Given that u is a d xx 1 vector such that norm(u) = 1 and for all d xx 1 vector x, how can we reason through the equation ? x^T(u u^T)x=(x^Tu)(u^Tx)=(⟨x,u⟩)(⟨x,u⟩)=((⟨x,u⟩))^2 >=0

Dangelo Rosario

Dangelo Rosario

Answered question

2022-10-01

Given that u is a d x 1 vector such that ||u|| = 1 and for all d x 1 vector x, how can we reason through the equation ? x T ( u u T ) x = ( x T u ) ( u T x ) = ( x , u ) ( x , u ) = ( ( x , u ) ) 2 0
In this case, x is a scalar(?). and ( u u T ) gives us 1(?) which mean 1 d x 1 x 1 , Then since ||u|| is 1 given. what can we say about each x 1 if x is a scalar, can we even transpose it ?

Answer & Explanation

Jarrett Pearson

Jarrett Pearson

Beginner2022-10-02Added 4 answers

x and u are vectors. Notice that if x = ( x 1 , . . . , x n ), then
x T x = i = 1 n x i 2 = x 2 .
Then,
( x T u ) ( u T x ) = ( u T x ) T ( u T x ) = u T x 2 0.

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