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Let $X$ be a real-valued square $n \times n$ matrix.

Is there decomposition $X = \Lambda \Lambda^\top$ where $\Lambda$ is a a real-valued $n \times k$ matrix, that always exists?

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    $\begingroup$ $\Lambda \Lambda^\top$ is always symmetric. $\endgroup$ – Lord Shark the Unknown Jul 23 '18 at 18:27
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    $\begingroup$ ..and positive semi-definite $\endgroup$ –  mheldman Jul 23 '18 at 18:32
  • $\begingroup$ Oh of course, Thanks! $\endgroup$ – stollenm Jul 23 '18 at 18:32
  • $\begingroup$ Interestingly, if $X$ is complex and symmetric, then there exists a complex $\Lambda$ so that $X = \Lambda \Lambda^T$ $\endgroup$ – Omnomnomnom Jul 23 '18 at 18:43
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Every real symmetric matrix can be written in the form

$$ X = Q D Q^T $$

where $Q$ is formed from an orthonormal set of eigenvectors of $X$ and the diagonal of $D$ contains the corresponding eigenvalues. If the eigenvalues are non-negative, then the real matrix $\Lambda = PD^{1/2}$ satisfies your condition.

$$ X = (PD^{1/2})(D^{1/2}P^T) = \Lambda \Lambda^T $$

Note that all matrices formed by the product $ \Lambda \Lambda^T $ are positive semidefinite, so the following statements are equivalent:

  1. $X$ is a real positive semidefinite matrix.
  2. There exists a real matrix $\Lambda$ such that $X = \Lambda \Lambda^T$.
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If $X \in \mathbb{C}^{m \times m}$ is positive definite and hermitian then there exists the Cholesky decomposition such that

$$ X = R^{*}R$$

where

$$ X = \underbrace{R_{1}^{*}R_{2}^{*} \cdots R_{m}^{*} }_{R^{*}}\underbrace{R_{m} \cdots R_{2} R_{1}}_{R}$$

with $$ X = R^{*}R , r_{jj} > 0 $$ which can be done as

enter image description here

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