Linear Equations

Basis
Last updated: Tags: Linear Algebra

Virtually every computational problem in science and engineering can be reduced to “find values satisfying these linear constraints.” The temperature distribution in a heated rod, the currents in an electrical circuit, the prices that clear a market — all become systems of linear equations. The matrix form of a linear system is the lens that makes these problems tractable.

Systems of linear equations

A system of mm linear equations in nn unknowns x1,,xnx_1, \ldots, x_n over a field FF has the form:

{a11x1+a12x2++a1nxn=b1a21x1+a22x2++a2nxn=b2am1x1+am2x2++amnxn=bm\begin{cases} a_{11} x_1 + a_{12} x_2 + \cdots + a_{1n} x_n = b_1 \\ a_{21} x_1 + a_{22} x_2 + \cdots + a_{2n} x_n = b_2 \\ \quad \vdots \\ a_{m1} x_1 + a_{m2} x_2 + \cdots + a_{mn} x_n = b_m \end{cases}

where the coefficients aijFa_{ij} \in F and the right-hand sides biFb_i \in F are given, and you want to find all tuples (x1,,xn)Fn(x_1, \ldots, x_n) \in F^n that satisfy all mm equations simultaneously. Such a tuple is called a solution, and the set of all solutions is the solution set.

Matrix form: Ax=bAx = b

Packaging the coefficients, unknowns, and right-hand sides into matrices turns the system into the compact equation

Ax = b, \tag{1}

where:

  • AMm,n(F)A \in M_{m,n}(F) is the coefficient matrix with (A)ij=aij(A)_{ij} = a_{ij},
  • x=(x1,,xn)Fnx = (x_1, \ldots, x_n)^\top \in F^n is the unknown vector (a column vector),
  • b=(b1,,bm)Fmb = (b_1, \ldots, b_m)^\top \in F^m is the right-hand-side vector.

The product AxAx gives back exactly the left-hand side of each equation. So solving the system is the same as finding all xFnx \in F^n for which AxAx equals bb.

The augmented matrix

Rather than manipulate the two sides of (1) separately, you combine them into a single matrix. The augmented matrix is

[Ab]Mm,n+1(F),[A \mid b] \in M_{m,n+1}(F),

formed by appending bb as an extra column to the right of AA. A vertical bar is drawn between the nn-th and (n+1)(n+1)-th columns as a visual reminder that the last column is the right-hand side.

Applying Gauss-Jordan elimination to [Ab][A \mid b] performs exactly the elementary row operations needed to simplify the system, while keeping track of both sides at once. The operations do not change the solution set, so the RREF of [Ab][A \mid b] gives an equivalent system from which solutions can be read off directly.

Reading off solutions from RREF

After reducing [Ab][A \mid b] to RREF, let [Rc][R \mid c] be the result. The variables corresponding to pivot columns of RR are basic variables (their values are forced once you choose the free variables), and variables corresponding to free columns are free variables (they can take any value in FF).

There are exactly three possible outcomes:

Outcome 1: Inconsistent (no solution)

The RREF [Rc][R \mid c] contains a row of the form

(0001),\begin{pmatrix} 0 & 0 & \cdots & 0 & \mid & 1 \end{pmatrix},

which represents the equation 0=10 = 1. This is impossible, so the system has no solution. Geometrically, the hyperplanes defined by the equations do not share a common point.

Outcome 2: Unique solution

The RREF has no all-zero rows in AA (each column of RR is a pivot column) and no pivot in the bb-column. Every variable is a basic variable — there are no free variables. You can read off the unique values x1,,xnx_1, \ldots, x_n directly from the last column of the RREF.

Outcome 3: Infinitely many solutions

At least one column of RR is a free column (at least one free variable), and there is no pivot in the bb-column. Each free variable can be set to an arbitrary value in FF, producing a distinct solution. The complete solution set is a particular solution plus any element of the kernel of AA (see Kernel):

x=xparticular+xhomogeneous,xhomogeneousker(A).x = x_{\text{particular}} + x_{\text{homogeneous}}, \qquad x_{\text{homogeneous}} \in \ker(A).

The particular solution is any single xx that satisfies Ax=bAx = b; the kernel part accounts for all the “freedom.”

The homogeneous system

The special case b=0b = \mathbf{0} gives the homogeneous system Ax=0Ax = \mathbf{0}. It is always consistent: x=0x = \mathbf{0} is always a solution (called the trivial solution). The solution set is exactly ker(A)\ker(A), the kernel of AA.

If the homogeneous system has only the trivial solution, the coefficient matrix has no free columns (all columns are pivot columns). If it has nontrivial solutions, then there is at least one free column, and the kernel is more than just {0}\{\mathbf{0}\}.

The homogeneous system plays a central role in understanding the structure of the solution set of the inhomogeneous system Ax=bAx = b: any two solutions of Ax=bAx = b differ by an element of ker(A)\ker(A).

Summary

  • A system of mm linear equations in nn unknowns is compactly written as Ax=bAx = b with AMm,n(F)A \in M_{m,n}(F).
  • The augmented matrix [Ab][A \mid b] packages both sides; applying Gauss-Jordan to it reduces the system without changing its solution set.
  • There are exactly three outcomes: inconsistent (a pivot appears in the bb-column), unique solution (no free variables), or infinitely many solutions (at least one free variable).
  • The homogeneous system Ax=0Ax = \mathbf{0} is always consistent; its solution set is ker(A)\ker(A), which is explored in Kernel.