A square matrix is diagonalizable when it has a basis of eigenvectors, i.e., when you can find enough linearly independent eigenvectors to fill the whole space.

Core idea

For an n×nn\times nn×n matrix AAA over a field FFF:

  • AAA is diagonalizable ⇔ there exists an invertible matrix PPP such that P−1AP=DP^{-1}AP=DP−1AP=D, where DDD is diagonal.
  • This happens ⇔ the vector space FnF^nFn has a basis consisting entirely of eigenvectors of AAA.

In words: you can “relabel” the coordinates so that the linear transformation just stretches or flips each coordinate axis without mixing them.

Precise criteria

Let the eigenvalues of AAA be λ1,…,λk\lambda_1,\dots,\lambda_k λ1​,…,λk​ (distinct values). For each eigenvalue λ\lambda λ:

  • Algebraic multiplicity : how many times λ\lambda λ appears as a root of the characteristic polynomial.
  • Geometric multiplicity : the dimension of the eigenspace Eλ=ker⁡(A−λI)E_\lambda =\ker(A-\lambda I)Eλ​=ker(A−λI).

Then:

  • AAA is diagonalizable ⇔ the sum of the dimensions of all its eigenspaces equals nnn.
  • Equivalently: for every eigenvalue λ\lambda λ, its geometric multiplicity equals its algebraic multiplicity, and the total of the geometric multiplicities is nnn.

So if you can find nnn linearly independent eigenvectors, the matrix is diagonalizable; if you cannot, it is not.

A very useful sufficient condition

There is a popular quick test:

  • If an n×nn\times nn×n matrix has n nn distinct eigenvalues (all different), then it is diagonalizable over that field.

This is sufficient but not necessary: a matrix can still be diagonalizable even if some eigenvalues repeat, as long as each repeated eigenvalue has “enough” eigenvectors (geometric multiplicity equals algebraic multiplicity).

Example sketch

  • A 3×33\times 33×3 matrix with eigenvalues 1,2,31,2,31,2,3 (all distinct) is automatically diagonalizable.
  • A 3×33\times 33×3 matrix with eigenvalues 1,1,21,1,21,1,2 is diagonalizable only if the eigenspace for λ=1\lambda =1λ=1 has dimension 2 (so you still get 3 independent eigenvectors total).

When a matrix is not diagonalizable

A matrix fails to be diagonalizable when:

  • At least one eigenvalue has geometric multiplicity strictly less than its algebraic multiplicity.
  • In Jordan-normal-form language: at least one Jordan block for some eigenvalue has size larger than 1×11\times 11×1, meaning there is a nontrivial nilpotent part.

For instance, the Jordan block

(λ10λ)\begin{pmatrix} \lambda&1\\ 0&\lambda \end{pmatrix}(λ0​1λ​)

has only one linearly independent eigenvector, so a 2×22\times 22×2 matrix similar to this is not diagonalizable.

Special important cases

Some common classes of matrices are always diagonalizable (over appropriate fields):

  • Real symmetric matrices are diagonalizable by an orthogonal matrix (spectral theorem).
  • More generally, normal matrices (those satisfying AA∗=A∗AAA^{}=A^{}AAA∗=A∗A) are diagonalizable by a unitary matrix over C\mathbb{C}C.
  • Projections (P2=PP^2=PP2=P) and involutions (A2=IA^2=IA2=I) are diagonalizable over suitable fields, with eigenvalues in {0,1}\{0,1\}{0,1} and {±1}\{\pm 1\}{±1} respectively.

Intuition in one line

You can think of a matrix as diagonalizable exactly when you can choose coordinates so that its action is just “stretch along axes, no mixing” — and that is possible iff eigenvectors span the whole space.

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