The Eigenvalue Equation And Singularity

From Scalar Action to a Homogeneous Equation

An eigenvalue describes a scalar action of a matrix on at least one nonzero vector. The key equation is \( \mathbf{A}\mathbf{x}=\lambda\mathbf{x} \): applying \(\mathbf{A}\) to \(\mathbf{x}\) has the same effect as multiplying \(\mathbf{x}\) by the scalar \(\lambda\). Moving the right-hand side to the left produces a homogeneous system. This change of form connects eigenvalues to a basic question about square matrices: does the associated homogeneous system have a nonzero solution?

Throughout this tutorial, \(\mathbf{A}\) is an \(n\times n\) real matrix, \(\lambda\in\mathbb{R}\), and vectors are columns in \(\mathbb{R}^n\). The identity matrix is \(\mathbf{I}_n\). Recall that \(\mathbf{x}\) is an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\) when \(\mathbf{x}\ne\mathbf{0}\) and \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). The condition \(\mathbf{x}\ne\mathbf{0}\) matters: the zero vector satisfies the equation for every scalar, but it cannot identify any eigenvalue.

Subtracting \(\lambda\mathbf{x}\) from both sides of the eigenvalue equation gives \(\mathbf{A}\mathbf{x}-\lambda\mathbf{x}=\mathbf{0}\). Since \(\lambda\mathbf{x}=(\lambda\mathbf{I}_n)\mathbf{x}\), this becomes \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\). The matrix \(\mathbf{A}-\lambda\mathbf{I}_n\) is square, so we can ask whether its homogeneous system has any nonzero solution. To connect that question rigorously to singularity, first establish the relevant criterion for square matrices.

Singularity and Nonzero Null Vectors

Proof. Suppose first that \(\operatorname{Nul}\mathbf{B}\) is nontrivial. Then there is a vector \(\mathbf{x}\ne\mathbf{0}\) such that \(\mathbf{B}\mathbf{x}=\mathbf{0}\). If \(\mathbf{B}\) had an inverse, multiplying the equation by \(\mathbf{B}^{-1}\) would give \(\mathbf{x}=\mathbf{B}^{-1}\mathbf{0}=\mathbf{0}\), a contradiction. Thus \(\mathbf{B}\) is singular.

For the converse, suppose \(\operatorname{Nul}\mathbf{B}\) is trivial, so the only solution of \(\mathbf{B}\mathbf{x}=\mathbf{0}\) is \(\mathbf{x}=\mathbf{0}\). Row-reduce \(\mathbf{B}\) to reduced row echelon form. Elementary row operations preserve the solutions of a homogeneous system: swapping equations, multiplying an equation by a nonzero scalar, or adding a multiple of one equation to another does not change which vectors satisfy all the equations. If a column of the reduced matrix had no pivot, its variable would be free. Setting that variable to \(1\), setting any other free variables to \(0\), and solving for the pivot variables would give a nonzero solution. The assumed triviality therefore means every column has a pivot. Since \(\mathbf{B}\) is \(n\times n\), its reduced row echelon form is then \(\mathbf{I}_n\). Thus a sequence of elementary row operations takes \(\mathbf{B}\) to \(\mathbf{I}_n\). Each row operation is multiplication on the left by an invertible elementary matrix; their product \(\mathbf{E}\) is invertible and satisfies \(\mathbf{E}\mathbf{B}=\mathbf{I}_n\). Hence \(\mathbf{B}=\mathbf{E}^{-1}\) is invertible. Therefore, if \(\mathbf{B}\) is singular, its null space cannot be trivial. This proves both directions. \(\square\)

The square-matrix hypothesis is essential to the row-reduction argument that full pivots give the identity matrix. The theorem concerns a matrix’s null space, not its row space or column space. Its practical content is direct: a nonzero solution to a homogeneous system rules out an inverse, while a trivial null space for a square matrix guarantees an inverse.

