Using Gram-Schmidt on independent vectors to find a special vector $w$

Question

Let $u,v \in \mathbb{R}^{n}$ be non-zero independent vectors. Show that there exists a vector $w \in \mathbb{R}^{n}$ such that $\langle u,w \rangle < 0$ and $\langle v,w \rangle > 0$.
I was guided to apply Grahm-Shmidt process on $\left\{ u,v \right\}$ and then write the vector $w$ in terms of the resulting vectors.
So I did Gram-Schmidt on $\left\{ u,v \right\}$ and got the vectors: $$s_{1} = \dfrac{u}{||u||}, s_{2} = \dfrac{v-\langle u,v \rangle \dfrac{u}{||u||^{2}}}{||v-\langle u,v \rangle \dfrac{u}{||u||^{2}}||}$$ Then I tried to write $w=as_{1}+bs_{2}$ for some $a,b \in \mathbb{R}$ and see what constraints I get on $a, b$ but it got ugly and too complicated.
If anyone have an idea I would glad to hear.

Answer 1

Hint: Look for a vector of the form $w = au + bv$ for $a,b \in \Bbb R$. Note that $$ \pmatrix{\langle w,u\rangle\\ \langle w,v \rangle} = M \pmatrix{a\\b} $$ for some $2 \times 2$ matrix $M$. Use the Cauchy-Schwarz inequality (and the specific condition under which equality holds) to conclude that $M$ is invertible. With that, it's easy to find $a,b$ that, for instance, make $\langle w,u \rangle = -1$ and $\langle w,v \rangle = 1$.

Alternatively, for a geometric approach: draw $u,v$ as two vectors emanating from the origin. Find points along the vectors $u$ and $v$ that are the same distance from the origin. Argue that the vector connecting the point along $u$ to the point along $v$ forms an obtuse angle with the vector $u$ but an acute angle with the vector $v$.