Diagonalization

Precise definition

A square matrix $A$ is diagonalizable over a field if $A=PDP^{-1}$ for an invertible $P$ and diagonal $D$. Equivalently, the vector space has a basis of eigenvectors of $A$. Columns of $P$ are those eigenvectors, and diagonal entries of $D$ are the corresponding eigenvalues in the same order.

Notation and mathematical language

Algebraic multiplicity is a root's multiplicity in the characteristic polynomial; geometric multiplicity is $\dim\ker(A-\lambda I)$. A matrix is diagonalizable iff the geometric multiplicities sum to $n$, equivalently each eigenvalue has enough independent eigenvectors. Distinct eigenvalues always yield independent eigenvectors.

Conceptual picture

In an eigenbasis, $A$ merely scales each coordinate, so repeated application is easy: $A^k=PD^kP^{-1}$. Diagonalization changes representation, not the underlying linear transformation.

Conditions and key results

The field matters: a real rotation matrix may diagonalize over $\mathbb C$ but not $\mathbb R$. Repeated eigenvalues do not automatically prevent diagonalization; the identity is diagonalizable despite one repeated eigenvalue. A defective matrix lacks a full eigenbasis.

A reliable strategy

1. Find and factor the characteristic polynomial over the stated field.
2. For each eigenvalue, solve $(A-\lambda I)v=0$ and count independent eigenvectors.
3. If there are $n$, place them as columns of $P$ and put matching eigenvalues on $D$.
4. Verify $AP=PD$ or $A=PDP^{-1}$ before using the form for powers or dynamics.

Fully worked example

Interpretation and application

Diagonalization simplifies recurrences, differential systems, Markov chains, and quadratic forms. If eigenvectors are nearly dependent, the formula can be numerically sensitive even when exact diagonalization exists.

Common mistakes

Verification and reasonableness

• Check each column equation $Av_j=\lambda_jv_j$.
• Verify $\det P\ne0$.
• Compare trace and determinant with the sum and product of diagonal entries of $D$.

Practice

1. Is every matrix with distinct eigenvalues diagonalizable?
2. Is $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ diagonalizable?
3. How is $A^k$ computed after diagonalization?

Further deduction

A real symmetric matrix is always orthogonally diagonalizable: $A=QDQ^T$ with $Q$ orthogonal and real $D$. This spectral theorem is stronger than ordinary diagonalizability because the eigenbasis can be chosen orthonormal. It also improves numerical stability since $Q^{-1}=Q^T$ preserves Euclidean lengths.

A projection matrix is diagonalizable because $P^2=P$ implies every eigenvalue satisfies $\lambda^2=\lambda$, hence is 0 or 1, and the space decomposes into kernel plus range. This example shows how an algebraic identity for a matrix can determine its possible spectrum and supply invariant subspaces without expanding a characteristic determinant.

Once $A=PDP^{-1}$, powers become $A^n=PD^nP^{-1}$, reducing repeated matrix multiplication to scalar powers on the diagonal. For a recurrence $x_{n+1}=Ax_n$, the eigen-coordinate components evolve independently as $\lambda_i^n$. Eigenvalues with magnitude below one decay, those above one grow, and those on the unit circle need closer analysis. This computational payoff depends on having a complete eigenbasis; a defective matrix requires another representation.

Related topics

• Eigenvalues
• Eigenvectors
• Basis