Definiteness of matrices formed by product of PD matrices and projections

Question

Let $ x \in S^{n-1}$ and a positive definite matrix $M \in \mathbb{R}^{n \times n}$. Define $$P := I - (x^T M^{-1} x)^{-1} M^{-1} xx^T$$ and $Q := M^{-1} P^T$. Is $Q$ positive semi definite?

(I figured out it is symmetric since $M P M^{-1} = P^T$ but I may be wrong)

So far

I noticed that for any vector $v$ perpendicular to $M^{-1} x $ we have

$$\begin{eqnarray} Qv &=& M^{-1} P^Tv \nonumber \\ &=& M^{-1} (I - (x^T M^{-1} x)^{-1}xx^T M^{-1} ) v \nonumber \\ &=& M^{-1}v \nonumber \end{eqnarray}$$

and then, since $M$ is positive definite, we have $v^T Qv = v^T M^{-1} v > 0$ This gives us at least $n-1$ linearly independent vectors for which $v^T Qv > 0$. If these were eigenvectors of $M^{-1}$ then it would have implied that we have $n-1$ positive eigenvalues, which leaves us with only one eigenpair to worry about. This is not the case, but I suppose we can somehow utilise this information.

Moreover, $Q$ as a non empty null space, as $ P^T x = 0$. And from that I infer that $x$ is not orthogonal to $M^{-1} x$.

Synthetic simulations suggests that $Q$ is indeed positive semi-definite with kernel dimension 1 (up to machine precision), but there's always the chance for a bug in the code and that some edge cases were missed.

Is there anything else that I'm missing? Is this statement possible to prove/ disprove given the information supplied?

Answer 1

Note that $Q=M^{-1}-(x^T M^{-1} x)^{-1} M^{-1}xx^TM^{-1}$ is symmetric. Also, by Cauchy-Schwarz inequality, \begin{align} v^TQv &=v^TM^{-1}v-(x^T M^{-1} x)^{-1}v^TM^{-1}xx^TM^{-1}v\\ &=\frac{(v^TM^{-1}v)(x^TM^{-1}x)-(v^TM^{-1}x)^2}{x^TM^{-1}x}\\ &=\frac{\|M^{-1/2}v\|^2\|M^{-1/2}x\|^2-\langle M^{-1/2}v,M^{-1/2}x\rangle^2}{\|M^{-1/2}x\|^2}\ge0\tag{1} \end{align} for any vector $v$. Therefore $Q$ is positive semidefinite. However, it is not positive definite, because $x^TQx=0$ (i.e. equality holds in $(1)$ when $v=x$).

Alternatively, note that $Q$ is the Schur complement of $x^TM^{-1}x$ in $$ A=\pmatrix{M^{-1}&M^{-1}x\\ x^TM^{-1}&x^TM^{-1}x}=\pmatrix{M^{-1/2}\\ x^TM^{-1/2}}\pmatrix{M^{-1/2}&M^{-1/2}x}. $$ Since $x^TM^{-1}x>0$ but $A$ is positive semidefinite and singular, $Q$ is positive semidefinite and singular too.

Answer 2

Here's the result by looking at the Signature of the matrix.
First $M$ evidently is real symmetric positive definite though the original post did not mention symmetry.

simplifications
let $B:= M^{-1}$ and $B$ is real symmetric positive definite.
$\alpha:= \mathbf x^T B \mathbf x\gt 0$

rescaling by a positive number doesn't change positive (semi)definiteness so it suffices to check whether $\alpha Q = \alpha B P^T$ is real symmetric positive semidefinite. By inspection this is symmetric (i.e. addition of two symmetric matrices), so it suffices to check the signs of the eigenvalues -- i.e. we want to know the matrix's Signature.

$\alpha Q = \alpha B P^T = \alpha B - B\mathbf {xx}^T B = B^{\frac{1}{2}} \big(\alpha I - B^{\frac{1}{2}}\mathbf x \mathbf x^T B^{\frac{1}{2}}\big) B^{\frac{1}{2}}$

and by Sylvester's Law of Inertia it suffices to check the Signature of
$\big(\alpha I - B^{\frac{1}{2}}\mathbf x \mathbf x^T B^{\frac{1}{2}}\big) $

now the main argument
$- B^{\frac{1}{2}}\mathbf x \mathbf x^T B^{\frac{1}{2}}$
is a real symmetric negative semidefinite matrix with rank one, so

$\lambda_k = 0$
$k \in \{1,2,..., n-1\}$
and
$\lambda_n = \text{trace}\Big(- B^{\frac{1}{2}}\mathbf x \mathbf x^T B^{\frac{1}{2}}\Big) = -\text{trace}\Big( B \mathbf x\mathbf x^T\Big) =-\text{trace}\Big( \mathbf x^T B \mathbf x\Big) = - \mathbf x^T B \mathbf x= - \alpha$

adding $\alpha I$ to this matrix shifts all eigenvalues by $\alpha$ with the result that

$\lambda_k = 0 +\alpha = \alpha \gt 0,\text{ for } k \in \{1,2,..., n-1\}$ but
$\lambda_n = -\alpha + \alpha = 0$

running this backwards we now know the signature of $Q$: it has $n-1$ eigenvalues that are positive, $0$ that are negative, and $1$ that is zero. So $Q$ is real symmetric positive semi-definite with rank $n-1$ as conjectured.