## Linear Algebra and Its Applications, Exercise 3.2.1

Exercise 3.2.1. a) Consider the vectors $a = \left( \sqrt{y}, \sqrt{x} \right)$ and $b = \left( \sqrt{x}, \sqrt{y} \right)$ where $x$ and $y$ are arbitrary positive real numbers. Use the Schwarz inequality involving $a$ and $b$ to derive a relationship between the arithmetic mean $\frac{1}{2}\left(x+y\right)$ and the geometric mean $\sqrt{xy}$.

b) Consider a vector from the origin to point $x$, a second vector of length $\|y\|$ from $x$ to the point $x+y$ and the third vector from the origin to $x+y$. Using the triangle inequality $\|x+y\| \le \|x\| + \|y\|$

derive the Schwarz inequality. (Hint: Square both sides of the inequality and expand the expression $\left( x+y \right)^T\left( x+y \right)$.)

Answer: a) From the Schwarz inequality we have $|a^Tb| \le \|a\| \|b\|$

From the definitions of $a$ and $b$, on the left side of the inequality we have $|a^Tb| = |\sqrt{y} \sqrt{x} + \sqrt{x} \sqrt{y}|$ $= |\sqrt{yx} + \sqrt{xy}| = 2|\sqrt{xy}| = 2\sqrt{xy}$

assuming we always choose the positive square root.

From the definitions of $a$ and $b$ we also have $\|a\|^2 = \left( \sqrt{y} \right)^2 + \left( \sqrt{x} \right)^2 = y + x = x+y$

and $\|b\|^2 = \left( \sqrt{x} \right)^2 + \left( \sqrt{y} \right)^2 = y + x = x+y$

so that the right side of the inequality is $\|a\|\|b\| = \sqrt{x+y} \sqrt{x+y}$ $= \sqrt{\left(x+y\right)^2} = x+y$

again assuming we choose the positive square root. (We know $x+y$ is positive since both $x$ and $y$ are.)

The Schwartz inequality $|a^Tb| \le \|a\| \|b\|$

then becomes $2\sqrt{xy} \le x+y$

or (dividing both sides by 2) $\sqrt{xy} \le \frac{1}{2}\left( x+y \right)$

We thus see that for any positive real numbers $x$ and $y$ the geometric mean $\sqrt{xy}$ is less than the arithmetic mean $\frac{1}{2}\left(x+y\right)$.

b) From the triangle inequality we have $\|x+y\| \le \|x\| + \|y\|$

for the vectors $x$ and $y$. Squaring the term on the left side of the inequality and using the commutative and distributive properties of the inner product we obtain $\|x+y\|^2 = \left( x+y \right)^T \left( x+y \right)$ $= x^T \left( x+y \right) + y^T \left( x+y \right)$ $= x^Tx + x^Ty + y^Tx + y^Ty$ $= x^Tx + 2x^Ty + y^Ty$ $= \|x\|^2 + 2x^Ty + \|y\|^2$

Squaring the term on the right side of the inequality we have $\left( \|x\| + \|y\| \right)^2 = \|x\|^2 + 2\|x\|\|y\| + \|y\|^2$

The inequality $\|x+y\| \le \|x\| + \|y\|$

is thus equivalent to the inequality $\|x\|^2 + 2x^Ty + \|y\|^2 \le \|x\|^2 + 2\|x\|\|y\| + \|y\|^2$

Subtracting $\|x\|^2$ and $\|y\|^2$ from both sides of the inequality gives us $2x^Ty \le 2\|x\|\|y\|$

and dividing both sides of the inequality by 2 produces $x^Ty \le \|x\|\|y\|$

Note that this is almost but not quite the Schwarz inequality: Since the Schwarz inequality involves the absolute value $|x^Ty|$ we must also prove that $-x^Ty \le \|x\|\|y\|$

(After all, the inner product $x^Ty$ might be negative, in which case the inequality $x^Ty \le \|x\|\|y\|$ would be trivially true, given that the term on the right side of the inequality is guaranteed to be positive.)

We have $-x^Ty = \left(-x\right)^Ty$. Since the triangle inequality holds for any two vectors we can restate it in terms of $-x$ and $y$ as follows: $\|\left(-x\right)+y\| \le \|-x\| + \|y\|$

Since $\|-x\| = \|x\|$ squaring the term on the right side of the inequality produces $\left(\|-x\| + \|y\|\right)^2 = \left(\|x\| + \|y\|\right)^2 = \|x\|^2 + 2\|x\|\|y\| + \|y\|^2$

as it did previously. However squaring the term on the left side of the inequality produces $\|\left(-x\right)+y\|^2 = \left( -x+y \right)^T \left( -x+y \right)$ $= \left(-x\right)^T \left( -x+y \right) + y^T \left( -x+y \right)$ $= \left(-x\right)^T\left(-x\right) + \left(-x\right)^Ty + y^T\left(-x\right) + y^Ty$ $= x^Tx - x^Ty - y^Tx + y^Ty = x^Tx - 2x^Ty + y^Ty$ $= \|x\|^2 - 2x^Ty + \|y\|^2$

The original triangle inequality $\|\left(-x\right)+y\| \le \|-x\| + \|y\|$

is thus equivalent to $\|x\|^2 - 2x^Ty + \|y\|^2 \le \|x\|^2 + 2\|x\|\|y\| + \|y\|^2$

or $-x^Ty \le \|x\|\|y\|$

Since we have both $-x^Ty \le \|x\|\|y\|$ and $x^Ty \le \|x\|\|y\|$ we therefore have $|x^Ty| \le \|x\|\|y\|$

which is the Schwarz inequality.

So the triangle inequality implies the Schwarz inequality.

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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