What an Eigenvalue Test Must Establish
Testing a proposed eigenvalue is not a matter of checking whether a determinant is nonzero or whether a number appears somewhere in a matrix. The test comes directly from the defining equation: a scalar \(\lambda\) is an eigenvalue of a square matrix \(\mathbf{A}\) if there is a nonzero vector \(\mathbf{v}\) such that \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). The requirement that \(\mathbf{v}\ne\mathbf{0}\) matters: the zero vector satisfies this equation for every scalar and therefore cannot establish that any scalar is an eigenvalue.
Throughout this tutorial, matrices and vectors have real entries and scalars are in \(\mathbb{R}\). An \(n\times n\) matrix acts on vectors in \(\mathbb{R}^n\), and \(\mathbf{I}_n\) denotes the \(n\times n\) identity matrix. The characteristic polynomial is \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_n)\). A root of this polynomial is a scalar that makes its value zero; a root is not itself an eigenvector.
The Determinant Test, With Its Conditions
Subtracting \(\lambda\mathbf{v}\) from both sides of the eigenvector equation gives \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\). Thus, a proposed scalar is an eigenvalue exactly when this homogeneous system has a nonzero solution. For a square matrix, the standard determinant test for a homogeneous system says that a nonzero solution exists exactly when the coefficient matrix has determinant zero.
Proof. Set \(\mathbf{B}=\mathbf{A}-\lambda\mathbf{I}_n\). By the definition of eigenvalue, \(\lambda\) is an eigenvalue exactly when \(\mathbf{B}\mathbf{v}=\mathbf{0}\) for some \(\mathbf{v}\ne\mathbf{0}\). Row operations preserve the solution set of a homogeneous system. They also preserve whether the determinant is zero: a row swap changes its sign, scaling a row by a nonzero scalar scales the determinant by that scalar, and adding a multiple of one row to another leaves the determinant unchanged. Reduce \(\mathbf{B}\) to RREF. If there is a pivot in every column, the square RREF is \(\mathbf{I}_n\), so its homogeneous system has only the zero solution and its determinant is nonzero. If there is not a pivot in every column, at least one variable is free; setting a free variable to \(1\), the other free variables to \(0\), and solving for the pivot variables gives a nonzero solution. In this case the square RREF has a zero row, so its determinant is zero. These two cases show that \(\mathbf{B}\) has a nonzero null vector exactly when \(\det\mathbf{B}=0\), proving the claim. \(\square\)
The direction of the test is a frequent source of mistakes. A zero determinant confirms that a candidate is an eigenvalue; a nonzero determinant rules it out. The test applies to square matrices because \(\mathbf{A}-\lambda\mathbf{I}_n\) must be a square matrix before its determinant is defined. The identity must have the same size as \(\mathbf{A}\).
Worked Examples: Applying the Test Correctly
Worked Example: A Nonzero Determinant Rules Out the Candidate
Test whether \(\lambda=2\) is an eigenvalue of \(\mathbf{A}=\begin{bmatrix}1&2\\0&3\end{bmatrix}\).
Since \(\mathbf{A}\) is \(2\times2\), use \(\mathbf{I}_2\). First calculate the shifted matrix and its determinant:
The determinant is not zero, so the theorem says \(2\) is not an eigenvalue. The system confirms the conclusion. If \(\mathbf{x}=(x_1,x_2)^{\top}\), then \((\mathbf{A}-2\mathbf{I}_2)\mathbf{x}=\mathbf{0}\) gives \(-x_1+2x_2=0\) and \(x_2=0\). Substitution into the first equation gives \(x_1=0\), so the only solution is \(\mathbf{x}=\mathbf{0}\).
Check: There is no nonzero vector satisfying \(\mathbf{A}\mathbf{x}=2\mathbf{x}\), in agreement with the nonzero determinant.
Worked Example: A Characteristic Root Is Not Automatically an Eigenvector
Find the eigenvectors corresponding to \(\lambda=2\) for \(\mathbf{A}=\begin{bmatrix}2&1\\0&2\end{bmatrix}\), and check whether \(\mathbf{e}_2=(0,1)^{\top}\) is one.
Subtract \(2\mathbf{I}_2\):
This matrix is already in RREF. Its homogeneous system is \(x_2=0\), with \(x_1\) free. Therefore the solutions are \(\mathbf{x}=(s,0)^{\top}\), where \(s\in\mathbb{R}\). The eigenvectors are the nonzero solutions, so they are \((s,0)^{\top}\) for \(s\ne0\). For instance, take \(\mathbf{v}=(1,0)^{\top}\).
Check: \(\mathbf{A}\mathbf{v}=(2,0)^{\top}=2(1,0)^{\top}=2\mathbf{v}\), so this is an eigenvector. In contrast, \(\mathbf{A}\mathbf{e}_2=(1,2)^{\top}\), whereas \(2\mathbf{e}_2=(0,2)^{\top}\). These are unequal, so \(\mathbf{e}_2\) is not an eigenvector. The scalar \(2\) is a characteristic root and an eigenvalue; it is not a vector to substitute as though it were an eigenvector.
