1
$\begingroup$

In the source question B is an element of $M_n(R)$ and is a symmetic matrix such that $v^tBv>0$.

Also $<.|.>$ is an inner product on $R^n$ called the $B$-inner product.

we are asked to prove that if BA is symmetric then $A$ is self-adjoint with respect to the $B$-inner product.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Homework needs work. $\endgroup$
    – Name
    Commented Apr 24, 2014 at 6:48
  • $\begingroup$ Im unsure of how you are supposed to prove something is self adjoint with respect to something else ? $\endgroup$
    – squibben
    Commented Apr 24, 2014 at 6:50
  • $\begingroup$ I have the following statements (1) since b is symmetric B=B^t also if BA is symmetric then BA=(BA)^t therefore BA=(BA)^t=B^t.A^t=B^t.A . also I have A is a linear map such that A:r^n->r^n and A* is a linear map such that A*:r^n->r^n now since A=A* A is self adjoint. but im not sure why it is self adjoint wrt the B-inner-product space ? $\endgroup$
    – squibben
    Commented Apr 24, 2014 at 7:01

0

Reset to default

Browse other questions tagged .