Diagnostic Problems on Fundamental Subspaces

A Four-Subspace Diagnostic

For a matrix \(\mathbf{A}\in\mathbb{R}^{m\times n}\), four subspaces capture different aspects of the same linear system. The column space describes which right-hand sides can be produced by \(\mathbf{A}\mathbf{x}\); the null space describes the solutions of \(\mathbf{A}\mathbf{x}=\mathbf{0}\). The row space records the independent equations, while the left null space consists of vectors that annihilate every column. A useful diagnostic is not just to calculate one of these spaces, but to find all four and check that their dimensions and vectors agree with the matrix.

In each problem, first find the RREF and count its pivots; call that number \(r\). Use the pivot-column indices to select columns from the original matrix for a basis of \(\operatorname{Col}\mathbf{A}\). The nonzero rows of the RREF give a basis of \(\operatorname{Row}\mathbf{A}\). Solve \(\mathbf{A}\mathbf{x}=\mathbf{0}\) for the null space, and solve \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\) for the left null space. The last two calculations are related, but their vectors have different numbers of entries.

Why the Standard Basis Choices Work

These basis rules follow from row reduction, but the column-space rule needs particular care: row operations generally change the columns themselves. The fact that the pivot columns of the original matrix work follows because row reduction applies an invertible operation to every column at once.

Proof. Row reduction consists of reversible elementary row operations, so \(\mathbf{R}=\mathbf{E}\mathbf{A}\) for an invertible matrix \(\mathbf{E}\). Write the columns of \(\mathbf{A}\) and \(\mathbf{R}\) as \(\mathbf{a}_j\) and \(\mathbf{r}_j\); then \(\mathbf{r}_j=\mathbf{E}\mathbf{a}_j\). In RREF, each nonpivot column is a linear combination of the pivot columns, as can be read from its entries in the pivot rows. Thus \(\mathbf{r}_j=\sum_k c_k\mathbf{r}_{p_k}\) for pivot indices \(p_k\). Multiplying by \(\mathbf{E}^{-1}\) gives \(\mathbf{a}_j=\sum_k c_k\mathbf{a}_{p_k}\), so the original pivot columns span \(\operatorname{Col}\mathbf{A}\). If a combination of the original pivot columns is zero, applying \(\mathbf{E}\) gives the same combination of the pivot columns of \(\mathbf{R}\). Those columns are independent: their entries in the distinct pivot rows form the identity pattern, so each coefficient must be zero. Hence the original pivot columns are a basis. For the row-space claim, each elementary row operation replaces rows by linear combinations of rows, and its inverse does the same; therefore row operations preserve row space. The nonzero rows of \(\mathbf{R}\) span it. They are independent because in a linear combination of those rows, the entry in each row’s pivot column is exactly that row’s coefficient. A zero combination therefore has every coefficient zero. \(\square\)

Proof. The pivot-basis result shows that the column space and row space each have dimension \(r\). In the homogeneous system \(\mathbf{A}\mathbf{x}=\mathbf{0}\), there are \(n\) variables and \(r\) pivot variables, so there are \(n-r\) free variables. Giving the free variables arbitrary values determines the pivot variables, and setting one free variable at a time to \(1\), with the others \(0\), gives \(n-r\) independent vectors spanning the solutions. Thus the nullity is \(n-r\). Apply the same free-variable argument to \(\mathbf{A}^{\top}\), which has \(m\) variables. Its rank is \(r\): the row space of \(\mathbf{A}^{\top}\) is the span of the transposes of the columns of \(\mathbf{A}\), so its dimension is \(\dim(\operatorname{Col}\mathbf{A})=r\). Hence \(\dim(\operatorname{Nul}\mathbf{A}^{\top})=m-r\). \(\square\)

One further check is orthogonality. For \(\mathbf{x}\in\mathbb{R}^n\), \(\mathbf{A}\mathbf{x}=\mathbf{0}\) exactly when the dot product of \(\mathbf{x}\) with every row of \(\mathbf{A}\) is zero. Thus \(\operatorname{Nul}\mathbf{A}=(\operatorname{Row}\mathbf{A})^\perp\). Likewise, \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\) exactly when \(\mathbf{y}\) is orthogonal to every column of \(\mathbf{A}\), so \(\operatorname{Nul}\mathbf{A}^{\top}=(\operatorname{Col}\mathbf{A})^\perp\). These statements follow directly from the matrix-vector product, and they provide a useful check on which ambient space each null space belongs to.

Worked Problem: A \(3\times4\) Matrix

Worked Problem: A \(4\times3\) Matrix

Worked Problem: A Rank-One \(2\times3\) Matrix

Common Pitfall: Pivot Columns of Which Matrix?

A frequent error is to take pivot columns from the RREF as a basis for the original column space. Row operations preserve row space and the solutions of a homogeneous system, but they do not generally preserve column space. In the first example, column 1 of the RREF is \((1,0,0)^\top\). It is not in \(\operatorname{Col}\mathbf{A}\): if it were \(\alpha(1,2,1)^\top+\beta(0,1,1)^\top\), the first and third entries would force \(\alpha=1\) and \(\beta=-1\), but the second entry would then be \(2-1=1\), not \(0\). The correction is to use the RREF only to identify pivot-column indices, then take those columns from the original matrix.

A second check is to keep the ambient dimensions visible. For an \(m\times n\) matrix, column-space and left-null vectors have \(m\) entries; row-space and null-space vectors have \(n\) entries. Finally, if there are \(r\) pivots, the dimension checks must read

These checks catch a wrong number of free variables or a left-null calculation performed with the wrong matrix. They do not replace direct substitution: each proposed null vector should still be multiplied by the relevant matrix, and each proposed row relation should be verified.