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If there is a positive semi-definite/symmetric matrix $B \in R^{n \times n}$ that has rank $r$.

How do I prove that there exists an $A \in R^{n \times r}$ such that $B = AA^{T}$

To be positive semidefinite,I know the following inequality has to hold:

$$x^TBx \geq 0$$

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By the spectral theorem $B=ODO^T$ for some orthogonal matrix $O$ and some diagonal matrix $D$, and $D$ has the same rank as $B$. Thus, $D$ has $r$ entries strictly greater than zero, and the rest equal $0$ (to see why, in the definition of positive semidefinite take $x$ to be a vector with one entry equals $1$ and the rest $0$). Let $C\in\mathbb R^{n\times r}$ be the "diagonal" matrix with entries on the diagonal equal to the square root of the non-zero entries of $D$, and you will see that $D=CC^T$. Thus $B=OCC^TO^T=AA^T$ where $A=OC$.

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