Zero Eigenvalues and Failure of Invertibility
An eigenvalue describes a scalar action of a matrix on at least one nonzero vector. For the scalar \(0\), the eigenvalue equation becomes \(\mathbf{A}\mathbf{x}=\mathbf{0}\). Thus, the question of whether zero is an eigenvalue asks whether this homogeneous system has a nonzero solution. For a square matrix, that question is also exactly the question of whether the matrix is invertible.
Throughout, let \(\mathbb{F}\) be either \(\mathbb{R}\) or \(\mathbb{C}\), and let \(\mathbf{A}\in\mathbb{F}^{n\times n}\), where \(n\ge 1\). The null space of \(\mathbf{A}\) is the set of all vectors \(\mathbf{x}\in\mathbb{F}^n\) satisfying \(\mathbf{A}\mathbf{x}=\mathbf{0}\). A square matrix is invertible if there is a matrix \(\mathbf{A}^{-1}\) with \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}_n\); otherwise it is singular.
The definition of eigenvalue requires a nonzero vector; the zero vector is included in an eigenspace because every vector space contains its zero vector. The null space \(\operatorname{Nul}\mathbf{A}\) also includes \(\mathbf{0}\). The key question is whether it contains any other vector.
The Null Space Tests Invertibility
Before connecting eigenvalues and singularity, we establish the square-matrix fact that makes the connection precise. Row operations preserve the solutions of a homogeneous system. In the reduced row echelon form (RREF), each variable without a pivot is free. In a square system, if every column has a pivot, there is a pivot in every row as well, so the RREF is the identity matrix.
Proof. First suppose \(\mathbf{A}\) 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 \(\mathbf{0}\).
Conversely, suppose \(\operatorname{Nul}\mathbf{A}=\{\mathbf{0}\}\). Row-reduce \(\mathbf{A}\) to its RREF. If any column had no pivot, the corresponding variable would be free in the homogeneous system; setting one free variable to \(1\) and the other free variables to \(0\) would produce a nonzero solution. Since row operations preserve the solution set, that would contradict the assumption. Hence every one of the \(n\) columns has a pivot. There are \(n\) rows, so the RREF is \(\mathbf{I}_n\). The row operations are reversible, so their combined action is multiplication by an invertible matrix \(\mathbf{E}\); therefore \(\mathbf{E}\mathbf{A}=\mathbf{I}_n\), and \(\mathbf{A}=\mathbf{E}^{-1}\) is invertible. \(\square\)
The square-matrix condition matters in this result. The proof uses the fact that \(n\) pivot columns in a matrix with \(n\) rows give a pivot in every row and hence RREF \(\mathbf{I}_n\). For a rectangular matrix, having no nonzero solution to \(\mathbf{A}\mathbf{x}=\mathbf{0}\) does not by itself mean that the matrix has a two-sided inverse.
Worked Example: A Singular Matrix and Its Null Space
Consider the square matrix \(\mathbf{A}=\begin{bmatrix}1&2&-1\\2&4&-2\\0&1&1\end{bmatrix}\). Find its null space and determine whether zero is an eigenvalue.
Row-reduce the matrix of the homogeneous system. First use \(R_2\to R_2-2R_1\), then exchange rows 2 and 3:
Now use \(R_1\to R_1-2R_2\):
The equations are \(x_1-3x_3=0\) and \(x_2+x_3=0\). Set the free variable \(x_3=t\). Then \(x_1=3t\), \(x_2=-t\), and
In particular, the null space contains a nonzero vector, so the matrix is singular by the theorem. Taking \(t=1\), direct substitution checks the claimed vector:
The vector is nonzero, so it is an eigenvector for eigenvalue \(0\). In fact, every vector in the displayed null space is in the zero-eigenspace.
The Equivalence and the Zero-Eigenspace
Proof. By definition, \(0\) is an eigenvalue of \(\mathbf{A}\) exactly when there is a vector \(\mathbf{x}\ne\mathbf{0}\) satisfying \(\mathbf{A}\mathbf{x}=0\mathbf{x}=\mathbf{0}\). This says exactly that \(\operatorname{Nul}\mathbf{A}\ne\{\mathbf{0}\}\). By the theorem above, for a square matrix this is equivalent to \(\mathbf{A}\) not being invertible, which is the definition of singularity. Each implication uses the same equivalences in reverse, so the statement holds in both directions. \(\square\)
Proof. By definition, \(E_0\) consists of the vectors \(\mathbf{x}\) satisfying \(\mathbf{A}\mathbf{x}=0\mathbf{x}\). Since \(0\mathbf{x}=\mathbf{0}\), this condition is precisely \(\mathbf{A}\mathbf{x}=\mathbf{0}\), which defines \(\operatorname{Nul}\mathbf{A}\). Thus the two sets are equal. The zero vector belongs to both sets, and the nonzero vectors in either set are exactly the eigenvectors for eigenvalue \(0\). \(\square\)
When \(\mathbf{A}\) is invertible, its null space is \(\{\mathbf{0}\}\), so there is no nonzero zero-eigenvector and \(0\) is not an eigenvalue. Some conventions still use \(E_0\) for the solution set of \(\mathbf{A}\mathbf{x}=\mathbf{0}\) in this case; that set is just \(\{\mathbf{0}\}\), and zero is not an eigenvalue because it has no nonzero eigenvector.
