Answer to Question #160248 in Linear Algebra for Nikhil Singh

Question

Let {u,v,w} be an orthornornal set of vectors in R^3. Show that they are linearly independent over R. Check whether u-v, u+v,w are orthogonal over R or not.

Answer 1

Let $\{u,v,w\}$ be an orthornornal set of vectors in $\mathbb R^3$. Let us show that they are linearly independent over $\mathbb R$. For this consider a linear combination $au+bv+cw=0$. Since this vectors are orthornormal, for their scalar products we have that $u\cdot u=v\cdot v=w\cdot w=1,\ \ u\cdot v=v\cdot u=u\cdot w=w\cdot u=v\cdot w=v\cdot w=0$.

Therefore, $(au+bv+cw)\cdot u=0\cdot u$ implies $a(u\cdot u)+b(v\cdot u)+c(w\cdot u)=0$, and thus $a=0.$ It follows from $(au+bv+cw)\cdot v=0\cdot v$ that $a(u\cdot v)+b(v\cdot v)+c(w\cdot v)=0$, and thus $b=0.$ By analogy, $(au+bv+cw)\cdot w=0\cdot w$ implies $a(u\cdot w)+b(v\cdot w)+c(w\cdot w)=0$, and thus $c=0.$

Consequently, the vectors $u,v,w$ are linearly independent over $\mathbb R$.

Let us show that $u-v, u+v,w$ are orthogonal over $\mathbb R$. Indeed, their scalar products are equal to zero: $(u-v)\cdot(u+v)=u\cdot u+u\cdot v-v\cdot u-v\cdot v=1+0-0-1=0$, $(u-v)\cdot w=u\cdot w - v\cdot w=0-0=0$ and $(u+v)\cdot w=u\cdot w + v\cdot w=0+0=0$.