2
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – Julien
    Jul 1, 2013 at 1:04

1 Answer 1

1
$\begingroup$

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 orthonormal basis must be symmetric semidefinite positive, and it looks like $$ M=\pmatrix{0&t&0\\ t&*&*\\ 0&*&*}\qquad t=\frac{\|y\|}{\|x\|}>0 $$ by blocks, where the first and the second row are the actual first and second rows of $M$, the remaining blocks being of the ad hoc size.

Can you see a contradiction?

Consider the restriction of the quadratic form $(Ax,x)$ to the span of $\{e_1,e_2\}$, whose matrix is the upper-left $2\times 2$ block of $M$. Since the restriction should be semidefinite positive as well, its eigenvalues, whence its determinant $-t^2<0$, must be nonnegative. Contradiction.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .