3.3 Homogeneous Systems of Linear Equations

Definition 3.3.1. A homogeneous system of linear equations is a system of the form; \begin {align*} a_{11}x_1+a_{12}x_2+\cdots +a_{1n}x_n &=0\\ a_{21}x_1+a_{22}x_2+\cdots +a_{2n}x_n &=0\\ \vdots \\ a_{m1}x_1+a_{m2}x_2+\cdots +a_{mn}x_n &=0 \end {align*}

In shortened notation, this is denoted by \(\sum \limits ^n_{j=1}a_{ij}x_j=0\) \((i=1,2,\dots , m)\)

The \(n\)-tuple \((0,0,0,\dots ,0)\) is always a solution of the homogeneous system. It is called the trivial solution of the system.

Example 3.3.2. The system of linear equations \begin {align*} x+y+z &=0\\ 2x-3y-z&=0\\ 4x+5y+2z &=0 \end {align*}

is a homogeneous system.

Theorem 3.3.3.

1.: If the number of equations is less than the number of variables in a homogeneous system of linear equations, then the system has a non-trivial solution.
2.: A system of \(n\) homogeneous linear equations in \(n\) variables has a non-trivial solution if and only if \(R\neq I_n\), where \(R\) is the reduced echelon matrix of the coefficient matrix of the system.

Proof. Both parts turn on counting leading elements. Reduce the augmented matrix \((A|0)\) to reduced echelon form. Row operations combine rows, so a column of zeros stays a column of zeros: the augmented column remains \(0\) throughout, and the reduced augmented matrix is \((R|0)\) where \(R\) is the reduced echelon matrix of \(A\). In particular there is never a leading element in the last column, which by Theorem 3.2.3(3) confirms what we already knew — a homogeneous system is always consistent, the trivial solution being one.

Let \(r\) be the number of leading elements of \(R\). Each occupies its own row and its own column, so the \(n\) variables split into \(r\) leading variables and \(n-r\) free ones, and by Theorem 3.2.3(2) the original system is equivalent to the reduced one. Assigning arbitrary values to the free variables determines the leading variables uniquely, so: \[\text {the system has a non-trivial solution}\iff n-r>0 \iff r<n,\] since if \(n-r>0\) we may set one free variable to \(1\) and obtain a solution which is not all zeros, while if \(r=n\) every variable is leading and the reduced equations read \(x_1=0,\dots ,x_n=0\).

1.: Each leading element lies in its own row, so \(r\leq m\), the number of equations. If \(m<n\) then \(r\leq m<n\), so \(r<n\) and a non-trivial solution exists.
2.: Here \(R\) is \(n\) by \(n\). If \(R=I_n\) then \(r=n\) and only the trivial solution exists. If \(R\neq I_n\) then \(r<n\) — a square reduced echelon matrix with a leading element in every row is the identity — so a non-trivial solution exists.

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Remark. Part (1) is the formal version of a fact worth carrying: more unknowns than equations always leaves room to move. It is the same statement as Theorem 6.2.8(ii), seen from the matrix side rather than the linear-map side.

Example 3.3.4. Find the values of \(t\) for which the homogeneous system \begin {align*} x+5y+3z &=0\\5x+y-z &=0\\x+2y+tz &=0 \end {align*}

has a non-trivial solution.

Solution. \begin {align*} \begin {pmatrix} 1&5&3&|&0\\5&1&-1&|&0\\1&2&t&|&0\\ \end {pmatrix} & \hspace {0.3cm} \begin {matrix} r_2\rightarrow r_2-5r_1\\ \longrightarrow \\ r_3\rightarrow r_3-r_1\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&5&3&|&0\\0&-24&-16&|&0\\ 0&-3&t-3&|&0\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_2\rightarrow -\frac {1}{8}r_2\\ \longrightarrow \\ r_3\rightarrow -r_3\\ \end {matrix}\\\\ & \hspace {0.3cm} \begin {pmatrix} 1&5&3&|&0\\ 0&3&2&|&0\\0&3&-t+3&|&0\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_3\rightarrow r_3-r_2\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&5&3&|&0\\ 0&3&2&|&0\\0&0&-t+1&|&0\\ \end {pmatrix} \end {align*}

\(\therefore \) The system has a non-trivial solution when \(t=1\).

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