Worked Example: Matching the Identity Matrix to the Dimension
Test \(\lambda=2\) for \(\mathbf{A}=\begin{bmatrix}1&0&0\\0&2&1\\0&0&4\end{bmatrix}\), and find one eigenvector.
The matrix has three rows and three columns, so the shifted matrix must use \(\mathbf{I}_3\). Subtracting gives
Its determinant is \((-1)(0)(2)=0\), so \(2\) is an eigenvalue. To find an eigenvector, solve the homogeneous system. The equations are \(-x_1=0\), \(x_3=0\), and \(2x_3=0\). Thus \(x_1=0\), \(x_3=0\), and \(x_2\) is free. Taking \(x_2=1\) gives \(\mathbf{v}=(0,1,0)^{\top}\).
Check: \(\mathbf{A}\mathbf{v}=(0,2,0)^{\top}=2(0,1,0)^{\top}=2\mathbf{v}\). By contrast, subtracting \(2\mathbf{I}_2\) would be invalid: a \(3\times3\) matrix and a \(2\times2\) matrix cannot be subtracted, and that expression has no determinant to evaluate.
Why the Eigenspace and the Eigenvectors Are Different
Proof. A vector \(\mathbf{x}\) belongs to \(\operatorname{Nul}(\mathbf{A}-\lambda\mathbf{I}_n)\) exactly when \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\). Expanding the left-hand side gives \(\mathbf{A}\mathbf{x}-\lambda\mathbf{x}=\mathbf{0}\), which is equivalent to \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). If \(\mathbf{x}\ne\mathbf{0}\), this is precisely the definition of an eigenvector. If \(\mathbf{x}=\mathbf{0}\), it is still in the null space but is not an eigenvector. Hence the eigenspace is the full null space, and its nonzero vectors are the eigenvectors. \(\square\)
This distinction helps keep two tasks separate. The determinant test decides whether the candidate scalar is an eigenvalue. If the determinant is zero, solving the homogeneous system finds its eigenspace; one must then choose a nonzero solution to give an eigenvector. A zero determinant does not make every vector an eigenvector, and the zero vector cannot serve as the required witness.
Common Errors and Their Corrections
Reversing the determinant test. In the first example, the incorrect conclusion “\(\det(\mathbf{A}-2\mathbf{I}_2)=-1\ne0\), therefore \(2\) is an eigenvalue” reverses the criterion. The correction is to check whether the determinant is zero: because \(-1\ne0\), the homogeneous system has only the zero solution, so \(2\) is not an eigenvalue.
Using an identity matrix of the wrong size. The expression \(\mathbf{A}-\lambda\mathbf{I}\) is only defined when the two matrices have the same dimensions. For an \(n\times n\) matrix, use \(\mathbf{I}_n\), not an identity matrix chosen from the size of some vector or from a remembered example. A rectangular matrix also has no eigenvalues under this definition: it maps between vector spaces of different dimensions, so \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) does not compare vectors in the same space.
Confusing a root, an eigenvalue, and an eigenvector. A root of \(p_{\mathbf{A}}\) is a scalar. It is an eigenvalue when the determinant test holds. An eigenvector is a nonzero column vector found by solving \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\). After proposing a vector, check it directly by computing both \(\mathbf{A}\mathbf{v}\) and \(\lambda\mathbf{v}\); equality is required.
| What is being tested? | Correct check | What it establishes |
|---|---|---|
| Candidate scalar \(\lambda\) | \(\det(\mathbf{A}-\lambda\mathbf{I}_n)=0\) | \(\lambda\) is an eigenvalue |
| Candidate vector \(\mathbf{v}\) | \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) and \(\mathbf{v}\ne\mathbf{0}\) | \(\mathbf{v}\) is an eigenvector |
| All vectors associated with \(\lambda\) | Solve \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\) | The solution set is \(E_\lambda\), including \(\mathbf{0}\) |
The reliable order is to confirm that the matrix is square, use an identity matrix of the matching size, and apply the determinant test to the proposed scalar. If the determinant is zero, solve the homogeneous system and select a nonzero vector. Finally, substitute that vector into the defining equation. Each step addresses a different question, and keeping them separate prevents a characteristic root from being mistaken for an eigenvector.
Check Your Understanding
For each proposed eigenvalue, first apply the determinant test. When asked for a vector, verify it by substitution.
- For \(\mathbf{A}=\begin{bmatrix}4&0\\0&1\end{bmatrix}\), is \(\lambda=1\) an eigenvalue? If so, give one eigenvector.
- For \(\mathbf{B}=\begin{bmatrix}1&1\\0&2\end{bmatrix}\), is \(\lambda=1\) an eigenvalue? Explain what the determinant tells you.
- For \(\mathbf{C}=\begin{bmatrix}3&0\\0&3\end{bmatrix}\), test whether \((1,1)^{\top}\) is an eigenvector for \(\lambda=3\).
- Why is \(\mathbf{I}_2\) the wrong identity matrix to use when forming \(\mathbf{D}-\lambda\mathbf{I}\) for a \(3\times3\) matrix \(\mathbf{D}\)?
- Explain why a nonzero vector is required in the definition of an eigenvector, even though the zero vector belongs to every eigenspace.
