---
tag: D7E9SZ4E82
kind: statement
role: theorem
name: Eigendecomposition
msc: [15A18, 15A20, 15A23]
uses: [0JXZKX526R, 0BQAA2BK3Q, 0N6960W0BV, 0CRMGPAH83, 0EB9N0S33C, 044JNC7ZAA]
about: [07SQCNV233]
sources:
  - url: https://en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix
same_as:
  wikidata: Q194919
  wikipedia: Eigendecomposition of a matrix
---

## Statement

Let $F$ be a field and let $A$ be an $n \times n$ [[square matrix|0JXZKX526R]] over $F$.
Suppose that $A$ has $n$ [[linearly independent|0CRMGPAH83]] [[eigenvectors|0N6960W0BV]] $q_1, \dots, q_n$ in $F^n$, with $A q_i = \lambda_i q_i$.
Let $Q$ be the matrix with columns $q_1, \dots, q_n$, and let $\Lambda$ be the [[diagonal matrix|044JNC7ZAA]] with diagonal entries $\lambda_1, \dots, \lambda_n$.
Then $Q$ is [[invertible|0EB9N0S33C]] and
$$A = Q \Lambda Q^{-1}.$$
Conversely, let $A = Q \Lambda Q^{-1}$ with $Q$ invertible and $\Lambda$ diagonal.
Then the columns of $Q$ are linearly independent eigenvectors of $A$.
The diagonal entries of $\Lambda$ are the corresponding [[eigenvalues|0BQAA2BK3Q]].

## Remarks

The columns of $AQ$ are the vectors $A q_i = \lambda_i q_i$, and these are also the columns of $Q \Lambda$.
Thus $AQ = Q\Lambda$, and $Q$ is invertible because its columns are independent.
For example, $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ has the eigenvector $(1, 0)^T$ for $2$ and $(1, 1)^T$ for $3$.
Thus $A = Q \Lambda Q^{-1}$ with $Q = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $\Lambda = \operatorname{diag}(2, 3)$.
The hypothesis is necessary: $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ has only one independent eigenvector, and it has no eigendecomposition.
When $A$ is real symmetric or complex [[normal|0MR0BYF0K2]], $Q$ can be chosen orthogonal or unitary, by the [[spectral theorem|06J4PXRG22]].

## Explanations

### L1 human

A matrix with enough eigenvectors acts, in the basis of those eigenvectors, by stretching each axis by its eigenvalue.
The eigendecomposition writes this as: change of basis, stretch, change back.
Not every matrix has one; exactly the [[diagonalisable|07SQCNV233]] matrices do.

### L2 human

The decomposition makes powers and functions of $A$ easy: $A^k = Q \Lambda^k Q^{-1}$, and the same holds for the exponential.
It is the matrix form of the criterion that [[a matrix is diagonalisable iff it has a basis of eigenvectors|056SW1E0AC]].
Distinct eigenvalues give enough eigenvectors, by [[this statement|0HNMFG8F8K]].
When eigenvectors are missing, the [[Jordan normal form|085HQ5G5TE]] or the [[Schur decomposition|V1JG4C7R7B]] takes its place.