What is the Cauchy-Schwarz inequality?

by | Oct 3, 2023 | Linear Algebra

Cauchy Schwarz Inequality

If u and v are two vectors, then as per the Cauchy-Schwarz inequality, the magnitude of the dot product between these two vectors is no longer than the product of the norms of these two vectors.

In other words,

|\vec{u} \cdot \vec{v}|\leq ||\vec{u}||||\vec{v}||

Now, if both u and v are n-dimensional vectors, then we know that

|\vec{u} \cdot \vec{v}|=\sum_{i=1}^{n}{u_i v_i}\\ =(u_1v_1+u_2v_2+ ... +u_nv_n)

and the magnitude of an n-dimensional vector u is:

||\vec{u}||=\sqrt{\sum_{i=1}^{n}{{u_i}^2}}=\sqrt{({u_1}^2+{u_2}^2+ ... +{u_n}^2)}

So,

||\vec{u}||||\vec{v}||=\sqrt{\sum_{i=1}^{n}{{u_i}^2}}\sqrt{\sum_{i=1}^{n}{{v_i}^2}} \\ =\sqrt{({u_1}^2+{u_2}^2+ ... +{u_n}^2)}\sqrt{({v_1}^2+{v_2}^2+ ... +{v_n}^2)}

Now, we know that:

(ab+cd)^2=a^2b^2+2abcd+c^2d^2 \leq (a^2+c^2)(b^2+d^2)

So, we can write:

(u_1v_1+u_2v_2+ ... +u_nv_n)^2 \leq ({u_1}^2+{u_2}^2+ ... +{u_n}^2)({v_1}^2+{v_2}^2+ ... +{v_n}^2)

or,

(\sum_{i=1}^{n}{u_i v_i})^2 \leq \sum_{i=1}^{n}{{u_i}^2}\sum_{i=1}^{n}{{v_i}^2}

 

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Amrita Mitra

Author

Ms. Amrita Mitra is an author, who has authored the books “Cryptography And Public Key Infrastructure“, “Web Application Vulnerabilities And Prevention“, “A Guide To Cyber Security” and “Phishing: Detection, Analysis And Prevention“. She is also the founder of Asigosec Technologies, the company that owns The Security Buddy.

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