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Let $Y \in \mathbb{R}^{n \times n}$ be a symmetric, positive semidefinite matrix such that $Y_{kk} = 1$ for all $k$.

This matrix is supposed to be factorized as $Y = V^T V$, where $V \in \mathbb{R}^{n \times n}$.

  • Does this factorization/decomposition have a name?
  • How is it possible to compute $V$?
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  • $\begingroup$ > Does this factorization/decomposition have a name? ANSWER. At some sources $Y=V^T V$ (in real case) or $Y=V^* V$ (in complex case) is called just PSD decomposition. Also, the fact that $Y=V^T V=W^T W$ iff $C=QW$ for an orthogonal $Q$ is sometimes called orthogonal freedom. And the fact that $Y=V^* V=W^* W$ iff $C=QW$ for an unitary $Q$ is sometimes called unitary freedom. $\endgroup$
    – V. Mikaelian
    Feb 7, 2022 at 20:20

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If $\rm Y$ is symmetric, then it is diagonalizable, its eigenvalues are real, and its eigenvectors are orthogonal. Hence, $\rm Y$ has an eigendecomposition $\rm Y = Q \Lambda Q^{\top}$, where the columns of $\rm Q$ are the eigenvectors of $\rm Y$ and the diagonal entries of diagonal matrix $\Lambda$ are the eigenvalues of $\rm Y$.

If $\rm Y$ is also positive semidefinite, then all its eigenvalues are nonnegative, which means that we can take their square roots. Hence,

$$\rm Y = Q \Lambda Q^{\top} = Q \Lambda^{\frac 12} \Lambda^{\frac 12} Q^{\top} = \underbrace{\left( Q \Lambda^{\frac 12} \right)}_{=: {\rm V}} \left( Q \Lambda^{\frac 12} \right)^{\top} = V^{\top} V$$

Note that the rows of $\rm V$ are the eigenvectors of $\rm Y$ multiplied by the square roots of the (nonnegative) eigenvalues of $\rm Y$.

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    $\begingroup$ You are right, the matrix is symmetric, I have not noticed it. I've added it to the problem description. I try to decompose it as you said. Thanks $\endgroup$
    – nlassaux
    May 26, 2016 at 22:10
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    $\begingroup$ @JohnK You can. I assumed positive definiteness because $\rm V$ is square. $\endgroup$
    – Rodrigo de Azevedo
    Apr 18, 2017 at 12:51
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    $\begingroup$ @JohnK I edited my answer. I can live with $\rm V$ having rows full of zeros. $\endgroup$
    – Rodrigo de Azevedo
    Apr 18, 2017 at 12:59
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    $\begingroup$ Is anything known about the uniqueness of this decomposition? If I have a symmetric psd matrix C such that $C=AA^T=BB^T$ then can we say something about relation between A and B? $\endgroup$
    – avk255
    Nov 29, 2019 at 16:28
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    $\begingroup$ @avk255 It should not be unique. One can always permute the columns of $\rm Q$ and the corresponding diagonal entries of $\Lambda$. $\endgroup$
    – Rodrigo de Azevedo
    Nov 29, 2019 at 18:36

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