Proof. By definition, \(\lambda\) is an eigenvalue exactly when there exists \(\mathbf{x}\ne\mathbf{0}\) with \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). Subtracting \(\lambda\mathbf{x}\) from both sides gives \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\). Conversely, if this last equation holds for some \(\mathbf{x}\ne\mathbf{0}\), adding \(\lambda\mathbf{x}\) to both sides gives \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\), so \(\lambda\) is an eigenvalue. Thus the eigenvalue condition is equivalent to \(\operatorname{Nul}(\mathbf{A}-\lambda\mathbf{I}_n)\) being nontrivial. The preceding theorem says this is equivalent to \(\mathbf{A}-\lambda\mathbf{I}_n\) being singular. \(\square\)

Working with the Shifted Matrix

The matrix \(\mathbf{A}-\lambda\mathbf{I}_n\) is formed by subtracting \(\lambda\) from each diagonal entry of \(\mathbf{A}\); entries off the diagonal remain unchanged. To test a proposed \(\lambda\), form this shifted matrix and solve its homogeneous system. If there is a nonzero solution, that solution is an eigenvector. If the only solution is zero, the shifted matrix is nonsingular, and the proposed scalar is not an eigenvalue. Row reduction is useful because it shows both the pivot pattern and the solution conditions.

Common Pitfall: The Zero Solution Is Not Enough

A homogeneous system always has the zero solution. It is a mistake to conclude that a scalar is an eigenvalue merely because \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\) has a solution; the required solution must be nonzero. For example, take \(\mathbf{A}=\begin{bmatrix}2&0\\0&5\end{bmatrix}\) and \(\lambda=2\). Then \(\mathbf{A}-2\mathbf{I}_2=\begin{bmatrix}0&0\\0&3\end{bmatrix}\). The equations give \(3x_2=0\), so \(x_2=0\), while \(x_1\) is free. Choosing \(\mathbf{x}=\begin{bmatrix}1\\0\end{bmatrix}\) gives a nonzero solution and verifies the eigenvalue condition: \(\mathbf{A}\mathbf{x}=\begin{bmatrix}2\\0\end{bmatrix}=2\mathbf{x}\).

A related wrong conclusion is: “The shifted system has the solution \(\mathbf{x}=\mathbf{0}\), so \(\lambda\) is an eigenvalue.” That argument fails because the zero vector solves every homogeneous system, including those whose matrices are nonsingular. The correction is to row-reduce and check whether at least one variable is free. A free variable can be chosen nonzero to produce a nonzero solution; if every variable is a pivot variable, the homogeneous system has only the zero solution.

Why This Equivalence Is Useful

The eigenvalue equation starts as a condition on a scalar and an unknown vector. Rewriting it as a homogeneous system makes familiar matrix tools available: row reduction reveals whether a nonzero solution exists, and the singularity criterion translates that answer into a statement about invertibility. The direction of the test is important: an eigenvalue makes \(\mathbf{A}-\lambda\mathbf{I}_n\) singular, not nonsingular. Conversely, singularity of that shifted matrix guarantees a nonzero vector on which \(\mathbf{A}\) acts by the scalar \(\lambda\).

This result also gives a useful conceptual distinction. The matrix \(\mathbf{A}\) itself need not be singular for \(\lambda\) to be an eigenvalue. What must be singular is the shifted matrix \(\mathbf{A}-\lambda\mathbf{I}_n\). In the first example, the relevant matrix was \(\mathbf{A}-2\mathbf{I}_2\), and its nonzero null vector supplied the eigenvector. In the third example, the shift by \(1\) had only the zero null vector, ruling out that scalar.

For a proposed scalar, a reliable check is therefore: form the shifted matrix by changing only its diagonal entries, solve its homogeneous system, and verify any nonzero candidate by substituting it into \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). This procedure tests precisely the required condition without confusing the ever-present zero solution with an eigenvector.