Eigenspaces of Symmetric Matrices

From an Eigenvalue to an Eigenspace

An eigenvalue identifies a particular way that a matrix acts by scaling: for a scalar \(\lambda\), a nonzero vector \(\mathbf{x}\) is an eigenvector when \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). Often there is more than one such vector, and the collection of all solutions—including the zero vector—is the eigenspace. To find it, move all terms to one side and solve a homogeneous system. This turns an eigenvector question into a familiar task with a precise goal: find a basis for the null space of \(\mathbf{A}-\lambda\mathbf{I}\).

Here we work over \(\mathbb{R}\), with \(\mathbf{A}\in\mathbb{R}^{n\times n}\). A matrix is symmetric if \(\mathbf{A}^{\top}=\mathbf{A}\). We will use the method for specified real eigenvalues of symmetric matrices. The computation below does not require a result about whether every eigenvalue of a symmetric matrix is real: whenever an eigenvalue is used here, it is real and either given or verified by finding a nonzero solution.

The equality in the definition follows by subtracting \(\lambda\mathbf{x}\) from both sides of \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). This translation is useful because row reduction preserves the solution set of a homogeneous system. In practice, form \(\mathbf{A}-\lambda\mathbf{I}_n\), row-reduce it, and express every solution using free variables. The vectors multiplying those free parameters are candidates for a basis.

Proof. By definition, \(E_{\lambda}\) consists of the vectors \(\mathbf{x}\) satisfying \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). The zero vector satisfies this equation because \(\mathbf{A}\mathbf{0}=\mathbf{0}=\lambda\mathbf{0}\), so \(E_{\lambda}\) is nonempty. If \(\mathbf{u},\mathbf{v}\in E_{\lambda}\) and \(a,b\in\mathbb{R}\), then linearity of matrix multiplication gives \(\mathbf{A}(a\mathbf{u}+b\mathbf{v})=a\mathbf{A}\mathbf{u}+b\mathbf{A}\mathbf{v}=a\lambda\mathbf{u}+b\lambda\mathbf{v}=\lambda(a\mathbf{u}+b\mathbf{v})\). Thus \(a\mathbf{u}+b\mathbf{v}\in E_{\lambda}\). The set contains zero and is closed under linear combinations, so it is a subspace. \(\square\)

How Row Reduction Produces a Basis

Recall that a basis is an ordered list of vectors that is both linearly independent and spans the space. In a homogeneous system in RREF, each free variable can be assigned independently; the pivot variables are then determined by the equations. Grouping the resulting solution by free parameters gives a spanning list. The positions of the free variables also show that this list is independent.

Proof. Row operations preserve the solutions of a homogeneous system, so it is enough to consider the RREF system. Every solution is obtained by assigning values \(t_1,\dots,t_q\) to its free variables and then using the equations to determine its pivot variables. By hypothesis, this gives \(\mathbf{x}=\sum_{k=1}^{q}t_k\mathbf{v}_k\). Therefore every solution is a linear combination of the listed vectors, and each listed vector is itself a solution by setting its parameter to \(1\) and all other parameters to \(0\). The list spans the null space. To prove independence, suppose \(\sum_{k=1}^{q}c_k\mathbf{v}_k=\mathbf{0}\). In free-variable position \(j_k\), the vector \(\mathbf{v}_k\) has entry \(1\), while every \(\mathbf{v}_{\ell}\) for \(\ell\ne k\) has entry \(0\): each vector was formed by setting one free variable to \(1\) and the others to \(0\). Thus the \(j_k\)-th entry of the sum is \(c_k\), and it must equal zero. This holds for every \(k\), so all coefficients vanish and the list is independent. If \(q=0\), there are no free variables, every variable is a pivot variable, and the homogeneous equations force every variable to be zero. The null space is then \(\{\mathbf{0}\}\). \(\square\)

For an eigenspace, apply this result to \(\mathbf{M}=\mathbf{A}-\lambda\mathbf{I}_n\). The result gives a basis, not just a collection of eigenvectors: it certifies that all vectors in the eigenspace can be built from the basis vectors and that none of those vectors is redundant.

Worked Eigenspace Calculations

Checking the Whole Answer

Substitution confirms that each proposed basis vector is an eigenvector, but a successful substitution alone does not prove that the list is a basis for the entire eigenspace. A list might contain valid eigenvectors while missing other solutions, or it might contain redundant vectors. The row-reduced system addresses both issues: its free variables produce a formula for every solution, while the independent free-variable choices produce an independent list.

It is also useful to distinguish the eigenspace from a basis for it. The eigenspace is a subspace and contains infinitely many vectors whenever it contains a nonzero vector over \(\mathbb{R}\), since every scalar multiple of a nonzero eigenvector is still in the eigenspace. A basis is a finite list that represents all those vectors without redundancy. For the second example, neither displayed basis vector is the whole eigenspace; each is one member of a two-vector list whose span gives the whole plane of solutions.

Common Sign Error

In the third example the equation is \(x_1+2x_2=0\). A common mistake is to write \(x_1=2x_2\), losing the minus sign, and then report \(\begin{bmatrix}2\\1\\0\end{bmatrix}\) as an eigenvector. Substitution exposes the error: \(\mathbf{A}\begin{bmatrix}2\\1\\0\end{bmatrix} =\begin{bmatrix}8\\10\\0\end{bmatrix}\), whereas \(2\begin{bmatrix}2\\1\\0\end{bmatrix} =\begin{bmatrix}4\\2\\0\end{bmatrix}\). These are not equal. The correct equation gives \(x_1=-2x_2\), and the corrected vector \(\begin{bmatrix}-2\\1\\0\end{bmatrix}\) satisfies the eigenvector equation. Checking by multiplication is a direct way to catch sign errors before accepting a computed basis.