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February 18, 2011

Two interesting proofs of the Cauchy-Schwarz inequality (complex case)

Filed under: mathematics — ckrao @ 10:06 am

In a previous post I gave two proofs of the Cauchy-Schwarz inequality for a real inner product space. Here I look at the case of a complex inner product space:

\displaystyle \left|\langle{x,y\rangle}\right| \leq \left\|x\right\| \left\|y\right\| \quad...(1)

One commonly used proof in the discrete case is via Lagrange’s identity. Here are two other nice proofs.

1. For any complex number \alpha we use the non-negativity of the square \displaystyle \left\|x-\alpha y\right\|^2:

\begin{array}{lcl} 0 &\leq & \left\|x-\alpha y\right\|^2\\&=& \left\|x|\right\|^2 + |\alpha|^2\left\|y\right\|^2 - \langle{x,\alpha y\rangle} - \langle{\alpha y,x\rangle}\\&=& \left(\begin{array}{cc} 1 & \overline{\alpha}\end{array}\right)\left( \begin{array}{cc} \left\|x\right\|^2 & \langle{x,y\rangle}\\ \langle{y,x\rangle} & \left\|y\right\|^2\end{array}\right)\left(\begin{array}{c} 1\\ \alpha \end{array}\right).\end{array}

Let \displaystyle A=\left( \begin{array}{cc} \left\|x\right\|^2 & \langle{x,y\rangle}\\ \langle{y,x\rangle} & \left\|y\right\|^2\end{array}\right). The above equation shows that if the 2 by 1 vector z = \left(\begin{array}{c} 1\\ \alpha \end{array}\right) has non-zero first component, then z^*A z \geq 0. This non-negativity also holds when z has zero first component as for \displaystyle z = \left(\begin{array}{c}0 \\ \alpha \end{array}\right) we have z^*A z = |\alpha|^2 \left\|y\right\|^2\geq 0. We conclude that A is positive semidefinite, so has non-negative determinant

\displaystyle 0 \leq \det A = \left\|x\right\|^2\left\|y\right\|^2 - \left|\langle{x,y\rangle} \right|^2,

from which (1) follows.

2. For convenience let \hat{y} = y/\left\|y\right\|. Consider the projection of x onto y, given by \langle{x,\hat{y}\rangle}\hat{y}. We compute the squared length of their difference.

\begin{array}{lcl} 0 & \leq & \left\|x- \langle{x,\hat{y}\rangle}\hat{y}\right\|^2\\&=& ||x||^2 + \left|\langle{x,\hat{y}\rangle} \right|^2 - 2|\langle{x,\hat{y}\rangle}|^2\\&=& \left\|x\right\|^2 - \left|\langle{x,\hat{y}\rangle}\right|^2 \\&=& \left\|x\right\|^2 - \frac{\left|\langle{x,y\rangle}\right|^2}{\left\|y\right\|^2},\end{array}

from which (1) easily follows. Equality holds iff the projection of x onto y is the zero vector: in other words, they are parallel or one of the vectors is zero.

Proof 2 shows that the Cauchy-Schwarz inequality is a consequence of Gram-Schmidt orthogonalisation applied to two vectors, also mentioned in [1].

Reference

[1] J. M. Steele, The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities, MAA Problem Books Series, 2004.

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2 Comments »

  1. […] thing in the complex case, I ran into some trouble. First, there are proofs here, here, here, and here, but none of them do it the way I’m thinking […]

    Pingback by Cauchy-Schwarz in complex case, using discriminant - MathHub — May 7, 2016 @ 10:50 pm | Reply


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