## 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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