Diagonalization

Diagonalization is the process of finding a basis in which a linear operator is represented by a diagonal matrix. Algebraically, it means the vector space has a basis of eigenvectors of the operator.

This page explains the finite-dimensional structure. The numerical problem of computing eigenvalues and eigenvectors with floating-point algorithms is treated separately in Matrix Diagonalization.

Definition

Let $A:V\to V$ be a linear operator on a finite-dimensional vector space. The operator is diagonalizable if there is a basis

$$\mathcal F=(v_1,\ldots,v_n)$$

of $V$ such that every $v_i$ is an eigenvector of $A$:

$$Av_i=\lambda_i v_i.$$

In that eigenbasis, the matrix of $A$ is diagonal:

$$A_{\mathcal F} = \operatorname{diag}(\lambda_1,\ldots,\lambda_n).$$

The diagonal entries are eigenvalues, repeated according to the chosen eigenbasis.

Matrix Form

Suppose an old basis $\mathcal B$ has already been chosen, and let $A_{\mathcal B}$ be the matrix of $A$ in that basis. If $P$ is the matrix whose columns are the old-basis coordinate columns of the eigenvectors,

$$P = \begin{pmatrix} [v_1]_{\mathcal B} & \cdots & [v_n]_{\mathcal B} \end{pmatrix},$$

then the change-of-basis formula gives

$$D = P^{-1}A_{\mathcal B}P,$$

where

$$D = \operatorname{diag}(\lambda_1,\ldots,\lambda_n).$$

Equivalently,

$$A_{\mathcal B} = PDP^{-1}.$$

This is the matrix form most often written as $A=PDP^{-1}$. The columns of $P$ are eigenvectors; $P^{-1}$ converts old coordinates into eigenbasis coordinates.

For the underlying coordinate convention, see Change of Basis.

When Diagonalization Exists

Diagonalization requires enough linearly independent eigenvectors to form a basis. Over $\mathbb C$, the characteristic polynomial always splits into linear factors, but that alone is not enough.

Let $\lambda$ be an eigenvalue. Its algebraic multiplicity is its multiplicity as a root of the characteristic polynomial. Its geometric multiplicity is

$$\dim\ker(A-\lambda I),$$

the dimension of its eigenspace.

A finite-dimensional complex matrix is diagonalizable exactly when, for every eigenvalue,

$$\text{geometric multiplicity} = \text{algebraic multiplicity}.$$

Equivalently, the direct sum of all eigenspaces is the whole vector space.

A Matrix That Fails

The matrix

$$J = \begin{pmatrix} \lambda & 1\\ 0 & \lambda \end{pmatrix}$$

has only one eigenvalue, $\lambda$. But

$$J-\lambda I = \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}$$

has a one-dimensional kernel. The eigenspace is spanned by $(1,0)^T$, so it cannot supply a basis for $\mathbb C^2$. This Jordan block is not diagonalizable.

The failure is not that the eigenvalue is degenerate. Degenerate eigenvalues are harmless when the eigenspace has the matching dimension. The problem is a shortage of independent eigenvectors.

Unitary Diagonalization

If $A$ has an orthonormal eigenbasis, then the diagonalizing matrix can be chosen unitary. If $U$ has the normalized eigenvectors as its columns, then

$$D = U^\dagger A U, \qquad A = U D U^\dagger.$$

Hermitian matrices always admit such a unitary diagonalization, with real diagonal entries. Unitary matrices also admit unitary diagonalization in finite dimension, with eigenvalues on the unit circle. More generally, finite-dimensional normal operators are unitarily diagonalizable.

This is why the finite-dimensional spectral theorem is so central in quantum mechanics: observables, ideal finite Hamiltonians, and many symmetry operators can be put into bases where their action is read directly from eigenvalues.

Worked Example

Consider

$$A = \begin{pmatrix} 2 & 1\\ 0 & 3 \end{pmatrix}.$$

The eigenvalues are $2$ and $3$. For $\lambda=2$, an eigenvector is

$$v_1 = \begin{pmatrix} 1\\ 0 \end{pmatrix}.$$

For $\lambda=3$, an eigenvector is

$$v_2 = \begin{pmatrix} 1\\ 1 \end{pmatrix}.$$

Put these eigenvectors into the columns of $P$:

$$P = \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}, \qquad P^{-1} = \begin{pmatrix} 1 & -1\\ 0 & 1 \end{pmatrix}.$$

Then

$$P^{-1}AP = \begin{pmatrix} 2 & 0\\ 0 & 3 \end{pmatrix}.$$

The original matrix was not diagonal in the standard basis. It became diagonal in the eigenbasis.

