Proof: Distinct Eigenvalues Provide a Full Eigenbasis

From Distinct Eigenvalues to a Basis

Diagonalization is possible when a matrix has enough independent eigenvectors to form a basis. A repeated eigenvalue can have an eigenspace with too few independent vectors, so merely counting eigenvalues with repetition is not enough. Distinct eigenvalues, however, force independence: eigenvectors chosen for different eigenvalues cannot combine to give a nontrivial dependence relation. This fact gives a useful guarantee. If an \(n\times n\) real matrix has \(n\) distinct real eigenvalues, choosing one eigenvector for each eigenvalue produces a basis of \(\mathbb{R}^n\), and that basis gives a diagonalization.

Throughout, matrices and vectors are over \(\mathbb{R}\). For an \(n\times n\) matrix \(\mathbf{A}\), a real number \(\lambda\) is an eigenvalue if there is a vector \(\mathbf{v}\ne\mathbf{0}\) such that \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Such a vector is an eigenvector for \(\lambda\). The eigenspace \(E_{\lambda}\) consists of \(\mathbf{0}\) together with all eigenvectors for \(\lambda\); equivalently, \(E_{\lambda}=\operatorname{Nul}(\mathbf{A}-\lambda\mathbf{I}_n)\).

The order condition in the definition matters. If the \(j\)th column of \(\mathbf{P}\) is an eigenvector for \(\lambda_j\), then the \(j\)th diagonal entry of \(\mathbf{D}\) must be \(\lambda_j\). The identity \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) expresses this column by column: each column of \(\mathbf{P}\) is multiplied by its own eigenvalue.

Why Distinct Eigenvalues Force Independence

The central step is a statement about any finite number of eigenvectors, not just \(n\) of them. It applies to vectors chosen from distinct eigenspaces, so it does not assume that each eigenspace is one-dimensional.

Proof. We use induction on \(m\). When \(m=1\), the list consists of the nonzero vector \(\mathbf{v}_1\), so its only linear combination equal to \(\mathbf{0}\) is \(c_1\mathbf{v}_1=\mathbf{0}\) with \(c_1=0\). Now suppose \(m>1\), and suppose the result holds for lists of \(m-1\) eigenvectors belonging to distinct eigenvalues. Assume that \(c_1\mathbf{v}_1+\cdots+c_m\mathbf{v}_m=\mathbf{0}\). Apply the matrix \(\mathbf{A}-\lambda_m\mathbf{I}\) to both sides. Since \(\mathbf{A}\mathbf{v}_i=\lambda_i\mathbf{v}_i\), the result is \(\sum_{i=1}^{m-1}c_i(\lambda_i-\lambda_m)\mathbf{v}_i=\mathbf{0}\); the term involving \(\mathbf{v}_m\) is zero. The first \(m-1\) vectors belong to distinct eigenvalues, so the induction hypothesis says they are independent. Therefore \(c_i(\lambda_i-\lambda_m)=0\) for every \(i<m\). Since \(\lambda_i\ne\lambda_m\), each \(c_i=0\) for \(i<m\). The original relation now reduces to \(c_m\mathbf{v}_m=\mathbf{0}\). Because \(\mathbf{v}_m\ne\mathbf{0}\), \(c_m=0\). All coefficients vanish, which proves independence. \(\square\)

The argument works because applying \(\mathbf{A}-\lambda_m\mathbf{I}\) removes the last eigenvector from the relation while multiplying each earlier eigenvector by a nonzero scalar. Distinctness is essential at exactly that step: if \(\lambda_i=\lambda_m\), then \(\lambda_i-\lambda_m=0\), and the argument gives no information about \(c_i\).

Why \(n\) Independent Vectors Give a Diagonalization

For a matrix with \(n\) distinct real eigenvalues, choose one eigenvector for each eigenvalue. The theorem proves that the resulting list of \(n\) vectors is independent. To connect this conclusion to diagonalization, we use the fact that \(n\) independent vectors in \(\mathbb{R}^n\) form a basis. Here is the matrix reason. Put those vectors into the columns of a square matrix \(\mathbf{P}\). Independence means \(\mathbf{P}\mathbf{x}=\mathbf{0}\) has only the zero solution. Thus its reduced row echelon form has a pivot in every column. Since there are \(n\) columns and \(n\) rows, it is \(\mathbf{I}_n\). Row operations are reversible, so applying them to any right-hand side shows that \(\mathbf{P}\mathbf{x}=\mathbf{b}\) is solvable for every \(\mathbf{b}\in\mathbb{R}^n\). Its columns therefore span \(\mathbb{R}^n\), as well as being independent; together, these properties mean they are a basis.

Proof. List the distinct eigenvalues as \(\lambda_1,\dots,\lambda_n\), and choose a nonzero eigenvector \(\mathbf{v}_i\) for each \(\lambda_i\). By the preceding theorem, these \(n\) vectors are independent. By the square-matrix basis argument above, they form a basis of \(\mathbb{R}^n\). Let \(\mathbf{P}\) be the matrix whose columns, in order, are \(\mathbf{v}_1,\dots,\mathbf{v}_n\), and let \(\mathbf{D}\) be the diagonal matrix with entries \(\lambda_1,\dots,\lambda_n\) in that order. The basis property makes \(\mathbf{P}\) invertible. For every column \(i\), the eigenvector equation gives \(\mathbf{A}\mathbf{v}_i=\lambda_i\mathbf{v}_i\), so the columns of \(\mathbf{A}\mathbf{P}\) are exactly the columns of \(\mathbf{P}\mathbf{D}\). Hence \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). Multiplying on the right by \(\mathbf{P}^{-1}\) gives \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), as required. \(\square\)

Worked Constructions

Common Mistake: Pairing the Wrong Eigenvalue With a Column

A common error is to find the right eigenvectors but place the eigenvalues in a different order on the diagonal. For the last example, the first column of \(\mathbf{Q}\) is an eigenvector for \(1\), not \(3\). If we incorrectly use \(\mathbf{D}_{\mathrm{wrong}}=\begin{bmatrix}3&0&0\\0&1&0\\0&0&5\end{bmatrix}\), then the first column of \(\mathbf{Q}\mathbf{D}_{\mathrm{wrong}}\) is \(\begin{bmatrix}3\\-3\\0\end{bmatrix}\). But the first column of \(\mathbf{B}\mathbf{Q}\) is \(\mathbf{B}\mathbf{u}_1=\begin{bmatrix}1\\-1\\0\end{bmatrix}\). Thus \(\mathbf{B}\mathbf{Q}\ne\mathbf{Q}\mathbf{D}_{\mathrm{wrong}}\), and the claimed diagonalization fails. The correction is to keep each eigenvalue beside its corresponding eigenvector: the order \(\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3\) requires diagonal entries \(1,3,5\).

The condition of \(n\) distinct real eigenvalues is sufficient, not necessary, for real diagonalizability. A repeated eigenvalue does not itself prevent diagonalization; what matters in general is whether there are enough independent eigenvectors. The present result is valuable because distinct eigenvalues settle that independence question immediately: selecting one nonzero eigenvector for each of the \(n\) eigenvalues gives the full eigenbasis without needing to compare dimensions of eigenspaces.