A Diagonalizable Matrix With Repeated Eigenvalues

Repeated Does Not Mean Undiagonalizable

When a characteristic polynomial has a repeated root, the corresponding eigenvalue occurs more than once in the diagonal matrix one would expect from a full eigenbasis. That repetition does not, by itself, decide whether the matrix is diagonalizable. The key question is whether the eigenspace for that eigenvalue supplies enough linearly independent eigenvectors to contribute all of its needed columns to an eigenbasis.

Throughout, matrices and eigenvectors are over \(\mathbb{R}\). For a real \(n\times n\) matrix \(\mathbf{A}\), an eigenvalue \(\lambda\) is a scalar for which \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) for some nonzero vector \(\mathbf{v}\). Its eigenspace is \(E_{\lambda}=\{\mathbf{v}\in\mathbb{R}^n:\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\}\), together with the zero vector. The characteristic polynomial is \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_n)\). The algebraic multiplicity \(\operatorname{am}(\lambda)\) is the number of times \(\lambda\) occurs as a root of that polynomial, counting repetitions. The geometric multiplicity \(\operatorname{gm}(\lambda)\) is \(\dim E_{\lambda}\).

Thus a repeated diagonal entry is not a difficulty by itself. If the same eigenvalue appears twice in \(\mathbf{D}\), the corresponding two columns of \(\mathbf{P}\) must be two linearly independent vectors from the same eigenspace. One eigenvector cannot be placed in two columns and still give an invertible \(\mathbf{P}\).

Why Enough Eigenspace Dimensions Give a Diagonalization

Proof. Suppose first that \(\mathbf{v}_1,\dots,\mathbf{v}_n\) are linearly independent eigenvectors, with respective eigenvalues \(\lambda_1,\dots,\lambda_n\). Put these vectors into the columns of \(\mathbf{P}\), and put their eigenvalues in the corresponding order on the diagonal of \(\mathbf{D}\): \(\mathbf{P}=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_n\end{bmatrix}\) and \(\mathbf{D}=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\). Independence of the \(n\) columns makes \(\mathbf{P}\) invertible. The eigenvector equations give \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\), column by column. Multiplying on the right by \(\mathbf{P}^{-1}\) gives \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\).

Conversely, suppose \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), where \(\mathbf{P}\) is invertible and \(\mathbf{D}\) is diagonal. Then \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). For each column \(\mathbf{p}_j\) of \(\mathbf{P}\), this equation says \(\mathbf{A}\mathbf{p}_j=d_{jj}\mathbf{p}_j\). Every column is nonzero because \(\mathbf{P}\) is invertible, so the columns are eigenvectors. They are linearly independent because they are the columns of an invertible matrix. Therefore \(\mathbf{A}\) has \(n\) linearly independent eigenvectors. \(\square\)

Proof. Choose a basis for each eigenspace \(E_{\lambda_i}\). We first show that the combined list of all these basis vectors is linearly independent. Suppose \(\mathbf{v}_1+\cdots+\mathbf{v}_s=\mathbf{0}\), where \(\mathbf{v}_i\in E_{\lambda_i}\); this form includes any proposed linear relation by grouping its terms according to eigenspace. Fix an index \(k\), and apply the matrix \(\prod_{i\ne k}(\mathbf{A}-\lambda_i\mathbf{I}_n)\) to the equation. For \(i\ne k\), the product includes the factor \(\mathbf{A}-\lambda_i\mathbf{I}_n\), which sends \(\mathbf{v}_i\) to \(\mathbf{0}\). On \(\mathbf{v}_k\), each factor acts by multiplication by \(\lambda_k-\lambda_i\). Hence the result is \(\bigl(\prod_{i\ne k}(\lambda_k-\lambda_i)\bigr)\mathbf{v}_k=\mathbf{0}\). The scalar product is nonzero because the eigenvalues are distinct, so \(\mathbf{v}_k=\mathbf{0}\). This holds for every \(k\), and the chosen basis vectors are independent within each eigenspace, so the combined list is linearly independent. By the dimension assumption it contains \(n\) vectors. The first theorem now gives a diagonalization. \(\square\)

This theorem is useful when an eigenvalue is repeated: its eigenspace can contribute several columns, provided its dimension is large enough. In applications, the algebraic and geometric multiplicities help count how many columns are needed and how many independent eigenvectors are available. The examples below check the eigenspaces directly and then verify the resulting matrix products.

Worked Example: A Repeated Eigenvalue With a Two-Dimensional Eigenspace

Worked Example: Checking the Eigenspace Before Building the Basis

Worked Example: Choosing Another Basis for the Repeated Eigenspace

Multiplicity as a Check, Not a Shortcut

For a diagonalization, each eigenvalue must appear in \(\mathbf{D}\) as many times as the number of eigenvectors selected for that eigenvalue. The examples show why both multiplicities matter: algebraic multiplicity records how often the eigenvalue is a root of the characteristic polynomial, while geometric multiplicity counts the independent eigenvectors actually available in its eigenspace. In the examples, the repeated eigenvalue has algebraic multiplicity \(2\) and geometric multiplicity \(2\), so it can supply two columns.

A practical check is to list the distinct eigenvalues, find a basis for each eigenspace, and count the total number of basis vectors. If that total is \(n\), the theorem above gives a basis of eigenvectors and therefore a diagonalization. If the total is less than \(n\), this method has not produced enough eigenvectors to diagonalize the matrix. Do not replace the eigenspace calculation with a guess based only on the repeated factor in the characteristic polynomial.

Common mistake. A student sees \(p_{\mathbf{A}}(\lambda)=(2-\lambda)^2(5-\lambda)\) and concludes, “The eigenvalue \(2\) is repeated, so \(\mathbf{A}\) cannot be diagonalized.” That conclusion is wrong for the matrix in the first worked example: solving \((\mathbf{A}-2\mathbf{I}_3)\mathbf{x}=\mathbf{0}\) gives \(x_3=0\), with \(x_1,x_2\) free. Its eigenspace therefore has dimension \(2\), enough to supply the two eigenvectors associated with the repeated entry. The correction is to calculate the eigenspace dimension, then check whether all eigenspaces together provide \(n\) independent eigenvectors.