear equations:

$\left\{\begin{matrix} 2x - y = 0\\ -x + 2y = 3 \end{matrix}\right.$

Matrix form:

  • [math]\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \end{bmatrix}[/math]

We can see this in two possible ways:

  • as a crossing of lines (hyperplanes) - the row picture
  • as vectors - the column picture

Row Picture

We have two lines:

  • $2x - y = 0$, $-x + 2y = 3$
  • plot them, and the point where they cross is the solution $\mathbf{x}$


The solution is $[x, y]^T = [1, 2]^T$

For 3 and more dimensions, we have (hyper)planes instead of lines.

Column Picture

We have two vectors:

  • [math]x \begin{bmatrix} 2 \\ -1 \end{bmatrix} + y \begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \end{bmatrix}[/math]

We need to combine first two vectors [math]v = \begin{bmatrix} 2 \\ -1 \end{bmatrix}[/math] and [math]u = \begin{bmatrix} -1 \\ 2 \end{bmatrix}[/math] so they form [math]\begin{bmatrix} 0 \\ 3 \end{bmatrix}[/math]

I.e. we want to find a Linear Combination of these columns.

  • From the vector picture we know that the solution is $[1, 2]^T$, so let's take
  • b17368e5e7774f8ca58e054c7b98183e.png
  • so we take 1 of vector $\mathbf u$ and 2 of vector $\mathbf v$ and end up at exactly $\mathbf b$!

Python Code

Python code to reproduce the figures  
import matplotlib.pylab as plt
import numpy as np

class Line:
    def __init__(self, slope, intercept):
        self.slope = slope
        self.intercept = intercept

    def calculate(self, x1):
        x2 = x1 * self.slope + self.intercept
        return x2

line1 = Line(2, 0)
line2 = Line(0.5, 1.5)

a = np.array([-6, 6])
plt.plot(a, line1.calculate(a), color='red', marker='')
plt.plot(a, line2.calculate(a), color='blue', marker='')
plt.scatter([1], [2])
plt.ylim([-1, 3])
plt.xlim([-1, 3])

import matplotlib.pylab as plt

plt.quiver([0, 0, 0], [0, 0, 0], 
           [-1, 2, 0], [2, -1, 3], color=['red', 'blue', 'black'], 
           scale_units='xy', scale=1)
plt.quiver([2], [-1], 
           [-2], [4], color='red', width=0.005, scale=1,
           angles='xy', scale_units='xy')
plt.ylim([-2, 5])
plt.xlim([-4, 4])

Solving the System

How we can solve the system $A \mathbf x = \mathbf b$?

Vector Solution: the solution found with the column picture

Matrix Solution

  • if $A$ is non-singular and invertible, then to find $\mathbf x$ multiply $\mathbf b$ by the inverse of $A$: $\mathbf x = A^{-1} \mathbf b$
  • if $\mathbf b \in C(A)$ - Column Space of $A$
    • this is especially important when the number of columns $n$ is less than the number of rows $m$

Solving $A \mathbf x = \mathbf 0$

Such systems are called homogeneous - see Homogeneous Systems of Linear Equations

Complete Solution to $A \mathbf x = \mathbf b$

Let $A$ be $n \times m$ matrix of rank $r$

  • we know that the system has a solution if $\mathbf b \in C(A)$


  • reduce $A$ to Row Reduced Echelon Form
  • set all free variables to 0 and solve - get $\textbf x_p = \textbf x_\text{particular}$
  • then solve $A \mathbf x_n = \mathbf 0$ - get all $\mathbf x_n$ - all $\mathbf x$ that solve the homogeneous system
  • Then find all other solutions: they are $\mathbf x = \textbf x_p + \mathbf x_n$
  • this solution is called the complete solution
  • why? $A \mathbf x_p = \mathbf b$ and $A \mathbf x_n = \mathbf 0$. Add them and get $A \cdot (\mathbf x_p + \mathbf x_n) = \mathbf b + \mathbf 0 = \mathbf b$
  • so we can the solution as the Nullspace $C(A)$ but shifted away from the origin by $x_p$
  • note that this solution doesn't form a subspace

General Case, $A \mathbf x = \mathbf b$

Let $A$ be $m \times n$ matrix of rank $r$

  • $m$ - rows, $n$ - cols
  • $r \leqslant m$, $r \leqslant n$

Full Column Rank

Full rank = $r$ is as big as it can be

  • suppose that $n \leqslant m$, i.e. the number of columns is smaller than the number of rows
  • so full column rank matrix = $r = n$
  • there's a pivot in every column, $n$ pivots $\Rightarrow$ no free variables


Nullspace $N(A)$?

  • there are no free variables to consider
  • so $N(A) = \{ \ \mathbf 0 \ \}$

Solution to $A \mathbf x = \mathbf b$:

  • if the solution exists (i.e. $\mathbf b \in C(A)$) then this solution is unique
  • so we have either 0 solutions or 1

Gaussian Elimination and LU Decomposition

Solving $A\mathbf x = \mathbf b$

  • As metioned earlier - need to reduce it to RREF
  • Can do that with Gaussian Elimination via LU Decomposition
  • Let $LU = A$
  • then $A \mathbf x = \mathbf b$ becomes $LU \mathbf x = \mathbf b$
  • Now let $U \mathbf x = \mathbf y$ and $L \mathbf y = \mathbf x$
  • With this substitution we need to solve two equations now, but they are simple because $L$ is lower-diagonal and $U$ is upper-diagonal

No Solution

No solution?

Full Row Rank

  • Now suppose that $m \leqslant n$ and $r = m$
  • so every row has a pivot, but only $r$ columns have pivots, the remaining $n - r$ don't
  • so there are $r$ pivot columns, and $n - r$ free columns


for which $\mathbf b$ we can solve $A \mathbf x = \mathbf b$?

  • there are no zero rows, so can solve the system for any $\mathbf b$

Full Rank

  • If our $A$ is square, $m = n = r$
  • then there's always a solution, and $A$ is called invertible
  • $N(A) = \{ \ \mathbf 0 \ \}$ - there's only one unique solution

$r < m$ and $r < n$


  • $A \mathbf x = \mathbf 0$ always have a solution - there's always something in the Nullspace $N(A)$ of $A$ apart from the zero-vector
  • reason: there are always free variables and we can assign any non-zero values to them and solve the homogeneous system

Zero or $\infty$ solutions


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