Diagonalization and When It Fails
~6 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable finding eigenvalues and eigenvectors of a matrix and understanding the distinction between algebraic and geometric multiplicity from the previous lecture.
Why this matters
Diagonalization is the bridge between abstract linear operators and their simplest possible representation.
In physics, we frequently seek a basis in which an operator acts as simply as possible—scaling each basis vector independently. This is the essence of diagonalization. While it is always possible for Hermitian operators (a topic for the next lecture), it is not guaranteed for general linear operators. Understanding precisely when diagonalization fails is crucial for analyzing non-Hermitian systems, defective matrices, and the limitations of modal decomposition in classical mechanics.
The Core Idea: When Can We Simplify?
Imagine a linear transformation \(T\) acting on a vector space \(V\). If we can find a basis of \(V\) consisting entirely of eigenvectors of \(T\), then the matrix representation of \(T\) in that basis is diagonal. This means \(T\) simply scales each basis vector by its corresponding eigenvalue, with no mixing between components. This is the most "decoupled" form an operator can take.
However, this is not always possible. A matrix might have fewer linearly independent eigenvectors than the dimension of the space. In such cases, no basis of eigenvectors exists, and the operator cannot be diagonalized. Instead, we must use a Jordan canonical form, which introduces "super-diagonal" entries representing the failure of complete decoupling. The key criterion is whether the eigenvectors span the entire space.
The Diagonalizability Criterion
The fundamental result governing diagonalization is the Eigenbasis Criterion (D-081). For a linear operator \(T\) on a finite-dimensional vector space \(V\) with \(\dim V = n\), represented by a matrix \(A\), \(T\) is diagonalizable if and only if its eigenvectors span \(V\).
This condition can be checked algebraically. The characteristic polynomial \(p(\lambda) = \det(A - \lambda I)\) must split into linear factors over the field (typically \(\mathbb{C}\) in physics). Furthermore, for every eigenvalue \(\lambda_i\), the geometric multiplicity \(g_i\) (the dimension of the eigenspace \(E_{\lambda_i}\)) must equal the algebraic multiplicity \(a_i\) (the number of times \(\lambda_i\) appears as a root of \(p(\lambda)\)).
If this condition holds, the vector space decomposes into a direct sum of eigenspaces. If it fails for any eigenvalue, the operator is defective and cannot be diagonalized.
Defective Matrices and Jordan Blocks
When diagonalization fails, we turn to the Jordan canonical form. A defective matrix has at least one eigenvalue where \(g_i < a_i\). In the Jordan basis, the matrix is block diagonal, with each block corresponding to an eigenvalue. These blocks are upper triangular with the eigenvalue on the diagonal and 1s on the super-diagonal.
For example, consider a \(2 \times 2\) matrix with a single eigenvalue \(\lambda\) of algebraic multiplicity 2 but geometric multiplicity 1. Its Jordan form is \(\begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}\). The "1" on the super-diagonal represents a generalized eigenvector, indicating that the operator does not fully decouple the components. This structure is essential in analyzing systems with repeated eigenvalues, such as degenerate oscillators or non-Hermitian quantum systems.
Physical Implications of Failure
In classical mechanics, diagonalization corresponds to finding normal modes. Undamped oscillators with symmetric mass and stiffness matrices are never defective, so they always have a full set of normal modes. Defective behaviour appears only when the equations are not symmetric, for example with some damping, gyroscopic or feedback terms, at a point where two modes coalesce; there the motion cannot be written as a superposition of independent oscillations.
In quantum mechanics, non-Hermitian Hamiltonians (arising in open quantum systems or effective descriptions) may have defective spectra. This results in non-orthogonal eigenvectors and a breakdown of the usual spectral decomposition. The time evolution of such systems can exhibit transient growth or decay that is not captured by simple exponential modes, a phenomenon known as "non-normality" or "pseudospectral" behavior.
The derivation
Diagonalizability and the Eigenbasis Criterion
Let \(T\) be a linear operator on a finite-dimensional vector space \(V\) with \(\dim V = n\), represented in some basis by a matrix \(A\). Then \(T\) is diagonalizable (similar to a diagonal matrix, \(P^{-1}AP = D\)) if and only if its eigenvectors span \(V\) — equivalently, if and only if the characteristic polynomial splits over the field and, for every eigenvalue \(\lambda_i\), the geometric multiplicity equals the algebraic multiplicity, \(g_i = a_i\). In that case \(V = E_{\lambda_1}\oplus\cdots\oplus E_{\lambda_k}\) is the direct sum of the eigenspaces.
A common misconception is that distinct eigenvalues guarantee diagonalizability. While it is true that eigenvectors corresponding to distinct eigenvalues are linearly independent, this only ensures diagonalizability if the total number of distinct eigenvalues equals the dimension of the space. If an eigenvalue is repeated, you must still check the geometric multiplicity. A matrix with all distinct eigenvalues is always diagonalizable, but a matrix with repeated eigenvalues may or may not be, depending on the eigenspace dimensions. C
Worked example
Problem. Consider the matrix \(A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}\). Determine if \(A\) is diagonalizable. If not, find its Jordan canonical form.
- Find the eigenvalues by solving \(\det(A - \lambda I) = 0\). The characteristic polynomial is \((2-\lambda)^2 = 0\), so \(\lambda = 2\) with algebraic multiplicity \(a = 2\).
- Find the eigenvectors by solving \((A - 2I)\mathbf{v} = 0\). This gives \(\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\), so \(v_2 = 0\). The eigenspace is spanned by \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\), so geometric multiplicity \(g = 1\).
- Compare \(g\) and \(a\). Since \(g = 1 < a = 2\), the eigenvectors do not span \(\mathbb{R}^2\). Therefore, \(A\) is not diagonalizable.
- Since \(A\) is already in Jordan form (upper triangular with \(\lambda\) on diagonal and 1 on super-diagonal), the Jordan canonical form is \(J = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}\).
Units check. The matrix entries are dimensionless, as expected for a linear operator on a dimensionless vector space. The Jordan form preserves the eigenvalues and the structure of the defect.
Three quick tests
- Is the matrix \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) diagonalizable?
Answer
Yes. It is already diagonal. The eigenvalue \(\lambda=1\) has algebraic multiplicity 2 and geometric multiplicity 2 (the entire space is the eigenspace). - Can a \(3 \times 3\) matrix with eigenvalues \(1, 1, 2\) be diagonalizable?
Answer
Only if the eigenspace for \(\lambda=1\) has dimension 2. If the geometric multiplicity of \(\lambda=1\) is 1, it is not diagonalizable. - What is the physical significance of a defective Hamiltonian in quantum mechanics?
Answer
It indicates that the system cannot be fully decomposed into independent stationary states. This often occurs in open quantum systems or effective non-Hermitian descriptions, leading to non-orthogonal states and complex time evolution.
Problem set for PU-104 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.