Homogeneous Systems in Echelon Form

Homogeneous Systems and Their Solutions

A system of linear equations is homogeneous when every equation has zero on the right-hand side. In matrix form it is \(\mathbf{A}\mathbf{x}=\mathbf{0}\), where \(\mathbf{A}\in\mathbb{R}^{m\times n}\), \(\mathbf{x}\in\mathbb{R}^n\), and \(\mathbf{0}\in\mathbb{R}^m\). Unlike a system with a nonzero right-hand side, this system always has at least one solution: choosing every entry of \(\mathbf{x}\) to be zero makes \(\mathbf{A}\mathbf{x}=\mathbf{0}\).

The zero solution is guaranteed, but it need not be the only solution. To describe all solutions, reduce the coefficient matrix to echelon form and use back substitution. A pivot variable corresponds to a pivot column; a variable corresponding to a column without a pivot is a free variable. In the homogeneous case, assigning values to the free variables leaves equations with zero right-hand sides, which determine the pivot variables.

We use echelon form in the following sense: all zero rows, if any, are at the bottom; the first nonzero entry of each nonzero row is its leading entry; and each leading entry is to the right of the leading entry in the row above. A row with leading entry in column \(j\) gives an equation in which the variable \(x_j\) can be solved for using variables to its right.

Why Row Reduction Preserves the Solutions

For a homogeneous system, row-reducing \(\mathbf{A}\) means applying row operations to its equations. The right-hand side remains zero under each operation. The reason this is valid is that each elementary row operation replaces equations by equivalent equations: it does not change which vectors satisfy the whole system.

Proof. It is enough to consider one row operation at a time. Swapping two rows only changes the order in which two equations are written, so it preserves their simultaneous solutions. Multiplying one row by a nonzero scalar replaces an equation by a nonzero multiple of that equation; a vector satisfies one equation exactly when it satisfies its nonzero multiple. Replacing a row by itself plus a scalar multiple of another row also preserves solutions: any vector satisfying both original equations satisfies their stated combination, and the original changed row can be recovered by subtracting the same multiple of the unchanged row. Thus each operation preserves the solution set. Applying this argument successively proves the claim for any finite sequence of row operations. \(\square\)

Consequently, we can solve the reduced system instead of the original one. In a homogeneous system, a zero row gives \(0=0\), which imposes no additional condition. The nonzero rows supply the equations used in back substitution.

Reading Every Solution from Echelon Form

Proof. Start with the last nonzero row. Its leading variable has a nonzero coefficient, and every other variable appearing in that row is to its right. Those variables are free variables, or pivot variables that occur later in the order of the rows. Once their values are set, the equation determines the leading variable uniquely by division by its nonzero coefficient. Move to the row above and repeat: the variables to the right have already been assigned, so that row uniquely determines its leading variable. Continuing upward determines every pivot variable. This constructs a solution for the chosen free-variable values. It also proves uniqueness: each pivot variable is forced by its row after the variables to its right are fixed. Any zero rows impose only \(0=0\), so they do not obstruct the construction. \(\square\)

This result gives a practical method: mark the pivot variables, name the free variables with parameters, and substitute upward from the bottom row. The resulting parameterized vector describes all solutions, because every solution must use some values for its free variables, and those values determine the pivot variables uniquely.

Common Mistakes and a Reliable Routine

Mistake: assuming a homogeneous system has only the zero solution. In the first worked example, the system is homogeneous, but \((15,-7,1)^{\top}\) is a nonzero solution. “Homogeneous” means that the right-hand side is zero; it does not, by itself, say that all entries of every solution must be zero. To decide what the solutions are, use the equations in echelon form.

Mistake: reversing a sign during back substitution. In the first example, the last equation is \(x_2+7x_3=0\). A wrong calculation is \(x_2=7x_3\). Moving \(7x_3\) to the other side gives \(x_2=-7x_3\), not \(7x_3\). Checking in the equation catches the error: \(7t+7t=14t\), which is generally not zero, whereas \(-7t+7t=0\).

Mistake: assigning arbitrary values to pivot variables. In the second example, \(x_1\) and \(x_2\) are constrained by the equations. The parameters belong to \(x_3\) and \(x_4\), the variables in columns without pivots. Once those values are chosen, back substitution fixes the pivot variables.

The check is useful even when the row reduction is correct: it verifies both the signs in the parameter formulas and that the formulas satisfy the original equations. Keep the free parameters in the final answer, since setting them all to zero records only the zero solution rather than describing the full solution set.