Question

If u, v are nx1 matrices and A is an nxn matrix, then, (v^tA^t Au)^2<=(u^tA^tAu)(v^tA^tAv). Let me re-write in attachment

Answer 1

\[(v^tA^tAu),=(u^tA^tAu)(v^tA^tAv)\]

Answer 2

sorry, the first term should be square

Answer 3

what info is given about u?

Answer 4

you mean
$$ (v^tA^tAu)^2 \le (u^tA^tAu)(v^tA^tAv) $$

Answer 5

yes. my bad. wrong typo

Answer 6

looks like I serious eye check up.

Answer 7

looks like you need to make Use of Cauchy-Swartz inequality.

Answer 8

let B = AA^t be a symmetric matrix.
\[ (v^tA^tAu)^2 \le (u^tA^tAu)(v^tA^tAv) \\
\left( \sum_{i=1}^{n} \sum_{i=1}^{n} v_i b_{ij}u_j\right)^2 \le \left( \sum_{i=1}^{n}\sum_{j=1}^{n}u_i b_{ij} u_j \right) \left( \sum_{i=1}^{n}\sum_{j=1}^{n}v_i b_{ij} v_j \right)\]

Answer 9

http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality#Rn

Answer 10

can I break it down. I realize that it 's 4 terms not 3, v^t, A^t, A, and u in the first term of the problem