# Decomposition of a positive semidefinite matrix

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 $V$?
• > 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. Feb 7 at 20:20

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 $$\rm Y$$.

• 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 May 26, 2016 at 22:10
• @JohnK You can. I assumed positive definiteness because $\rm V$ is square. Apr 18, 2017 at 12:51
• @JohnK I edited my answer. I can live with $\rm V$ having rows full of zeros. Apr 18, 2017 at 12:59
• 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? Nov 29, 2019 at 16:28
• @avk255 It should not be unique. One can always permute the columns of $\rm Q$ and the corresponding diagonal entries of $\Lambda$. Nov 29, 2019 at 18:36