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 $V$?

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,

$$\rm Y = Q \Lambda Q^{\top} = Q \Lambda^{\frac 12} \Lambda^{\frac 12} Q^{\top} = (Q \Lambda^{\frac 12}) (Q \Lambda^{\frac 12})^{\top}$$

Let $\rm V := (Q \Lambda^{\frac 12})^{\top}$. Thus, $\rm Y = V^{\top} V$. Note that the rows of $\rm V$ are the eigenvectors of $\rm Y$ multiplied by the square roots of the (nonnegative) eigenvalues of $\rm Y$.

  • $\begingroup$ 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 '16 at 22:10
  • $\begingroup$ Why can you not take the square root when the matrix is positive semidefinite? $\endgroup$ – JohnK Apr 18 '17 at 12:46
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    $\begingroup$ @JohnK You can. I assumed positive definiteness because $\rm V$ is square. $\endgroup$ – Rodrigo de Azevedo Apr 18 '17 at 12:51
  • $\begingroup$ @JohnK I edited my answer. I can live with $\rm V$ having rows full of zeros. $\endgroup$ – Rodrigo de Azevedo Apr 18 '17 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 '19 at 16:28

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