Extended Cauchy-Schwarz Inequality Proof

Extended Cauchy-Schwarz inequality is given as

$$(b^{T}Bb)(d^T{B}^{-1}d)\ge (b^{T}d{)}^2$$

With Cauchy-Schwarz inequality derived in the last post, the extended Cauchy-Schwarz inequality can be proved as well. This inequality is only applicable when the matrix B is a positive definite matrix.

As always, I shall list out some prerequisite information.

1. The matrix $B$ is positive definite, which means that it is also a symmetric matrix in which all eigenvalues are greater than or equal to 0.

$$B=PQ{P}^T=(P{Q}^{\frac{1}{2}}{P}^T)(P{Q}^{\frac{1}{2}}{P}^T)$$

2. A symmetric matrix can be decomposed into two square root matrices.

$$\Vert b{\Vert }^2\Vert d{\Vert }^2\ge (bd{)}^2$$

3. The Cauchy-Schwarz Inequality we already knew.

We can then state the Cauchy-Schwarz Inequality as follows such that $b^{\ast }=B^{\frac{1}{2}}b$ and $d^{\ast }=B^{-\frac{1}{2}}d$.

$$(b^{\ast T}{b}^{\ast })(d^{\ast T}{d}^{\ast })\ge (b^{\ast T}{d}^{\ast }{)}^2$$

References:

• Applied Multivariate Statistical Analysis (2014)
• Extended Cauchy-Schwarz Inequality from the youtube channel statisticmatt