Functions of a Diagonalizable Operator

If

$$A=PDP^{-1},$$

then powers are easy:

$$A^k = P D^k P^{-1}.$$

For a function $f$ defined on the eigenvalues,

$$f(A) = P f(D) P^{-1},$$

where

$$f(D) = \operatorname{diag} \left( f(\lambda_1),\ldots,f(\lambda_n) \right).$$

For a time-independent finite-dimensional Hamiltonian,

$$e^{-iHt/\hbar} = P \operatorname{diag} \left( e^{-iE_1t/\hbar},\ldots,e^{-iE_nt/\hbar} \right) P^{-1}$$

when $H=PDP^{-1}$ with $D=\operatorname{diag}(E_1,\ldots,E_n)$. In the Hermitian case, $P$ can be chosen unitary.

The projector-based version of this idea is Spectral Decomposition. The power-series and exponential viewpoint is Matrix Functions and Exponentials.

Physical Interpretation

Diagonalization finds a representation in which an operator acts independently on basis directions. In an energy eigenbasis, a time-independent Hamiltonian multiplies each energy component by its energy. In an observable eigenbasis, measurement projectors and outcome probabilities become transparent. In coupled finite systems, diagonalization often identifies normal modes or decoupled combinations.

The diagonal basis is not a different physical system. It is a basis in which the operator’s structure is easier to read.

Common Mistakes

• Assuming every matrix is diagonalizable.
• Confusing algebraic multiplicity with geometric multiplicity.
• Treating a repeated eigenvalue as a problem even when its eigenspace has the correct dimension.
• Forgetting that eigenvectors used as columns of $P$ must be linearly independent.
• Using $P^{-1}$ where $P$ belongs in $A=PDP^{-1}$.
• Applying diagonalization formulas to defective matrices.
• Treating numerical eigenvectors as exact without checking residuals and orthogonality.

Cross-Links

• Eigenvalues and Eigenvectors
• Change of Basis
• Matrices as Linear Maps
• Hermitian Operators
• Unitary Operators
• Normal Operators
• Spectral Decomposition
• Matrix Functions and Exponentials
• Matrix Diagonalization
• Spectral Decomposition
• Functions of Operators
• Time-Independent Hamiltonians

References

• S. Axler, Linear Algebra Done Right, 3rd ed., Springer, 2015.
• G. Strang, Linear Algebra and Its Applications, 4th ed., Brooks/Cole, 2006.
• P. R. Halmos, Finite-Dimensional Vector Spaces, 2nd ed., Springer, 1974.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.

Exercises

1. Diagonalize

$$A = \begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix}.$$

Solution

The eigenvalues are $2$ and $0$. Normalized eigenvectors are

$$u_1 = \frac{1}{\sqrt2} \begin{pmatrix} 1\\ 1 \end{pmatrix}, \qquad u_2 = \frac{1}{\sqrt2} \begin{pmatrix} 1\\ -1 \end{pmatrix}.$$

With

$$U = \frac{1}{\sqrt2} \begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix},$$

one obtains

$$U^\dagger A U = \begin{pmatrix} 2 & 0\\ 0 & 0 \end{pmatrix}.$$

2. Show that

$$J = \begin{pmatrix} 4 & 1\\ 0 & 4 \end{pmatrix}$$

is not diagonalizable.

Solution

There is only one eigenvalue, $4$, with algebraic multiplicity $2$. But

$$J-4I = \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}$$

has kernel spanned by $(1,0)^T$. The geometric multiplicity is $1$, so there are not enough eigenvectors to form a basis.

3. Suppose $A=PDP^{-1}$ and $D=\operatorname{diag}(2,5)$. What is $A^3$?

Solution

Use powers of the diagonal matrix:

$$A^3 = P D^3 P^{-1} = P \begin{pmatrix} 8 & 0\\ 0 & 125 \end{pmatrix} P^{-1}.$$

No direct multiplication of $A$ by itself is needed.