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The textbook I uses says that only when matrix $A \in \mathbb{R}^{n\times n}$ is positive-definite the Cholesky decomposition of $A$ is unique but it does not provide any proof.

I did Cholesky decomposition to a positive-semidefinite matrix $ A = \left( \begin{smallmatrix} 1&2\\ 2&4 \end{smallmatrix}\right)$ and I have the following $$ A = \begin{pmatrix} 1&2\\ 2&4 \end{pmatrix}= \begin{pmatrix} 1&0\\ 2&0 \end{pmatrix} \begin{pmatrix} 1&2\\ 0&0 \end{pmatrix} $$ I do not understand why this decomposition is not unique.

Could anyone figure out what mistake I made. Thank you in advance.

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Let's try to compute (all the possible) Cholesky decompositions of a $2 \times 2$ real positive-semidefinite matrix $$ \begin{pmatrix} d & e \\ e & f \end{pmatrix}. $$

We want $$ \begin{pmatrix} a & 0 \\ b & c \end{pmatrix} \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} = \begin{pmatrix} a^2 & ab \\ ab & b^2 + c^2 \end{pmatrix} = \begin{pmatrix} d & e \\ e & f \end{pmatrix}. $$

This forces $a^2 = d$ so $a = \sqrt{d}$. Then we have two cases:

  1. If $d \neq 0$ then $b = \frac{e}{\sqrt{d}}$ and $c = \sqrt{f - \frac{e^2}{d}}$. In this case, the decomposition is unique.
  2. If $d = 0$ then we can take any $b$ as long as $f - b^2 \geq 0$ and then take $c = \sqrt{f - b^2}$. In this case, unless $f = 0$, the decomposition won't be unique.

Your matrix falls into case $(1$) so indeed in your case the Cholesky decomposition is unique but in general it might not be unique. For an example of case $(2)$ we have,

$$ \begin{pmatrix} 0 & 0 \\ 0 & 4 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 2 & 0 \end{pmatrix} \begin{pmatrix} 0 & 2 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 2 \end{pmatrix} $$

(the first decomposition corresponds to taking $b = 2$ while the second corresponds to taking $b = 0$ and there are infinitely many other decompositions with $-2 \leq b \leq 2$).

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    $\begingroup$ If I understand you correctly, you mean when a matrix $A \in \mathbb{R}^{n \times n}$ is positive semi-definite, it is possible but not for sure the Cholesky decomposition is not unique. $\endgroup$ – Mr.Robot Nov 7 '17 at 8:18
  • $\begingroup$ @Mr.Robot: Precisely. $\endgroup$ – levap Nov 7 '17 at 8:22
  • $\begingroup$ I assume in case 2. you mean $c = \sqrt{f-b^2}$ (copy-paste error?). You can't divide by $d$ in case 2 anyways. $\endgroup$ – Marius Hofert Jul 1 '18 at 14:13
  • $\begingroup$ @MariusHofert: Yep, thanks for catching the typo. $\endgroup$ – levap Jul 1 '18 at 19:55

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