Decomposition of a positive semidefinite matrix

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 $\rm Y$.