TITLE: Show that $\langle v,w\rangle _1=c\langle v,w\rangle _2$ for some scalar $c$.
QUESTION [5 upvotes]: Let $V$ be a vector space over $F$ and $\langle ,\rangle _1$ and $\langle ,\rangle _2$ be two inner products defined on it.
It is given that $\langle v,w\rangle _1=0\iff \langle v ,w\rangle _2=0 \tag{H}.$
Show that $\langle v,w\rangle _1=c\langle  v,w\rangle _2$ for some scalar $c$.

Fix $w\in V$.
Define $f_1:V\to F$ by $f_1(v)=\langle v,w\rangle _1$.
and
Define $f_2:V\to F$ by $f_2(v)=\langle v,w\rangle _2$.
$v\in \ker f_1\iff v\in \ker f_2$ .
If $\ker f=V$ then choose $c=1$ and we are done.
If $\ker f\neq V$ then $\exists v_0\in V$ such that $f_1(v_0)\neq 0\implies f_2(v_0)\neq 0$.
How to choose $c$ in this case?
Please help me out.

REPLY [2 votes]: We don't need any assumption about the dimension or basis of the vector space. If $V=0$, we have nothing to prove. Suppose $V\ne0$. Pick a nonzero vector $u\in V$. By scaling the second inner product if necessary, we may assume that
$$
\langle u,u\rangle_1=\langle u,u\rangle_2.\tag{1}
$$
For any $v\in V$, let $x = v-\dfrac{\langle v,u\rangle_1}{\langle u,u\rangle_1}u$. Then $\langle x,u\rangle_1=0$. Hence $\langle x,u\rangle_2=0$, meaning that
$$
\langle v,u\rangle_2=\langle v,u\rangle_1.\tag{2}
$$
Now, for any $t\ne t_0=\dfrac{\langle u,v\rangle_1}{\langle u,u\rangle_1}$, define $v_t=v-tu$ so that $\langle u,v_t\rangle_1\ne0$. Let $y=u-\dfrac{\langle u,v_t\rangle_1}{\langle v_t,v_t\rangle_1}v_t$. Then $\langle y,v_t\rangle_1=0$. Hence $\langle y,v_t\rangle_2=0$, i.e., $\langle u,v_t\rangle_2=\dfrac{\langle u,v_t\rangle_1}{\langle v_t,v_t\rangle_1}\langle v_t,v_t\rangle_2$. Therefore, by $(2)$, we get
$$
\langle v_t,v_t\rangle_1=\langle v_t,v_t\rangle_2.\tag{3}
$$
As $(3)$ holds for every $t\ne t_0$, it must also hold for every $t\in F$. In particular, by putting $t=0$, we obtain $\langle v,v\rangle_1=\langle v,v\rangle_2$.
This is the same condition as $(1)$, except that the role of $u$ is now played by $v$. So, if we apply a similar argument to the above, we will get an analogous result to $(2)$, namely $\langle w,v\rangle_2=\langle w,v\rangle_1$ for any vector $w$. Since both $v$ and $w$ are arbitrary, the two inner products are identical to each other.