Linear Algebra and Its Applications, Exercise 3.2.6

Exercise 3.2.6. Suppose that $a$ and $b$ are unit vectors. Then a one-line proof of the Schwarz inequality is as follows:

$|a^Tb| = |\sum a_jb_j| \le \sum |a_j||b_j| \le \sum \frac{1}{2} \left(|a_j|^2+|b_j|^2\right) = \frac{1}{2} + \frac{1}{2} = \|a\|\|b\|$

What previous exercise justifies the middle step of this proof?

Answer: From exercise 3.2.1(a) we have $\sqrt{xy} \le \frac{1}{2}\left(x+y\right)$ for any positive $x$ and $y$ . We can easily extend this result to any non-negative $x$ and $y$ : If $x = 0$ then $\sqrt{xy} = \sqrt{0} = 0$ and $\frac{1}{2} \left(x+y\right) = \frac{1}{2} y \ge 0$ as long as $y \ge 0$ . So $\sqrt{xy} \le \frac{1}{2}\left(x+y\right)$ if $x = 0$ and $y \ge 0$ . A similar argument shows that $\sqrt{xy} \le \frac{1}{2}\left(x+y\right)$ if $x \ge 0$ and $y = 0$ . Combined with the previous result this shows that $\sqrt{xy} \le \frac{1}{2}\left(x+y\right)$ for all $x \ge 0$ and $y \ge 0$ .

Let $x = |a_j|^2 \ge 0$ and $y = |b_j|^2 \ge 0$ . We then have $\sqrt{|a_j|^2|b_j|^2} \le \frac{1}{2}\left(|a_j|^2+|b_j|^2\right)$ or $|a_j||b_j| \le \frac{1}{2}\left(|a_j|^2+|b_j|^2\right)$ . This is the middle step of the one-line proof of the Schwarz inequality shown above.

Note that exercise 3.2.1(a) actually used the Schwarz inequality to prove that $\sqrt{xy} \le \frac{1}{2}\left(x+y\right)$ , so strictly speaking the above proof is circular. To remedy this we can prove the result from exercise 3.2.1(a) without using the Schwarz inequality. The proof (adapted from Wikipedia) is as follows:

We know that $\left(x-y\right)^2 \ge 0$ for any $x$ and $y$ . Expanding the lefthand side we have $\left(x-y\right)^2 = x^2 -2xy + y^2$ . Substituting into the inequality we have $x^2 -2xy + y^2 \ge 0$ or $2xy \le x^2 + y^2$ . Adding $2xy$ to both sides of the inequality we have $2xy + 2xy \le x^2 + 2xy+ y^2$ or $4xy \le \left(x+ y\right)^2$ . If $x \ge 0$ and $y \ge 0$ then $xy \ge 0$ and we can take the square root of both sides of the inequality to obtain $2\sqrt{xy} \le \left(x+ y\right)$ or $\sqrt{xy} \le \frac{1}{2}\left(x+y\right)$ .

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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Linear Algebra and Its Applications, Exercise 3.2.6

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