Are all Vectors of a Basis Orthogonal?

Solution 1:

No. The set $\beta=\{(1,0),(1,1)\}$ forms a basis for $\Bbb R^2$ but is not an orthogonal basis. This is why we have Gram-Schmidt!

More general, the set $\beta=\{e_1,e_2,\dotsc,e_{n-1},e_1+\dotsb+e_n\}$ forms a non-orthogonal basis for $\Bbb R^n$.

To acknowledge the conversation in the comments, it is true that orthogonality of a set of vectors implies linear independence. Indeed, suppose $\{v_1,\dotsc,v_k\}$ is an orthogonal set of nonzero vectors and $$ \lambda_1 v_1+\dotsb+\lambda_k v_k=\mathbf 0\tag{1} $$ Then applying $\langle-,v_j\rangle$ to (1) gives $\lambda_j\langle v_j,v_j\rangle=0$ so that $\lambda_j=0$ for $1\leq j\leq k$.

The examples provided in the first part of this answer show that the converse to this statement is not true.

Solution 2:

In general: No! Any vectors ${\bf u} = (u_1,u_2,u_3)$, ${\bf v} = (v_1,v_2,v_3)$ and ${\bf w} = (w_1,w_2,w_3)$ will span $\mathbb{R}^3$ if, and only if, the following determinant is non-zero: $$\left|\begin{array}{ccc} u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\ u_3 & v_3 & w_3 \end{array}\right| \neq 0$$ This is equivalent to the single polynomial condition $$u_1(v_2w_3-w_2v_3) - v_1(u_2v_3 - w_2u_3) + w_1(u_2v_3-v_2u_3) \neq 0$$

For a set of non-zero vectors to be orthogonal, we need them all to be at right angles:

$$\langle {\bf u}, {\bf v}\rangle = \langle {\bf u}, {\bf w}\rangle = \langle {\bf v}, {\bf w}\rangle =0$$ For this to happen, we need three simultaneous polynomial conditions:

\begin{eqnarray*} u_1v_1 + u_2v_2 + u_3v_3 &=& 0 \\ \\ u_1w_1 + u_2w_2 + u_3w_3 &=& 0 \\ \\ v_1w_1 + v_2w_2 + v_3w_3 &=& 0 \end{eqnarray*}

There are many sets of linearly independent vectors $\{{\bf u},{\bf v},{\bf w}\}$ which are not orthogonal.