there is a proof for $Q$ being a $3x3$ real matrix, and $v,w \ in \mathbb{R}^3$ that shows $(Qv) \cdot (Qw) = v \cdot w$.
It goes like this:
$(Qv) \cdot (Qw) = v \cdot (Q^{T}Qw) = v \cdot (Iw) = v \cdot w$.
I don't see how the first first equality goes to the next, as in, how does the $Q$ from the front of $v$ go into the bracket with $Qw$?
Asked
Active
Viewed 4,342 times
1
Twenty-six colours
- 1,881
-
1You are using $(Qv)^T=v^T Q^T$; transposition of matrices is multiplication-reversing. – Angina Seng May 06 '17 at 06:34
-
Thanks, should''ve realised – Twenty-six colours May 06 '17 at 06:38
2 Answers
3
The dot product of two vectors $v$ and $w$ is given by $v\cdot w = v^T w$. Therefore, $$(Qv)\cdot (Qw) = (Qv)^T(Qw) = v^TQ^TQw = v^Tw = v\cdot w$$
florence
- 12,819
0
Kuldeep Singh's Linear Algebra: Step by Step (2013) pp 325-326 details this more. I don't affiliate Kuldeep Singh.
