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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 $V$?
Rodrigo de Azevedo
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nlassaux
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  • 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.

     – V. Mikaelian Feb 07 '22 at 20:20

1 Answers1

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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 $\rm Y$.

Rodrigo de Azevedo
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    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 – nlassaux May 26 '16 at 22:10
  • Why can you not take the square root when the matrix is positive semidefinite? – JohnK Apr 18 '17 at 12:46
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    @JohnK You can. I assumed positive definiteness because $\rm V$ is square. – Rodrigo de Azevedo Apr 18 '17 at 12:51
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    @JohnK I edited my answer. I can live with $\rm V$ having rows full of zeros. – Rodrigo de Azevedo Apr 18 '17 at 12:59
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    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? – avk255 Nov 29 '19 at 16:28
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    @avk255 It should not be unique. One can always permute the columns of $\rm Q$ and the corresponding diagonal entries of $\Lambda$. – Rodrigo de Azevedo Nov 29 '19 at 18:36
  • Wikipedia says: "Y is positive definite if and only if such a decomposition exists with V invertible. " Does this mean, in the above case V is also invertible? (just to confirm)

    source: https://en.wikipedia.org/wiki/Definite_symmetric_matrix#Decomposition

     – Ufos Apr 03 '21 at 19:36
  • @Ufos If $\rm Y$ is only positive semidefinite, $\Lambda$ will have zero diagonal entries and, thus, $\rm V$ cannot be invertible. – Rodrigo de Azevedo Apr 03 '21 at 19:41
  • @RodrigodeAzevedo yeah, right, the article also specifies the matrix must be hermitian, of which symmetric is a special case. So, in such case, V still will be invertible? – Ufos Apr 03 '21 at 19:48
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    @Ufos The zero square matrix is also Hermitian. The OP mentioned that the matrix is real, and I am not so familiar with complex matrices. – Rodrigo de Azevedo Apr 03 '21 at 19:55
  • Using house holder, even $Q$ is positive definite right? But wouldn't that mean $\sigma=Y$ – D. Sikilai Apr 13 '24 at 19:54