Let $\langle, \rangle$ and $||,||$ an inner product and a norm over $\mathbb{R}^n$. Suppose for all $a\in \mathbb{R}$ the inequality
$$
||u||\le ||u+av||
$$
holds. How can I prove that $\langle u,v\rangle=0$? Any help would be appreciated.
Prove that a inner product is zero
inner-productsnormed-spaces
Related Solutions
Untangling the notation
Let $W = V \times V$ with addition defined by $(v_1,w_1) + (v_2,w_2) = (v_1+v_2,w_1+w_2)$.
- Each element of $W$ is a pair of elements from $V$. So an element of $W$ looks like $w = (v_,v_2)$ where $v_1,v_2 \in V$. Above, all of $v_1,v_2,w_1,w_2$ are in $V$. The addition is componentwise.
and scalar multiplication be defined by $(a+bi)(v,w) = (av-bw,aw+bv)$.
Although not mentioned, $W$ is going to be designed as a complex vector space. So, we have to explain how we multiply a complex number, which is $(a+bi)$, with some element of $W$, say $(v,w)$ where $v,w \in V$. One of course must check the usual properties of multiplication for this , but I won't do it.
and inner product $\langle \cdot,\cdot \rangle' : W \times W \to \mathbb C$ defined by : $$ \langle (v_1,w_1),(v_2,w_2)\rangle' = (\langle v_1,v_2\rangle + \langle w_1,w_2\rangle) + i(-\langle v_1,w_2\rangle + \langle w_1,v_2 \rangle) $$
Over here, you are defining a complex inner product on $W$, taking values in $\mathbb C$, so you are specifying explicitly the real and imaginary parts of the result. Again, $v_1,v_2,w_1,w_2$ are vectors in $V$, and all terms on the right hand side involve the inner product on $V$.
Now, here's the point : we have not shown yet that the above function is an inner product, so the author calling it an inner product is "incorrect writing"(but not an incorrect statement), especially when he wants us to prove that it is positive definite first.
Solving the problem
So we have the real vector space $V$, and it has an inner product $\langle \cdot , \cdot \rangle$. This inner product has nothing to do with the usual $\mathbb R^n$-inner product, or the absolute value $|\cdot|$ : an inner product is one that satisfies (conjugate) symmetry, linearity in the first argument and positive definiteness, and while operating you must assume only these three properties and no more.
For positive definiteness, we need to prove exactly this : for every $w \in W$, we have $\langle w,w\rangle' \geq 0$(NOTE : This involves the fact that $\langle w,w\rangle'$ must in fact be a real number because no order is defined on the complex numbers , but we will see that this is the case here), with equality exactly when $w = 0$.
For this, first since $w \in W$, we will write it as $w=(w_1,w_2)$ where $w_1,w_2 \in V$. Then we use the formula to write down $\langle w,w\rangle'$ : $$ \langle w,w\rangle' = \langle (w_1,w_2),(w_1,w_2)\rangle \\ = \langle w_1,w_1\rangle + \langle w_2,w_2\rangle - i\langle w_1,w_2\rangle + i\langle w_1,w_2\rangle \\ = \langle w_1,w_1 \rangle + \langle w_2,w_2\rangle $$
is therefore a real number, and can be compared with $0$.
However, now things are very simple : because the inner product on $V$ is positive definite, certainly the last line is a sum of two non-negative terms and is hence non-negative.
Furthermore, if $\langle w,w\rangle' = 0$, then the last line is zero, which by non-negativity forces each of $\langle w_1,w_1\rangle$ and $\langle w_2,w_2\rangle$ to be zero, which forces both $w_1=w_2=0$ because the inner product on $V$ is positive definite. Of course, then $w =(0,0)$ is the zero element of $W$.
Thus, we show that positive definiteness holds.
Since I have answered the question, you should now strengthen your grip by proving that the function $\langle \cdot,\cdot\rangle'$ satisfies all the other properties of an inner product on $W$ as well.
Best Answer
Assume $\langle u,v \rangle \neq 0$. Square both sides of your inequality to get $$\|u\|^2 \leq \|u\|^2 + 2a\langle u,v \rangle + a^2\|v\|^2$$ or $$2a\langle u,v \rangle + a^2\|v\|^2 \geq 0$$ for all $a \in \mathbb{R}$. Choosing $a = -\frac{\langle u,v \rangle}{\|v\|^2}$ yields a contradiction.