Show that $\langle v,w\rangle _1=c\langle v,w\rangle _2$ for some scalar $c$.

Question

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.

Answer 1

I suppose here that $V$ is having a finite dimension equal to $n$.

Then it exists an orthonormal basis $(e_1, \dots, e_n)$ for $\langle v,w\rangle _1$. As a consequence of the hypothesis, $(e_1, \dots, e_n)$ is an orthogonal basis for $\langle v,w\rangle _2$

It is sufficient to prove that the $\langle e_i,e_i\rangle _2$ are all equal.

So take $i \in \{2, \dots, n\}$. We have

$$\langle e_1 - e_i, e_1 + e_i\rangle_1 = \langle e_1, e_1\rangle_1 -\langle e_i, e_i\rangle_1 = 0 = \langle e_1 - e_i, e_1 + e_i\rangle_2=\langle e_1, e_1\rangle_2 -\langle e_i, e_i\rangle_2.$$

According to hypothesis $(H)$.

Therefore the $\langle e_i,e_i\rangle _2$ are all equal to let's say $1/c$ and we get the desired result.

Answer 2

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.

Answer 3

Let $w\in V\setminus\{0\}$, $T_i:V\longrightarrow F$ be given by $T_i(v) = {\langle v, w\rangle}_i$ for $i \in \{1,2\}$ and $T = T_1 + T_2$. Notice that $T, T_1$ and $T_2$ are linear.

$\quad(1)$ $\;$ If $T\equiv 0$ we can take $c=-1$.

$\quad(2)$ $\;$ Otherwise, since $T$ is not identically $0$, $\ker T\subsetneq V$. We may hence write

$$V = (\ker T) \oplus W,\tag{1}$$

where $\dim W > 0$.

$\quad\quad(2.1)$ $\;$If $\dim W = 1$, say $W=\text{span}(u)$, we can take $c = T_1(u)/T_2(u)$.

$\quad\quad(2.2)$ $\;$If $\dim W \geqslant 2$, we may choose a linearly indepenent set $\{u, v\} \subset W$ with $T(u) = T(v) = 1$. Then $T_i(u), T_i(v)\neq 0$ for $i=1,2$ and

\begin{align} &\left\{\begin{array}{c} T_1(u)+T_2(u) = 1\\ T_1(v)+T_2(v) = 1 \end{array}\right. \\\implies &\left\{\begin{array}{c} \frac{T_1(u)}{T_2(u)}= \frac1{T_2(u)} - 1\\ \frac{T_1(v)}{T_2(v)}= \frac1{T_2(v)} - 1 \end{array}\right. \end{align}

If we had $T_1 = c\,T_2$, then necessarily

$$\frac1{T_2(u)} - 1 = \frac1{T_2(v)} - 1 \implies T_2(u) = T_2(v) \,\text{ and }\, T_1(u) = T_1(v)$$

In particular, we'd have $u-v \in \ker T_1 = \ker T_2\subset \ker T$. But since the sum in $(1)$ is direct, this would mean $u-v = 0$, contradicting the linear independence of $\{u,v\}$. It follows that either $\dim W \geqslant 2$ can never happen, or else $T_1\neq cT_2$ when $\dim W \geqslant 2$.

Notice that because of the rank-nullity theorem, this issue can only ever arise when $V$ is infinite dimensional. Indeed, when $V$ is finite dimensional, the theorem guarantees that

$$\dim W = \dim V - \dim \ker T = \dim \text{Im } T \leqslant 1.$$