{"id":2009,"date":"2024-12-10T21:40:07","date_gmt":"2024-12-10T21:40:07","guid":{"rendered":"https:\/\/www.ethanepperly.com\/?p=2009"},"modified":"2024-12-15T21:48:13","modified_gmt":"2024-12-15T21:48:13","slug":"my-favorite-proof-of-the-cauchy-schwarz-inequality","status":"publish","type":"post","link":"https:\/\/www.ethanepperly.com\/index.php\/2024\/12\/10\/my-favorite-proof-of-the-cauchy-schwarz-inequality\/","title":{"rendered":"My Favorite Proof of the Cauchy\u2013Schwarz Inequality"},"content":{"rendered":"\n

The Cauchy\u2013Schwarz inequality<\/a> has many proofs. Here is my favorite, taken from Chapter 3 of The Schur complement and its applications<\/a><\/em>; the book is edited by Fuzhen Zhang<\/a>, and this chapter was contributed by him as well. Let \"x,y \in \real^n\" be vectors, assemble the matrix \"B = \onebytwo{x}{y}\", and form the Gram matrix<\/a>

  <\/span>   <\/span>\"\[A = B^\top B = \onebytwo{x}{y}^\top \onebytwo{x}{y} = \twobytwo{\norm{x}^2}{y^\top x}{x^\top y}{\norm{y}^2}.\]\"<\/p>Since<\/a> \"A\" is a Gram matrix, it is positive semidefinite<\/a>. Therefore, its determinant<\/a> is nonnegative:

  <\/span>   <\/span>\"\[\det(A) = \norm{x}^2\norm{y}^2 - |x^\top y|^2 \ge 0.\]\"<\/p>Rearrange to obtain the Cauchy\u2013Schwarz inequality

  <\/span>   <\/span>\"\[|x^\top y| \le \norm{x}\norm{y}.\]\"<\/p>Equality occurs if and only if \"A\" is a rank<\/a>-one matrix, which occurs if and only if \"x\" and \"y\" are scalar multiples.<\/p>\n\n\n\n

I like this proof because it is perhaps the simplest example of the (block) matrix technique for proving inequalities. Using this technique, one proves inequalities about scalars (or matrices) by embedding them in a clever way into a (larger) matrix. Here is another example of the matrix technique, adapted from the proof of Theorem 12.9 of these lecture notes<\/a> by Joel Tropp<\/a>. Jensen’s inequality<\/a> is a far-reaching and very useful inequality in probability theory. Here is one special case of the inequality.<\/p>\n\n\n\n

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Proposition<\/strong> (Jensen’s inequality for the inverse): Let \"x_1,\ldots,x_n\" be strictly positive numbers. Then the inverse of their average is no bigger than the average of their inverses:

  <\/span>   <\/span>\"\[\left( \frac{1}{n} \sum_{i=1}^n x_i \right)^{-1} \le \frac{1}{n} \sum_{i=1}^n \frac{1}{x_i}.\]\"<\/p><\/p>\n<\/blockquote>\n\n\n\n

To prove this result, embed each \"x_i\" into a \"2\times 2\" positive semidefinite matrix \"\twobytwo{x_i}{1}{1}{1/x_i}\". Taking the average of all such matrices, we observe that

  <\/span>   <\/span>\"\[A = \twobytwo{\frac{1}{n} \sum_{i=1}^n x_i}{1}{1}{\frac{1}{n} \sum_{i=1}^n \frac{1}{x_i}}\]\"<\/p>is positive semidefinite as well. Thus, its determinant is nonnegative:

  <\/span>   <\/span>\"\[\det(A) = \left(\frac{1}{n} \sum_{i=1}^n x_i\right) \left(\frac{1}{n} \sum_{i=1}^n \frac{1}{x_i}\right) - 1 \ge 0.\]\"<\/p> Rearrange to obtain

  <\/span>   <\/span>\"\[\left( \frac{1}{n} \sum_{i=1}^n x_i \right)^{-1} \le \frac{1}{n} \sum_{i=1}^n \frac{1}{x_i}.\]\"<\/p>Remarkably, we have proven a purely scalar inequality by appeals to matrix theory.<\/p>\n\n\n\n

The matrix technique for proving inequalities is very powerful. Check out Chapter 3 of The Schur complement and its applications<\/a><\/em> for many more examples.<\/p>\n\n\n\n

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