Thread: Cauchy Schwarz question

I have no idea to start. I'm struggling so much in this class. Can someone help me out?

(a) Given that x² + y² + z² = 1, use the cauchy-schwarz inequality to find the largest possible value of the expression 2x + 3y + 6z

(b) Let A be a diagonalizable matrix, all of whose eigenvalues are either 0 or 1. Show that A² = A

thanks in advance!

If A = (2,3,6) and B = (x,y,z), then

$\displaystyle \vec A \cdot \vec B = 2x + 3y + 6z
$
is the expression you want to maximize.

Cauchy Schwarz says
$\displaystyle (\vec A \cdot \vec B)^2 <= (\vec A \cdot \vec A) (\vec B \cdot \vec B)
$

$\displaystyle B^2 = x^2 + y^2 + z^2 = 1
$
$\displaystyle A^2 = 2^2 + 3^2 + 6^2 = 49
$

so
$\displaystyle (\vec A \cdot \vec B)^2 <= 49
$

or
$\displaystyle (\vec A \cdot \vec B) <= 7
$

For the 2nd, try
$\displaystyle A = S^{-1} D S
$

$\displaystyle A^2 = S^{-1} D S S^{-1} D S = S^{-1} D^2 S = S^{-1} D S = A
$

since D has only diagonal elements 0 or 1.

(b)Since A is a diagonalizable matrix, there exist invertable matrix B and diagonal matrix D( whose diagonal elements are 0 or 1), such that
$\displaystyle BAB^{-1}=D$.
since $\displaystyle D^2=D$, that is,
$\displaystyle BA^2B^{-1}=BAB^{-1}$
combined with B is invertable, gives the result.