Question

How to show that for any vector in R3, (v dot (Av) is greater than or equal to 0

Answer 1

sorry I don't know how to do this

Answer 2

do you know linear algebra?
I have a super hard take home test im trying to do

Answer 3

give me a question and ill see if I can do it

Answer 4

have you taken a linear algebra course? i'm typing one right now

Answer 5

Let A and B be nxn maricies. Suppose v is an eigenvector of A associated with the eigenvalue x and is also an eigenvector of B associated with the eigenvalue y. Show that x+y is an eigenvalue of A+B

Answer 6

nope never seen anything like it sorry

Answer 7

yess, currently in linear algebra, haven't read the question yet. Lemme see

Answer 8

Are you sure about that problem statement? Because it isnt true. As a matter of fact take v = (1,1,1) and\[A = \left[\begin{matrix}-1 &0 & 0 \\ 0 & -1 & 0\\ 0 & 0 & -1\end{matrix}\right]\]

Answer 9

unless, of course, you're referring to an eigenvector by using v, but then it wouldn't be any vector in R3

Answer 10

Because v is an eigenvector of A associated with the eigenvalue x, then :
\[Av=xv\]
and because v is an eigenvector of B associated with the eigenvalue y, then :
\[Bv=yv\]
Hence :
\[(A+B)v=Av+Bv=xv+yv=(x+y)v\]
Therefor x+y is an eigenvalue of A+B