Diagonalization Tool

Math Linear Algebra • Linear Transformations and Eigenvalues

Written by STEM Calculators Team

Diagonalize \(A\) as \(A=PDP^{-1}\) when possible. Eigenvalues come from \(\det(A-\lambda I)=0\), columns of \(P\) are eigenvectors, and \(D\) contains eigenvalues on the diagonal. Includes a diagonalizable check (geometric multiplicities), reconstruction error, spectrum plot, and optional fast power \(A^k\).

Matrix size

Tolerance

Display decimals

Steps

Allow complex diagonalization (\(\mathbb{C}\)) Normalize eigenvectors Scale so first nonzero component is 1 Show spectrum plot Show \(A^k\) via diagonalization

Matrix \(A\)

A is 2×2

Inputs accept -3.5, 2e-4, fractions like 7/3, and constants pi, e.

Results

Characteristic polynomial

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Eigenvalues (algebraic mult.)

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Diagonalizable?

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Reconstruction error \(\|A-PDP^{-1}\|_F\)

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Factorization

If diagonalizable, \(A=PDP^{-1}\). Columns of \(P\) are eigenvectors.

Eigenspaces summary

Geometric multiplicity = \(\dim(\mathrm{Null}(A-\lambda I))\).

Ready

Enter matrix \(A\), then click “Calculate”.

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Theory

Diagonalization \(A=PDP^{-1}\)

A square matrix \(A\in\mathbb{R}^{n\times n}\) is diagonalizable if there exists an invertible matrix \(P\) and a diagonal matrix \(D\) such that \[ A = P D P^{-1}. \] This is a similarity transformation, meaning \(A\) and \(D\) represent the same linear map but in different bases.

How \(P\) and \(D\) are built

Eigenvalues and eigenvectors satisfy \(Av=\lambda v\). If you can find \(n\) linearly independent eigenvectors \(v_1,\dots,v_n\), then define \[ P = [v_1\ v_2\ \cdots\ v_n],\qquad D = \mathrm{diag}(\lambda_1,\lambda_2,\dots,\lambda_n). \] Then \(AP=PD\). Multiplying by \(P^{-1}\) gives \(A=PDP^{-1}\).

In words: diagonalization is possible when eigenvectors form a full basis of \(\mathbb{R}^n\) (or \(\mathbb{C}^n\)).

Diagonalizable test: algebraic vs geometric multiplicity

Each eigenvalue \(\lambda\) has:

Algebraic multiplicity = how many times \(\lambda\) appears as a root of \(\det(A-\lambda I)=0\).
Geometric multiplicity = \(\dim(\mathrm{Null}(A-\lambda I))\), the dimension of its eigenspace.

Always \(1\le\) geometric multiplicity \(\le\) algebraic multiplicity. The matrix is diagonalizable if the sum of geometric multiplicities equals \(n\).

Why diagonalization matters: fast powers \(A^k\)

If \(A=PDP^{-1}\), then \[ A^k = (PDP^{-1})^k = P D^k P^{-1}, \] and \(D^k\) is easy because it just raises diagonal entries: \[ D^k = \mathrm{diag}(\lambda_1^k,\dots,\lambda_n^k). \] This is a standard technique for computing repeated applications of a linear transformation.

Defective cases and Jordan form (university note)

If an eigenvalue repeats but does not have enough independent eigenvectors, the matrix is defective and cannot be diagonalized. In that case, one uses generalized eigenvectors and Jordan normal form: \(A=PJP^{-1}\) where \(J\) has Jordan blocks (nearly diagonal).