Diagonalizable Linear Map

Overview

Diagonalization replaces a linear transformation with independent scalar multiplications along basis directions. It simplifies repeated application, polynomial expressions in an operator, and linear recurrences. It is a property over a specified field, and the presence of eigenvalues alone does not guarantee it.

Prerequisites

Use finite-dimensional vector spaces, bases, and linear endomorphisms. A nonzero vector $v$ is an eigenvector of $T$ with eigenvalue $\lambda$ when $Tv=\lambda v$. The zero vector is excluded from this definition because it satisfies the equation for every scalar.

Definition and conventions

An endomorphism $T:V\to V$ of a finite-dimensional $F$-vector space is diagonalizable over $F$ if some $F$-basis consists of eigenvectors. In matrix language, its matrix $A$ satisfies

for some invertible matrix $P$ with entries in $F$. The columns of $P$ are the eigenvectors in the original coordinates. Similarity, rather than arbitrary row reduction, is the relevant change of matrix.

Intuition

Eigenvectors identify directions preserved by the operator. A complete eigenbasis lets each vector be split into independently evolving components. Distinct eigenvalues supply independent directions, but a repeated eigenvalue can have too few eigenvectors to fill the space.

Examples and non-examples

Take

Vectors $(1,0)$ and $(1,1)$ have eigenvalues $2$ and $3$. With $P$ formed from these columns, $P^{-1}AP=\operatorname{diag}(2,3)$, so $A^m=P\operatorname{diag}(2^m,3^m)P^{-1}$ for nonnegative integers $m$.

The matrix

is not diagonalizable over any field: its eigenspace for $1$ has dimension one although its characteristic polynomial is $(x-1)^2$. By contrast the identity matrix has the same characteristic polynomial and is diagonalizable. Repeated roots therefore do not decide the issue.

Rotation by a right angle on $\mathbb R^2$ has characteristic polynomial $x^2+1$. It is not diagonalizable over $\mathbb R$, but becomes diagonalizable over $\mathbb C$.

Important properties

Eigenvectors for distinct eigenvalues are linearly independent. One proof applies $T-\lambda I$ to a shortest supposed dependence, reducing its length and obtaining a contradiction.

More generally, $T$ is diagonalizable exactly when its minimal polynomial is a product of distinct linear factors over $F$. Both splitting and absence of repeated factors matter. A self-adjoint real operator on a finite-dimensional inner product space is diagonalizable with an orthonormal eigenbasis, but arbitrary real operators do not satisfy this spectral theorem.

Connections and further reading

Diagonalization is the simplest case of decomposing a space as a module over $F[x]$, where $x$ acts by $T$. Jordan form describes failures when the characteristic polynomial splits. Read Hefferon, Chapter Five, Section II for the eigenvalue definitions and diagonalizability criterion.