# Decomposing Square Matrix Into Two Matrices Which Are Transposes of Each Other

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?

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

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 $X = \Lambda \Lambda^T$.

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 