symmetric matrix problem

[matt_jon]

If A is a symmetric matric with A' =A, and any vector u, v
Do we have this result (u-v)'Av= v'A(u-v), and how to prove it?
thanks

 

[Steven G]

If A is a symmetric matric with A' =A, and any vector u, v
Do we have this result (u-v)'Av= v'A(u-v), and how to prove it?
thanks

We know that (AB)'= B'A' and (C')' = C

[(u-v)'Av]' = [{(u-v)'A}v]' = v'{(u-v)'A}' = v'{A'(u-v)} = v'A'(u-v). So the answer to your question is no. Note that we proved that (ABC)' = C'B'A'

 

[Ishuda]

If A is a symmetric matric with A' =A, and any vector u, v
Do we have this result (u-v)'Av= v'A(u-v), and how to prove it?
thanks

I'm assuming the ' means transpose.

One of the reasons the forum guide lines
"http://www.freemathhelp.com/forum/threads/41536-Read-Before-Posting!!
are as they are is that, unless, for example, you show what you have tried to answer the problem, the volunteer here doesn't know where to start helping.

For example, suppose you know the distributive laws for transpose and matrix multiplication, then we have
(u-v)'Av = (u' - v')Av = u'Av - v'Av
and
v'A(u-v) = v'Au - v'Av
So, what you are being asked to prove or disprove is
If A' = A then u'Av = v'Au

So, could you show us what you have attempted so far?