Take the 2-minute tour ×  Let $V$ be a finite-dimensional inner
product space and let $x,y\in V$ be nonzero vectors. If there is
a self-adjoint operator $A:V\rightarrow V$ such that $A(x)=y$
and $\langle A(v),v\rangle\geq0$ for all $v\in V$, then $\langle
x,y\rangle>0$.  I think we can conclude the following inequality
$$\langle x,y \rangle=\langle x,A(x) \rangle=\langle A^*(x),x
\rangle=\langle A(x),x\rangle\geq 0$$  but I'm not able to show
that strict inequality holds (or, in other words, that equality
is not possible). Can someone give me a hint?  share|improve
this question I added the operator theory tag since the result
is true more generally (and can be proved the same way) on a
Hilbert space, with $A$ a bounded self-adjoint positive
operator. –  1015 Jul 1 '13 at 1:04  1 Answer 1  up vote 1 down
vote accepted  Assume for a contradiction that $(x,y)=0$. Then
take an orthonormal basis $(e_j)$ of $V$ starting with $$
e_1=\frac{x}{\|x\|}\qquad e_2=\frac{y}{\|y\|}. $$ The matrix $M$
of $A$ in this orthono