Common Error: Assuming Every Matrix Is Diagonalizable

Eigenvalues Do Not Guarantee Diagonalization

Diagonalization requires more than finding eigenvalues. A matrix is diagonalizable when there is a basis consisting of its eigenvectors. Even if its characteristic polynomial has roots and each root gives an eigenspace, those eigenspaces may not contain enough independent vectors to make a basis. Matrices with this shortfall are called defective.

The scalar field matters when asking whether a matrix has eigenvalues. A real matrix need not have a real eigenvalue: for example, \(\begin{bmatrix}0&-1\\1&0\end{bmatrix}\) has characteristic polynomial \(\lambda^2+1\), which has no real root. Over \(\mathbb{C}\), every square matrix has an eigenvalue; this existence statement follows from the Fundamental Theorem of Algebra, whose proof belongs to the later study of complex scalars. Neither statement says that a matrix has enough independent eigenvectors to diagonalize. The examples here are real matrices with real eigenvalues, so the failure is specifically a shortage of independent eigenvectors over \(\mathbb{R}\).

Each column of \(\mathbf{P}\) in a diagonalization is an eigenvector, and the matching diagonal entry of \(\mathbf{D}\) is its eigenvalue. Since an invertible matrix has independent columns, diagonalization supplies \(n\) independent eigenvectors. Conversely, if \(n\) independent eigenvectors form a basis, placing them in the columns of \(\mathbf{P}\) gives a diagonalization. Thus, the relevant question is not merely how many eigenvalues appear, but whether their eigenvectors can form a basis.

Proof. Suppose first that \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), where \(\mathbf{P}\) is invertible and \(\mathbf{D}\) is diagonal. From \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\), the \(j\)th column \(\mathbf{p}_j\) of \(\mathbf{P}\) satisfies \(\mathbf{A}\mathbf{p}_j=d_{jj}\mathbf{p}_j\). Each column is nonzero because the columns of an invertible matrix are independent. Therefore the \(n\) columns are independent eigenvectors. Conversely, suppose \(\mathbf{p}_1,\dots,\mathbf{p}_n\) are independent eigenvectors with eigenvalues \(d_1,\dots,d_n\). Set \(\mathbf{P}=\begin{bmatrix}\mathbf{p}_1&\cdots&\mathbf{p}_n\end{bmatrix}\) and \(\mathbf{D}=\operatorname{diag}(d_1,\dots,d_n)\). Independence makes \(\mathbf{P}\) invertible, and the eigenvector equations give \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). Multiplying on the right by \(\mathbf{P}^{-1}\) yields \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). \(\square\)

A Repeated Eigenvalue Can Have Too Few Eigenvectors

The example separates two counts that are easy to confuse. The repeated root has algebraic multiplicity two, but its eigenspace has dimension one (its geometric multiplicity). Repetition in the characteristic polynomial does not itself provide additional eigenvectors.

When Multiplicity Counts Do Give Enough Vectors

A repeated eigenvalue is not automatically a problem. The following result explains precisely how a mismatch between algebraic and geometric multiplicity prevents diagonalization. Here \(\operatorname{am}(\lambda)\) is the multiplicity of \(\lambda\) as a root of the characteristic polynomial, and \(\operatorname{gm}(\lambda)=\dim E_\lambda\).

Proof. If \(\mathbf{A}\) is diagonalizable, write \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) with \(\mathbf{D}\) diagonal. Similar matrices have the same characteristic polynomial, since determinant multiplicativity gives \(\det(\mathbf{A}-t\mathbf{I})=\det(\mathbf{P})\det(\mathbf{D}-t\mathbf{I})\det(\mathbf{P}^{-1})=\det(\mathbf{D}-t\mathbf{I})\). If \(\lambda\) occurs \(k\) times on the diagonal of \(\mathbf{D}\), then the product \(\det(\mathbf{D}-t\mathbf{I})\) has exactly \(k\) factors \(\lambda-t\), so \(\operatorname{am}(\lambda)=k\). Also, for a diagonal matrix, \((\mathbf{D}-\lambda\mathbf{I})\mathbf{x}=\mathbf{0}\) permits nonzero coordinates only in positions where the diagonal entry is \(\lambda\). Its eigenspace therefore has dimension \(k\). Similarity carries this eigenspace to \(E_\lambda\): \(\mathbf{A}(\mathbf{P}\mathbf{x})=\lambda\mathbf{P}\mathbf{x}\) exactly when \(\mathbf{D}\mathbf{x}=\lambda\mathbf{x}\), and invertibility of \(\mathbf{P}\) preserves dimension. Hence \(\operatorname{gm}(\lambda)=k=\operatorname{am}(\lambda)\). The contrapositive proves the final claim. \(\square\)

Common Mistake: Counting Roots Instead of Directions

A tempting but incorrect argument for \(\mathbf{A}=\begin{bmatrix}2&1\\0&2\end{bmatrix}\) is: “The characteristic polynomial has degree two and its only root is repeated twice, so there must be two eigenvectors.” The repetition counts the root in the characteristic polynomial, not independent directions in its eigenspace. The correction is to solve the eigenvector equation:

It forces \(y=0\), leaving only one independent direction. The geometric multiplicity is therefore one, while the algebraic multiplicity is two, and the proved obstruction rules out diagonalization.

When checking a proposed diagonalization, keep three questions separate: what scalar field is being used, which eigenvalues occur, and how many independent eigenvectors are available. The first determines which eigenvalues count; the last determines whether those eigenvalues can be arranged into a basis. An eigenvalue is evidence of an eigenvector, not evidence that an entire eigenbasis exists.