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 $V$?
  • $\begingroup$ > 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. $\endgroup$ Feb 7 at 20:20

1 Answer 1


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 $\rm Y$.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @JohnK You can. I assumed positive definiteness because $\rm V$ is square. $\endgroup$ Apr 18, 2017 at 12:51
  • 1
    $\begingroup$ @JohnK I edited my answer. I can live with $\rm V$ having rows full of zeros. $\endgroup$ Apr 18, 2017 at 12:59
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ @avk255 It should not be unique. One can always permute the columns of $\rm Q$ and the corresponding diagonal entries of $\Lambda$. $\endgroup$ Nov 29, 2019 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.