Prove the matrix to be symmetric positive definite

Question

Let $M$ be a $q \times p$ real matrix. I want to prove the follwoing: if $I_p - M^\mathrm{T} M$ is positive definite, then the $(p + q) \times (p + q)$ matrix $$ S = \begin{pmatrix} I_p & M^\mathrm{T} \\ M & I_q \\ \end{pmatrix} $$ is symmetric positive definite.

How do I go about doing this?

Answer 1

Hint: Try the block diagonal form of the matrix $S$. \begin{equation} S = \begin{pmatrix} I - M^{'}M & M^{'} \\ 0 & I \end{pmatrix} \begin{pmatrix} I & 0 \\ M & I \end{pmatrix} \end{equation}

\begin{equation} S = \begin{pmatrix} I & M^{'} \\ 0 & I \end{pmatrix} \begin{pmatrix} I - M^{'}M & 0 \\ 0 & I \end{pmatrix} \begin{pmatrix} I & 0 \\ M & I \end{pmatrix} \end{equation}

and using the fact that if $A$ is invertible and $S$ is positive definite $A^{'}SA$ is positive definite as well.

Answer 2

Recall that the eigenvalues of a positive definite matrix are positive and that a symmetric matrix is positive definite if and only all its eigenvalues positive.

Since both $M^TM$ and $I_p - M^TM$ are positive definite, this implies that the eigenvalues of $M^TM$ must satisfy $0 < \mu < 1$. Let $(x,y)^T$ be an eigenvector of $S$ associated to the eigenvalue $\lambda$. Then we have

$$ x + M^T y = \lambda x \implies M^T y = (\lambda - 1)x,\\ Mx + y = \lambda y \implies Mx = (\lambda - 1)y. $$

Hence,

$$ M^TM x = M^T (\lambda - 1)y = (\lambda - 1)^2 x. $$

If $x = 0$ then $\lambda = 1$. If $x \neq 0$, then $x$ is an eigenvector of $M^T M$ associated with the eigenvalue $(\lambda - 1)^2$ which implies that $0 < (\lambda - 1)^2 < 1$. Thus, $0 < \lambda < 2$. In any case, we have shown that the eigenvalues of $S$ are positive and hence $S$ is positive definite.

Answer 3

Use the fact that a square matrix $A$ is positive definite iff $z^\mathrm{T} A z > 0$ for all $z$, along with the factorization @karakfa gave for $S$.