Cauchy-Schwarz for positive definite matrix

Question

I've run across the following inequality, and haven't been able to come up with a proof for it (and potentially am not convinced it is true).

Show that $$|\langle a, Mb\rangle| \leq \sqrt{ a^\intercal M a b^\intercal M b}$$ for a positive definite matrix $M$ and arbitrary vectors $a$ and $b$.

Attempt at Solution

Using Cauchy Schwarz we get that $$|\langle a, Mb \rangle| \leq \Vert a \Vert \, \Vert b \Vert \\ = \sqrt{a^\intercal a (Mb)^\intercal Mb} \\ = \sqrt{a^\intercal a b^\intercal M M b}.$$

However, from here on out we can't just move around the $M$.

Answer 1

Presumably the symbol $\langle x,y\rangle$ means the Euclidean inner product of two vectors $x$ and $y$. If so, the inequality in question is precisely Cauchy-Schwarz inequality, not just something analogous to it.

Since $M$ is positive definite, $(x,y):=\langle x,My\rangle$ defines an inner product. The inequality in question is thus equivalent to $|(a,b)|\le\sqrt{(a,a)(b,b)}$, which is Cauchy-Schwarz inequality with respect to the new inner product $(\cdot,\cdot)$.

Alternatively, since $M$ is positive definite, it admits a unique positive definite square root $M^{1/2}$. Therefore the inequality in question can be rewritten as $$ |\langle M^{1/2}a,M^{1/2}b\rangle|\le\sqrt{\langle M^{1/2}a,M^{1/2}a\rangle\langle M^{1/2}b,M^{1/2}b\rangle}, $$ which is Cauchy-Schwarz inequality with respect to the old inner product $\langle\cdot,\cdot\rangle$.

Answer 2

Please notice that inner product is conjugate symmetric thus $|\langle x, y\rangle|=\overline{|\langle y, x\rangle|}$ in other words if we are speaking on complex field $\mathbb{C}$ for example, for two complex vector $x$ and $y$ we have $|\langle x, y\rangle|=x^*y$ and $|\langle y, x\rangle|=y^*x$ and thus $|\langle x, y\rangle|=|\langle y, x\rangle|^*$ where $^*$ means transpose-complex conjugate (Hermitian) operation. Since in your notation you are using $^\intercal$ so I assume we are speaking about real number field $\mathbb{R}$. In this case $|\langle x, y\rangle|=|\langle y, x\rangle|^\intercal$ or equivelently $x^\intercal y=y^\intercal x$. Now back to problem

$$|\langle a, Mb\rangle|=a^\intercal Mb=b^\intercal M^\intercal a=|\langle Mb, a\rangle|$$

and we define $\Vert x \Vert=\sqrt{|\langle x, x\rangle|}$ Thus the Cauchy-Schwarz inequality yields

$$|\langle a, Mb \rangle| \leq \Vert a \Vert \, \Vert Mb \Vert = \sqrt{a^\intercal a}\sqrt{ (Mb)^\intercal Mb}=\sqrt{a^\intercal a}\sqrt{ b^\intercal M^\intercal Mb}=\sqrt{a^\intercal a}\sqrt{ b^\intercal M Mb} \neq \sqrt{ a^\intercal M a b^\intercal M b}$$ $M$ is positive definite and hence symmetric thus $M^ \intercal=M$.

And to show that the last inequality is true, assume $M$ is full rank and we perform eigenvalue decomposition $M=U\Lambda U^\intercal$ where $U^\intercal U=UU^\intercal=I$, thus we can write $a$ and $b$ in terms of the eigen-basis of $M$, i.e. $U$. Thus for some vectors $c,d$ we have $a=Uc$ and $b=Ud$. Now we have

$$\sqrt{a^\intercal a b^\intercal M Mb}=\sqrt{c^\intercal U^\intercal Ucd^\intercal U^\intercal U\Lambda U^\intercal U \Lambda U^\intercal Ud}=\sqrt{c^\intercal cd^\intercal \Lambda^2d}=\sqrt{\left( \sum_i c_i^2 \right) \left( \sum_i \lambda_i^2d_i^2 \right)}$$ While $$\sqrt{ a^\intercal M a b^\intercal M b} = \sqrt{c^\intercal U^\intercal U\Lambda U^\intercal Ucd^\intercal U^\intercal U\Lambda U^\intercal Ud}= \sqrt{c^\intercal \Lambda cd^\intercal \Lambda d}=\sqrt{\left( \sum_i \lambda_i c_i^2 \right) \left( \sum_i \lambda_i d_i^2 \right)}$$