Eigenspaces of Distinct Eigenvalues Form a Direct Sum

How Distinct Eigenspaces Fit Together

An eigenvalue describes how a linear map scales certain vectors: if \(T(\mathbf{v})=\lambda\mathbf{v}\) for a nonzero vector \(\mathbf{v}\), then \(\lambda\) is an eigenvalue and \(\mathbf{v}\) is an eigenvector. The vectors satisfying this equation, together with the zero vector, make up the eigenspace \(E_{\lambda}\). When a map has several distinct eigenvalues, the corresponding eigenspaces might be added together. The key fact is that this addition has no ambiguity: a vector in their sum can be written as a sum of components from the different eigenspaces in only one way.

Let \(V\) be a vector space over a field \(\mathbb{F}\), and let \(T:V\to V\) be a linear map. The field matters because eigenvalues and vectors are taken in that field. For a matrix \(\mathbf{A}\in\mathbb{F}^{n\times n}\), the associated map is \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\). The eigenspace for an eigenvalue \(\lambda\) is \[ E_{\lambda}=\{\mathbf{v}\in V:T(\mathbf{v})=\lambda\mathbf{v}\}. \] It is a subspace: linearity gives \(T(\mathbf{0})=\mathbf{0}\), and if \(\mathbf{u},\mathbf{v}\in E_{\lambda}\) and \(a,b\in\mathbb{F}\), then \(T(a\mathbf{u}+b\mathbf{v})=a\lambda\mathbf{u}+b\lambda\mathbf{v}=\lambda(a\mathbf{u}+b\mathbf{v})\). For a matrix, this is the null space of \(\mathbf{A}-\lambda\mathbf{I}_n\).

Directness concerns uniqueness of the components; it does not by itself say that the sum equals \(V\). If it does equal \(V\), the subspaces together provide unique components for every vector in the whole space. If the sum is a proper subspace, the same uniqueness applies only to vectors in that sum.

The Zero-Sum Test for Directness

Proof. Suppose first the sum is direct and \(\mathbf{u}_1+\cdots+\mathbf{u}_r=\mathbf{0}\). The zero vector has the representation \(\mathbf{0}+\cdots+\mathbf{0}\), where each zero belongs to its corresponding subspace. Uniqueness of representation therefore gives \(\mathbf{u}_i=\mathbf{0}\) for every \(i\). Conversely, suppose every zero-sum relation with \(\mathbf{u}_i\in U_i\) has all components zero. If \(\mathbf{x}\) has two representations, say \(\mathbf{x}=\mathbf{u}_1+\cdots+\mathbf{u}_r=\mathbf{v}_1+\cdots+\mathbf{v}_r\), then \((\mathbf{u}_1-\mathbf{v}_1)+\cdots+(\mathbf{u}_r-\mathbf{v}_r)=\mathbf{0}\). Each difference belongs to its subspace, so the assumption gives \(\mathbf{u}_i-\mathbf{v}_i=\mathbf{0}\) for every \(i\). Thus the representations agree component by component, as required. \(\square\)

This test turns the question about unique representation into a zero-sum question. For eigenspaces, the linear map itself supplies a way to create such a relation: applying \(T-\lambda\operatorname{id}_V\) removes the component from \(E_{\lambda}\). Repeating that idea yields the directness result.

Proof. We prove that if \(\mathbf{v}_i\in E_{\lambda_i}\) and \(\mathbf{v}_1+\cdots+\mathbf{v}_r=\mathbf{0}\), then every \(\mathbf{v}_i=\mathbf{0}\). Use induction on \(r\). For \(r=1\), the equation is \(\mathbf{v}_1=\mathbf{0}\). Now suppose the claim holds for \(r-1\) distinct eigenvalues and consider \(r\) of them. Apply the linear map \(T-\lambda_r\operatorname{id}_V\) to the zero-sum equation. Since \(T(\mathbf{v}_i)=\lambda_i\mathbf{v}_i\), and \(T(\mathbf{v}_r)-\lambda_r\mathbf{v}_r=\mathbf{0}\), the result is \(\sum_{i=1}^{r-1}(\lambda_i-\lambda_r)\mathbf{v}_i=\mathbf{0}\). For \(i<r\), the vector \((\lambda_i-\lambda_r)\mathbf{v}_i\) lies in \(E_{\lambda_i}\), because a subspace is closed under scalar multiplication. The induction hypothesis applied to the first \(r-1\) eigenspaces gives \((\lambda_i-\lambda_r)\mathbf{v}_i=\mathbf{0}\) for each \(i<r\). Distinctness means \(\lambda_i-\lambda_r\ne0\), so multiplication by this nonzero field element can be reversed, giving \(\mathbf{v}_i=\mathbf{0}\) for each \(i<r\). The original equation then gives \(\mathbf{v}_r=\mathbf{0}\). The zero-sum test proves that the sum is direct. \(\square\)

In particular, choosing one nonzero eigenvector from each of several distinct eigenspaces gives a linearly independent list. Indeed, a relation among those vectors is a zero-sum relation with one component in each eigenspace; directness makes every component zero, and a nonzero eigenvector cannot be multiplied by a nonzero scalar to give zero. This conclusion does not assert that an arbitrary list chosen from the union of the eigenspaces is independent: vectors chosen from the same eigenspace may already be dependent.

Working with Direct Sums

What Directness Says About Eigenvectors

Proof. Suppose a linear combination of the combined list equals \(\mathbf{0}\). Group its terms by eigenspace, writing the equation as \(\mathbf{w}_1+\cdots+\mathbf{w}_r=\mathbf{0}\), where \(\mathbf{w}_i\) is the linear combination of the selected vectors in \(E_{\lambda_i}\). The direct-sum theorem gives \(\mathbf{w}_i=\mathbf{0}\) for every \(i\). By the assumed independence of the list within \(E_{\lambda_i}\), every coefficient in that group is zero. This holds for every group, so all coefficients in the original relation are zero. \(\square\)

Common Pitfalls and Why the Result Matters

Pitfall: treating directness as a claim about the whole ambient space. In the second example, \(E_1\oplus E_2\) is direct but does not contain \((0,0,1)^{\top}\). Directness means that components in the stated subspaces are unique; it does not mean those subspaces account for every vector in \(V\). To claim the sum equals \(V\), one must also show that every vector of \(V\) belongs to the sum.

Pitfall: treating every collection of eigenvectors as independent. In the third example, the incorrect calculation would be to label all four vectors “eigenvectors” and infer that the list is independent. The explicit calculation \(\mathbf{u}_1+\mathbf{u}_2-\mathbf{u}_3=\mathbf{0}\) disproves that claim. The direct-sum theorem separates contributions from distinct eigenvalues; it does not remove dependence among vectors within a single eigenspace.

The result is useful when constructing an eigenbasis. If independent lists are chosen inside several eigenspaces for distinct eigenvalues, the corollary guarantees that joining those lists preserves independence. One can therefore count contributions from different eigenspaces without worrying that one eigenspace's vectors cancel against another's. Whether those vectors together span the entire space is a separate question; directness guarantees uniqueness of a decomposition where one exists.