📂Matrix Algebra

Uniqueness Proof of Cholesky Decomposition

Theorem

$A>0$ has a unique Cholesky decomposition.

Explanation

Eigenvalue diagonalization, Singular value decomposition, Schur decomposition, LU decomposition, LDU decomposition all share the commonality of not being unique. This is because these methods either utilize the relationship between eigenvalues and eigenvectors, or because $1 = a \dfrac{1}{a}$ thus $L$ or $U$ can be divided into parts.

However, Cholesky decomposition doesn’t use the concept of eigenvalues and is expressed as $A=LL^{T}$, so $1$ can’t be split into two. Elaborating on this intuitive argument in a slightly meticulous and complicated manner immediately completes the proof.

Proof

$A$ is positive definite, hence an invertible matrix, and there exist a lower triangular matrix $L$ and diagonal matrix $D$ satisfying $A:=LDL^{T}$.

Multiplying $A=LDL^{T}$ by $\mathbf{x}^{T} \ne \mathbb{0}$ on the left and $\mathbf{x}$ on the right gives

$$\mathbf{x}^{T} A \mathbf{x} = \mathbf{x}^{T} LDL^{T} \mathbf{x} = (L^{T} \mathbf{x})^{T} D (L^{T} \mathbf{x}) >0$$

Therefore, $D$ is a positive definite matrix, and since the eigenvalues are all positive, the diagonal elements are all positive.

Thus,

$$D^{ 1/2 } := \text{diag} (\sqrt{d_{11}} , \sqrt{d_{22}} , \cdots , \sqrt{d_{nn}} )$$

can be defined, and it will be $D = D^{ 1/2 } D^{ 1/2 }$.

From $$A = L D L^{T} = L D^{ 1/2 } D^{ 1/2 } L^{T}$$,

defining $\overline{L} := LD^{ 1/2 }$ gives

$$A = \overline{L} \overline{L}^{T}$$

Such a $\overline{L}$ is unique, thus the Cholesky decomposition is also unique.

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