Worked Example: An Invertible Matrix Has No Zero Eigenvalue
Let \(\mathbf{B}=\begin{bmatrix}1&2\\3&5\end{bmatrix}\). Solve \(\mathbf{B}\mathbf{x}=\mathbf{0}\) and verify that \(\mathbf{B}\) is invertible.
Reduce the coefficient matrix using \(R_2\to R_2-3R_1\), \(R_2\to -R_2\), and \(R_1\to R_1-2R_2\):
The reduced homogeneous system gives \(x_1=0\) and \(x_2=0\). Therefore \(\operatorname{Nul}\mathbf{B}=\{\mathbf{0}\}\), and the square-matrix theorem says \(\mathbf{B}\) is invertible. An inverse is \(\mathbf{B}^{-1}=\begin{bmatrix}-5&2\\3&-1\end{bmatrix}\). Verify by multiplication on both sides:
There is no nonzero solution to \(\mathbf{B}\mathbf{x}=\mathbf{0}\), so \(0\) is not an eigenvalue. This illustrates why “zero is a scalar” is not enough: the eigenvalue definition also requires a nonzero vector on which the matrix acts as multiplication by that scalar.
Worked Example: A Larger Zero-Eigenspace
Let \(\mathbf{C}=\begin{bmatrix}1&2&3\\2&4&6\\0&0&0\end{bmatrix}\). Find the full zero-eigenspace and check two vectors that span it.
Use \(R_2\to R_2-2R_1\). The resulting homogeneous equations reduce to one equation:
Choose \(x_2=s\) and \(x_3=t\). Then \(x_1=-2s-3t\), so every solution is
Both spanning vectors satisfy the equation. Indeed,
Conversely, the row-reduced equation forces every solution to have the stated form, so these vectors describe the entire null space, not just two examples from it. Since the null space contains nonzero vectors, \(\mathbf{C}\) is singular and \(0\) is an eigenvalue; its zero-eigenspace is exactly the set of solutions just found.
Why the Connection Is Useful
The equivalence lets one answer two questions with one homogeneous system. If a nonzero solution to \(\mathbf{A}\mathbf{x}=\mathbf{0}\) exists, it is both evidence that \(\mathbf{A}\) is singular and an eigenvector for eigenvalue \(0\). If row reduction gives only the trivial solution, then a square \(\mathbf{A}\) is invertible and has no zero eigenvalue.
The null space also describes how solutions to an inhomogeneous system can fail to be unique. If \(\mathbf{x}_0\) solves \(\mathbf{A}\mathbf{x}=\mathbf{b}\) and \(\mathbf{v}\in\operatorname{Nul}\mathbf{A}\), then
Thus, when the null space contains a nonzero vector, any existing solution has a distinct companion solution. This does not guarantee that a solution exists for every \(\mathbf{b}\); it says that if one exists, a nontrivial null space prevents it from being unique. An invertible square matrix, by contrast, has zero null space and gives a unique solution for each right-hand side.
Common Pitfall: The Zero Vector Is Not an Eigenvector
A common incorrect argument is: “For every matrix \(\mathbf{A}\), \(\mathbf{A}\mathbf{0}=\mathbf{0}\), so every matrix has eigenvalue \(0\).” The calculation is true, but the conclusion does not follow because an eigenvector must be nonzero. For example, take \(\mathbf{A}=\mathbf{I}_2\). The equation \(\mathbf{I}_2\mathbf{x}=\mathbf{0}\) gives \(x_1=0\) and \(x_2=0\), so its only solution is the zero vector. The matrix is invertible, its null space is \(\{\mathbf{0}\}\), and it has no zero eigenvalue.
The correction is to check both parts of the eigenvalue definition: the equation \(\mathbf{A}\mathbf{x}=0\mathbf{x}\) must hold, and the vector \(\mathbf{x}\) must be nonzero. A nontrivial solution to the homogeneous system supplies such a vector; the trivial solution alone never does.
Check Your Understanding
Use the definition of eigenvalue and the square-matrix criterion to answer each question.
- Explain why a nonzero solution of \(\mathbf{A}\mathbf{x}=\mathbf{0}\) proves both that \(\mathbf{A}\) is singular and that \(0\) is an eigenvalue.
- For \(\mathbf{D}=\begin{bmatrix}1&0&0\\0&0&0\\0&0&2\end{bmatrix}\), find \(\operatorname{Nul}\mathbf{D}\) and give a nonzero zero-eigenvector.
- Suppose a square matrix has \(\operatorname{Nul}\mathbf{A}=\{\mathbf{0}\}\). What does this tell you about invertibility and the eigenvalue \(0\)?
- For \(\mathbf{C}\) in the third worked example, verify that \(\begin{bmatrix}-2\\1\\0\end{bmatrix}+\begin{bmatrix}-3\\0\\1\end{bmatrix}\) is in its zero-eigenspace.
- Identify the error in the claim “\(\mathbf{A}\mathbf{0}=\mathbf{0}\), so \(\mathbf{0}\) is an eigenvector for eigenvalue \(0\).”
