Linear Independence Of Eigenvectors For Distinct Eigenvalues

The Claim and Its Hypotheses

Eigenvectors are connected to one another through the action of the same matrix: each is multiplied by its own eigenvalue. When those eigenvalues are different, this shared structure prevents the eigenvectors from cancelling in a nontrivial linear combination. The key is to use the eigenvalue equations to create a new relation involving fewer vectors, which makes induction possible.

Let \(\mathbb{F}\) be either \(\mathbb{R}\) or \(\mathbb{C}\), let \(\mathbf{A}\in\mathbb{F}^{n\times n}\), and view \(\mathbf{A}\) as acting on column vectors in \(\mathbb{F}^n\). A scalar \(\lambda\in\mathbb{F}\) is an eigenvalue of \(\mathbf{A}\) if there is a nonzero vector \(\mathbf{v}\in\mathbb{F}^n\) such that \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Such a nonzero vector is an eigenvector associated with \(\lambda\). The requirement \(\mathbf{v}\ne\mathbf{0}\) is part of the definition.

Pairwise distinct means \(\lambda_i\ne\lambda_j\) whenever \(i\ne j\). The theorem does not require the eigenvectors to have any particular length or form, and the eigenspace for one eigenvalue can contain many different nonzero eigenvectors. It does require that the eigenvalues assigned to the selected vectors be distinct. The proof uses this condition when it divides out the scalar difference between two eigenvalues.

A Useful Fact About Eigenspaces

For a fixed eigenvalue \(\lambda\), its eigenspace is the set of all vectors satisfying \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\), including the zero vector. Before proving the theorem, it is useful to see how the eigenvalue equation restricts vectors that could belong to two different eigenspaces.

Proof. Suppose that \(\mathbf{v}\) satisfies both equations. Their left sides are equal, so \(\lambda\mathbf{v}=\mu\mathbf{v}\), and therefore \((\lambda-\mu)\mathbf{v}=\mathbf{0}\). Since \(\lambda\ne\mu\), the scalar \(\lambda-\mu\) is nonzero and has a multiplicative inverse in \(\mathbb{F}\). Multiplying by that inverse gives \(\mathbf{v}=\mathbf{0}\). Conversely, \(\mathbf{0}\) satisfies both equations, so it is indeed the only common vector. \(\square\)

This lemma concerns a vector that belongs to both eigenspaces. The independence theorem is stronger: it rules out a nontrivial linear combination involving one eigenvector from each of several distinct eigenspaces. The induction proof extends the same basic idea—using different eigenvalues to produce a nonzero scalar factor—to a relation with any finite number of terms.

Proof by Induction

We prove the theorem for every positive integer \(p\), the number of selected eigenvectors. The induction step starts by assuming there is a linear relation. Applying \(\mathbf{A}\) to it gives another relation, with the coefficient of each eigenvector multiplied by its eigenvalue. Subtracting a suitable multiple of the first relation eliminates one term. The induction hypothesis then forces all the remaining coefficients to vanish.

Proof. We use induction on \(p\). For \(p=1\), suppose \(c_1\mathbf{v}_1=\mathbf{0}\). Since \(\mathbf{v}_1\ne\mathbf{0}\), some coordinate of \(\mathbf{v}_1\) is nonzero. The corresponding coordinate of \(c_1\mathbf{v}_1\) can be zero only if \(c_1=0\). Thus the one-vector list is independent.

Now let \(p\ge2\), and assume the theorem holds for lists of \(p-1\) eigenvectors with pairwise distinct eigenvalues. Consider a relation among the \(p\) vectors: \[ c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}. \] Apply the matrix \(\mathbf{A}\) to both sides. Matrix multiplication is linear, and \(\mathbf{A}\mathbf{v}_i=\lambda_i\mathbf{v}_i\), so \[ c_1\lambda_1\mathbf{v}_1+\cdots+c_p\lambda_p\mathbf{v}_p=\mathbf{0}. \] Multiply the original relation by \(\lambda_p\) and subtract it from this new relation. The \(\mathbf{v}_p\) term cancels, leaving \[ c_1(\lambda_1-\lambda_p)\mathbf{v}_1+\cdots+ c_{p-1}(\lambda_{p-1}-\lambda_p)\mathbf{v}_{p-1}=\mathbf{0}. \] The first \(p-1\) eigenvalues are pairwise distinct, so the induction hypothesis says that \(\mathbf{v}_1,\dots,\mathbf{v}_{p-1}\) are linearly independent. Consequently, each coefficient in this last relation is zero: \[ c_i(\lambda_i-\lambda_p)=0\qquad (i=1,\dots,p-1). \] For each such \(i\), pairwise distinctness gives \(\lambda_i-\lambda_p\ne0\). Therefore \(c_i=0\) for every \(i=1,\dots,p-1\). Substituting these values into the original relation gives \(c_p\mathbf{v}_p=\mathbf{0}\). Since \(\mathbf{v}_p\ne\mathbf{0}\), it follows that \(c_p=0\). Every coefficient is zero, as required. This proves the statement for \(p\), and induction proves it for every positive integer \(p\). \(\square\)

Worked Examples

What the Distinctness Condition Does—and Does Not—Say

The theorem is about a list containing one nonzero eigenvector for each of several distinct eigenvalues. It does not say that every collection of eigenvectors of one matrix is independent. If two selected vectors have the same eigenvalue, the theorem's hypothesis fails, even if the vectors happen to be independent. Nor does it say that vectors for distinct eigenvalues must be orthogonal; independence and orthogonality are different properties.

A common mistake is to infer orthogonality from distinct eigenvalues. For a concrete counterexample, use the first worked example. Its eigenvectors satisfy \[ \mathbf{v}_1^{\top}\mathbf{v}_2 =\begin{bmatrix}1&0\end{bmatrix} \begin{bmatrix}1\\1\end{bmatrix} =1. \] The eigenvalues are distinct, but the dot product is \(1\), not \(0\). The correct conclusion is that these eigenvectors are linearly independent, as the theorem states—not that they are orthogonal.

Another error is to omit the nonzero condition. The zero vector satisfies \(\mathbf{A}\mathbf{0}=\lambda\mathbf{0}\) for every scalar \(\lambda\), but it is not an eigenvector. Including it would immediately produce a dependent list: the coefficient \(1\) on \(\mathbf{0}\), with zero coefficients on the other vectors, gives a nontrivial relation equal to \(\mathbf{0}\). The theorem avoids this issue by requiring each selected vector to be nonzero.

Finally, repeated eigenvalues do not invalidate independence in every case; they simply make this theorem unavailable. For instance, the identity matrix \(\mathbf{I}_2\) has eigenvalue \(1\), and \(\begin{bmatrix}1\\0\end{bmatrix}\) and \(\begin{bmatrix}0\\1\end{bmatrix}\) are independent eigenvectors for that same eigenvalue. By contrast, choosing \(\begin{bmatrix}1\\0\end{bmatrix}\) twice gives a dependent list. Both choices have a repeated eigenvalue, so independence must be checked by other reasoning.