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 $V$?
Best Answer
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 $\rm Y$.