Definition · applies to symmetric positive definite matrices · also called Cholesky factorization

Undergraduate

Consider a symmetric matrix $A$ whose eigenvalues are all positive (that is, $A$ is positive definite). The Cholesky decomposition of $A$ is the factorization

\[A = L L^{\top},\]

where $L$ is lower triangular with positive diagonal entries.

Examples

Worked examples

  • The matrix \(\begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}\) : None Undergraduate
    The matrix \(A = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}\) has no Cholesky decomposition, because $A$ is not symmetric.
  • The matrix \(\begin{bmatrix} 4 & 2 \\ 2 & 3 \end{bmatrix}\) : \[L = \begin{bmatrix} 2 & 0 \\ 1 & \sqrt{2} \end{bmatrix}\] Undergraduate
    The matrix \(A = \begin{bmatrix} 4 & 2 \\ 2 & 3 \end{bmatrix}\) is symmetric with positive eigenvalues $\frac{7 \pm \sqrt{17}}{2}$, so it has a Cholesky decomposition. Its factor $L$ is the LU factor $U$ with each row divided by the square root of its diagonal entry, transposed.
  • The matrix \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) : None Undergraduate
    The matrix \(A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) has no Cholesky decomposition, because $A$ is not symmetric.
  • The matrix \(A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\) : None Undergraduate
    The matrix \(A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\) has no Cholesky decomposition, because $A$ has the negative eigenvalue $-1$, so it is not positive definite. The matrix \(A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\) is symmetric but has no Cholesky decomposition, because its eigenvalue $-1$ is negative, so it is not positive definite.

Properties

  • A real square matrix $A$ has a Cholesky decomposition exactly when $A$ is symmetric positive definite, and then the decomposition is unique.
  • Computing a Cholesky decomposition takes about half the work of an LU decomposition, and it needs no row exchanges.
  • If $z$ is a vector of independent standard normal random variables and $\Sigma = L L^{\top}$, then $\mu + L z$ is normally distributed with mean $\mu$ and covariance $\Sigma$.

Notes

The related $L D L^{\top}$ decomposition, with $L$ lower triangular with ones on its diagonal and $D$ diagonal, avoids square roots and also applies to some symmetric matrices that are not positive definite.

For complex matrices, $A$ is Hermitian positive definite and $A = L L^*$.

Distinctions involving this term

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