One Question, Several Tests
For a square matrix, invertibility can be checked in several ways. Row reduction counts pivots; rank records that count; column independence asks whether the columns contain redundancy; and the null spaces test whether the matrix or its transpose has a nonzero vector that it sends to zero. These are not separate coincidences. For an \(n\times n\) matrix, each test gives the same answer, and the connections explain how to interpret a computation rather than merely read off a yes-or-no result.
Throughout, matrices and vectors have real entries. The rank of \(\mathbf{A}\in\mathbb{R}^{n\times n}\) is the dimension of its column space, the space spanned by its columns. The null space is \(\operatorname{Nul}\mathbf{A}=\{\mathbf{x}\in\mathbb{R}^n:\mathbf{A}\mathbf{x}=\mathbf{0}\}\). The left null space is \(\operatorname{Nul}\mathbf{A}^{\top}=\{\mathbf{y}\in\mathbb{R}^n:\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\}\). A matrix is invertible if there is a matrix \(\mathbf{B}\) such that \(\mathbf{A}\mathbf{B}=\mathbf{B}\mathbf{A}=\mathbf{I}_n\).
In row reduction, a pivot is a leading entry in the reduced row echelon form (RREF). The rank equals the number of pivots. For an \(n\times n\) matrix, having \(n\) pivots means that every column is a pivot column and the RREF is \(\mathbf{I}_n\). We will also use the Matrix Rank-Nullity Theorem, which states that \(\operatorname{rank}\mathbf{A}+\operatorname{nullity}\mathbf{A}=n\), and the standard result that row rank equals column rank, so \(\operatorname{rank}\mathbf{A}^{\top}=\operatorname{rank}\mathbf{A}\).
Proof. Suppose first that \(\mathbf{A}\) is invertible and \(\mathbf{A}\mathbf{x}=\mathbf{0}\). Multiplying by \(\mathbf{A}^{-1}\) gives \(\mathbf{x}=\mathbf{A}^{-1}\mathbf{0}=\mathbf{0}\). Thus the null space contains only the zero vector. Conversely, suppose \(\operatorname{Nul}\mathbf{A}=\{\mathbf{0}\}\). Its nullity is zero, so rank-nullity gives \(\operatorname{rank}\mathbf{A}=n\). Its RREF therefore has \(n\) pivots and is \(\mathbf{I}_n\). Each row operation is multiplication on the left by an invertible elementary matrix, so the sequence of row operations gives an invertible matrix \(\mathbf{E}\) with \(\mathbf{E}\mathbf{A}=\mathbf{I}_n\). Hence \(\mathbf{A}=\mathbf{E}^{-1}\), which is invertible. \(\square\)
The Equivalent Tests
- \(\mathbf{A}\) is invertible.
- The RREF of \(\mathbf{A}\) has a pivot in every column (equivalently, in every row).
- \(\operatorname{rank}\mathbf{A}=n\).
- The columns of \(\mathbf{A}\) are linearly independent.
- \(\operatorname{Nul}\mathbf{A}=\{\mathbf{0}\}\).
- \(\operatorname{Nul}\mathbf{A}^{\top}=\{\mathbf{0}\}\).
Proof. The lemma gives equivalence between invertibility and a trivial null space. By rank-nullity, \(\operatorname{Nul}\mathbf{A}=\{\mathbf{0}\}\) if and only if \(\operatorname{rank}\mathbf{A}=n\). Rank is the number of pivots, so rank \(n\) is equivalent to having \(n\) pivots in the \(n\) columns; in an \(n\times n\) matrix this is also a pivot in every row. The columns are linearly independent exactly when the equation \(\mathbf{A}\mathbf{x}=\mathbf{0}\), whose coefficients are the column weights, has only the solution \(\mathbf{x}=\mathbf{0}\). This is precisely the trivial-null-space condition. Finally, rank-nullity applied to \(\mathbf{A}^{\top}\) gives \(\operatorname{Nul}\mathbf{A}^{\top}=\{\mathbf{0}\}\) if and only if \(\operatorname{rank}\mathbf{A}^{\top}=n\). Since row rank equals column rank, \(\operatorname{rank}\mathbf{A}^{\top}=\operatorname{rank}\mathbf{A}\), so this is equivalent to \(\operatorname{rank}\mathbf{A}=n\). All six conditions are therefore equivalent. \(\square\)
The square-matrix hypothesis matters. For a rectangular matrix, having a pivot in every column and having a pivot in every row are different requirements, and the matrix cannot be invertible in the two-sided sense defined above. Here, the same number \(n\) of rows and columns allows full rank to mean both full column rank and full row rank.
Worked Tests
Worked Example: An Invertible Matrix
Test \(\mathbf{A}=\begin{bmatrix}1&1&0\\0&1&1\\1&0&1\end{bmatrix}\) using pivots, rank, column independence, and both null spaces.
