Orthonormalization with a matrix

Question

In my lecture notes we have a $n \times p$ block matrix $$ Y = \begin{pmatrix} I_p \\ Q \end{pmatrix} $$ and it is orthonormalized in the following way: $$ \hat{Y} = \begin{pmatrix} I_p \\ Q \end{pmatrix} \left(I + Q^*Q\right)^{-1/2}. $$ I can not see how the matrix on the right orthonormalizes neither do I know how to handle the inverse square root. Can someone elaborate? Any help is very much appreciated!

Answer 1

Presumably, when you say that we're "orthonormalizing" the matrix $Y$, you mean that the columns of $\hat Y$ will be an orthonormal basis for the column space of $Y$. I think our only assumption on $Q$ is that it is $(n-p)\times p$.

First, the $-1/2$: note that for any positive semidefinite matrix $A$, there exists a unique positive semidefinite square root $A^{1/2}$. If $A$ is invertible (positive definite), then we may find the inverse of this square root, namely $A^{-1/2}$. Notably, $A^{1/2}$ (and $A^{-1/2}$) necessarily commute with $A$.

To see that the columns of $\hat Y$ are orthonormal, it suffices to note that $(\hat Y)^* \hat Y = I$. Indeed, we have: $$ (\hat Y)^* \hat Y = [(I + Q^*Q)^{-1/2}]^* \pmatrix{I&Q^*}\pmatrix{I\\Q}(I + Q^*Q)^{-1/2} \\ = (I + Q^*Q)^{-1/2}(I + Q^*Q)(I + Q^*Q)^{-1/2} \\ = (I + Q^*Q)(I + Q^*Q)^{-1/2}(I + Q^*Q)^{-1/2} \\ = (I + Q^*Q)(I + Q^*Q)^{-1} = I $$ To see that the column space of $\hat Y$ is identical to that of $Y$, it suffices to note that $\hat Y = YM$ for an invertible matrix $M$.