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?

  • $\begingroup$ You just need to have $\langle u,u \rangle \ge 0$ for all $u$. It does not have to be positive definite, but only positive. $\endgroup$ – DisintegratingByParts Jul 11 '18 at 1:58

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 $$\langle u, u \rangle \langle v, v \rangle = u^\dagger M u v^\dagger M v = \lambda \mu (u^\dagger u)(v^\dagger v) < 0 \le |\langle u, v \rangle|^2.$$

  • $\begingroup$ The inequality may hold in the semidefinite cases, but the usual statement of the theorem includes the inequality only being saturated for parallel vectors. That part fails for singular $M$. $\endgroup$ – J.G. Jul 10 '18 at 15:21

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.


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