Subtract row 1 from row 3, add row 2 to the new row 3, scale row 3 by \(\tfrac12\), then clear the entries above the third and second pivots:
There are three pivots, so \(\operatorname{rank}\mathbf{A}=3\). Every column is a pivot column, and the columns are independent. The homogeneous system has only the zero solution, and the theorem also gives a trivial left null space. To check the invertibility directly, take
Multiplying row by row gives
Check: A left inverse of a square matrix is its inverse: in this case, the row reduction already showed that \(\mathbf{A}\) is invertible, and the displayed product verifies that the stated \(\mathbf{B}\) is its inverse. The null-space tests are also consistent: the RREF gives \(\mathbf{A}\mathbf{x}=\mathbf{0}\) only \(\mathbf{x}=\mathbf{0}\); the transpose has rank \(3\), so rank-nullity gives \(\operatorname{Nul}\mathbf{A}^{\top}=\{\mathbf{0}\}\).
Worked Example: A Singular Matrix and Its Two Null Spaces
Consider \(\mathbf{C}=\begin{bmatrix}1&2&3\\2&4&6\\0&1&1\end{bmatrix}\). Row reduction gives
There are two pivots, so the rank is \(2\), not \(3\), and column 3 is not a pivot column. The RREF system gives \(x_1=-x_3\) and \(x_2=-x_3\). Taking \(x_3=1\) produces \(\mathbf{x}=(-1,-1,1)^{\top}\) in the null space. Substitution checks it:
For the left null space, reduce the transpose, not \(\mathbf{C}\):
The equations are \(y_1+2y_2=0\) and \(y_3=0\). Setting \(y_2=1\) gives \(\mathbf{y}=(-2,1,0)^{\top}\). Indeed,
Thus both null spaces contain nonzero vectors, and the matrix is singular. Rank-nullity also confirms that each null space has dimension \(3-2=1\), using \(\operatorname{rank}\mathbf{C}^{\top}=\operatorname{rank}\mathbf{C}=2\).
Worked Example: A Parameter Changes the Test
For which real values of \(s\) is \(\mathbf{A}_s=\begin{bmatrix}1&1\\1&s\end{bmatrix}\) invertible?
Subtract row 1 from row 2:
If \(s\ne1\), the second row has a pivot. Scaling that row by \(1/(s-1)\) and subtracting the new row 2 from row 1 gives
Thus there are two pivots and the matrix is invertible. If \(s=1\), the reduced matrix is \(\begin{bmatrix}1&1\\0&0\end{bmatrix}\), which has rank \(1\), so it is singular. In that case, the equation gives \(x_1=-x_2\); choosing \(x_2=1\) gives the nonzero null vector \((-1,1)^{\top}\), and \(\mathbf{A}_1(-1,1)^{\top}=(0,0)^{\top}\).
Check: For \(s\ne1\), the claimed inverse is \(\frac{1}{s-1}\begin{bmatrix}s&-1\\-1&1\end{bmatrix}\). Multiplication yields
A Common Mistake: Confusing the Two Null Spaces
The null space and left null space answer different questions. A vector \(\mathbf{x}\) belongs to \(\operatorname{Nul}\mathbf{A}\) when it gives a dependence relation among the columns of \(\mathbf{A}\). A vector \(\mathbf{y}\) belongs to \(\operatorname{Nul}\mathbf{A}^{\top}\) when it gives a dependence relation among the rows of \(\mathbf{A}\), because \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\) is equivalent to \(\mathbf{y}^{\top}\mathbf{A}=\mathbf{0}^{\top}\).
A tempting but wrong calculation for the matrix \(\mathbf{C}\) above is to solve \(\mathbf{C}\mathbf{x}=\mathbf{0}\), obtain \((-1,-1,1)^{\top}\), and call it a left-null vector. It is not: direct substitution gives
The correction is to solve the transposed system. Its left-null vector \((-2,1,0)^{\top}\) does satisfy \(\mathbf{C}^{\top}(-2,1,0)^{\top}=\mathbf{0}\), as checked above. Invertibility makes both null spaces trivial, but for a singular matrix their vectors need not be the same.
Using the Tests Efficiently
For a square matrix, choose the test that best matches the information available. If the matrix is already reduced, count pivots. If the question concerns whether columns are redundant, solve the homogeneous system. If a nonzero null vector is easy to spot, it immediately proves singularity. To test the left null space, work with the transpose or use rank equality and rank-nullity. A single failed condition is enough to conclude that the matrix is not invertible; the theorem says all the others fail as well.
Keep the hypotheses in view: the equivalence here concerns square matrices. For a rectangular matrix, a trivial null space means its columns are independent, but does not by itself make the matrix invertible. The square setting is what lets full column rank, full row rank, and an inverse coincide.
Check Your Understanding
For each question, connect the computation to the equivalent conditions rather than relying on a single label.
- A \(4\times4\) matrix has three pivots. What are its rank and nullity, and can it be invertible?
- If the columns of a square matrix are linearly independent, what does rank-nullity imply about its null space?
- For a square matrix of rank \(n-1\), what is the dimension of its left null space?
- Explain why row-reducing \(\mathbf{A}\) tests \(\operatorname{Nul}\mathbf{A}\), while row-reducing \(\mathbf{A}^{\top}\) tests \(\operatorname{Nul}\mathbf{A}^{\top}\).
- For \(\begin{bmatrix}1&1\\1&s\end{bmatrix}\), identify the parameter value at which the second pivot disappears and give a nonzero null vector there.
