Calculating the kernel of a matrix - An example [dark version]

A 3×4 real matrix’s kernel is found by turning the equation A x = 0 into a solvable system of linear equations and then reading off the free parameters from row-reduced (triangular) form. In this worked example, the kernel is a two-dimensional subspace of R4, meaning there are two independent directions of vectors that the matrix sends to the zero vector.

The kernel of a matrix A is defined as the set of all vectors x in R4 (because A has four columns) such that A x equals the zero vector in R3. Concretely, this becomes the homogeneous linear system A x = 0. Solving it is done by applying row operations to the augmented system (with right-hand side zero). Row operations preserve the solution set, so the kernel can be computed by transforming the matrix into an echelon/triangular form where entries below the diagonal are zero.

The calculation proceeds by eliminating the first column’s lower entries using combinations of rows: the second row is adjusted by subtracting 3 times the first row, and the third row is adjusted by subtracting 4 times the first row. This creates zeros in the first column below the pivot position. Next, the second column is cleaned up by using the new second row to eliminate the third row’s second-column entry. The third row ends up becoming all zeros, which is consistent with it being a multiple of the second row.

Once the matrix reaches echelon form, the system splits into equations involving pivot variables and free variables. The free variables are X3 and X4 (the “bound variables” are those tied to pivots). From the second row, the homogeneous equation is -2·X2 − 2·X3 + X4 = 0. To parameterize solutions, X3 and X4 are set to real parameters: X3 = α and X4 = β. Solving the second-row equation gives X2 = −α + (1/2)β.

The first row then provides the remaining relationship among X1, X2, X3, and X4. Substituting X2 = −α + (1/2)β, X3 = α, and X4 = β into the first-row equation yields X1 = −(3/2)α − (1/2)β (equivalently, the transcript presents it as X1 = −3/2·α − 1/2·β). Therefore every kernel vector has the form (x1, x2, x3, x4) = (−3/2·α − 1/2·β, −α + 1/2·β, α, β), with α, β ranging over the real numbers.

Finally, the solution set is expressed as a linear combination of two vectors—one multiplying α and one multiplying β. The kernel is written as the span of two linearly independent vectors in R4, making the kernel exactly a two-dimensional subspace. The two basis vectors are extracted from the coefficients of α and β, and their linear independence confirms they form a basis for the kernel in this example.