Cauchy Schwarz question

**mapleleaf** · December 1st 2009, 08:57 AM

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!

**qmech** · December 1st 2009, 10:12 AM

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

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

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

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

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

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

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

$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.

**Shanks** · December 2nd 2009, 01:44 AM

(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
$BAB^{-1}=D$ .
since $D^2=D$ , that is,
$BA^2B^{-1}=BAB^{-1}$
combined with B is invertable, gives the result.

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