$B$ is self-adjoint s.t. $\langle Bx,x \rangle \geq 0$ Show that if $\langle Bv,v \rangle =0$ then $Bv=0$ for complex space.

Question

How to show if $B^*=B$ and $(Bx)^*x\geq 0$ for all $x\in \mathbb{C}^n$ then $(Bv)^*v=0 \Rightarrow Bv=0$

The book arrives at a statement $$\langle Bv,h \rangle + \langle Bh,v \rangle=0.\tag1$$

where $h$ is any vector. Then it says since $h$ was arbitrary we can substitute $ih$ for $h$ and gets:} $$ -i\langle Be,h \rangle+i\langle Bh,e \rangle=0\tag2$$ Somehow $(1)$ and $(2)$ show that

$$\langle Be,h \rangle=0$$

How since the $h$ in $(1)$ and $(2)$ are different?

Answer 1

You say $(1)$ holds "where $h$ is any vector."

I prefer to phrase it thus: "For every vector $h,$ line $(1)$ holds."

The word "where" should not be used as a quantifier. It should be used for such things as "where $h$ is Planck's constant", etc.

If a statement is true of every vector and $ih$ is a vector, then that statement is true of $ih.$

Dividing both sides of $(2)$ by $i,$ you get $$-\langle Be, h\rangle +\langle Bh,e\rangle=0.\tag3$$

The word "every" remains. For every vector $h,$ line $(1)$ holds, and for every vector $h,$ line $(3)$ holds. Thus they must hold when the same vector is put in both places.