When does the Cauchy-Schwarz inequality hold

Question

If we have a hermitian matrix $M$, and we define an inner product on $\mathbb{C}^n$ by $\left\langle u,v \right\rangle=u^{\dagger}Mv$, does it follow that $$\left|\left\langle u,v \right\rangle\right|^2\leq \left\langle u,u \right\rangle\left\langle v,v \right\rangle$$ or, are there additional conditions on $M$ in order for this to be true?

Answer 1

If you have positive definite matrix (i.e. all its eigenvalues are strictly positive), not just a Hermitian matrix, then $\langle u, v \rangle$ is an inner product; positive definiteness helps prove the positive definiteness axiom of inner product. The Cauchy-Schwarz inequality therefore holds, by the usual proofs.

If you have a negative definite matrix, then $-M$ is positive definite, hence $-\langle u, v \rangle$ forms an inner product. By the Cauchy-Schwarz inequality, we get $$|-\langle u, v \rangle|^2 \le (-\langle u, u, \rangle)(-\langle v, v \rangle),$$ which is equivalent to the inequality stated.

The definiteness in both cases can be relaxed to indefiniteness too. You won't get an inner product, but Cauchy Schwarz still holds. For example, in this proof, note that definiteness is never used. So, in a sense, we just require all the eigenvalues of $M$ to be greater than or equal to $0$, or all less than or equal to $0$.

If $M$ has eigenvalues that are each strictly positive and strictly negative, then the inequality doesn't hold. Take $u$ an eigenvalue corresponding to $\lambda > 0$ and $v$ corresponding to $\mu < 0$. Then $$\langle u, u \rangle \langle v, v \rangle = u^\dagger M u v^\dagger M v = \lambda \mu (u^\dagger u)(v^\dagger v) < 0 \le |\langle u, v \rangle|^2.$$

Answer 2

Cauchy-Schwarz inequality holds for any positive-definite bilinear form.

Let $\xi>0$ positive-definite square pattern, and $x_1,...,x_n\in V$

Define Gramian matrix $B_\xi(x_1,...x_n)_{ij}=\xi(x_i,x_j)$ and $G_\xi(x_1,...,x_n)=det(B_\xi(x_1,...x_n))$

We can show $G_\xi(x_1,...,x_n)\geq0$ given $\xi>0$, and the equality attained iff $x_1,...x_n$ are linearly dependent.(Using Sylvester's theorem and the fact $\xi>0$ )

Cauchy-Schwarz can now be derived as a special case of $n=2$ (Note that $n$ is not necessarily $dim(V)$)

$G_\xi(x,y)=det\begin{bmatrix}\xi(x,x)&\xi(x,y)\\\xi(y,x)&\xi(y,y)\end{bmatrix}= \xi(x,x)\xi(y,y)-\xi(x,y)\xi(y,x)\geq0$

If we arrange the inequality and use the symmetry property of $\xi$ we get

$\xi(x,x)\xi(y,y)\geq|\xi(x,y)|^2$ with equality iff x and y are linearly dependent.