Are all Vectors of a Basis Orthogonal? Assuming we have a basis for a set $\mathbb{R}^n$, would any set of linearly independent vectors that form a basis for $\mathbb{R}^n$ also be orthogonal to each other?
Take the trivial case of $(1,0)$ and $(0,1)$. Now any set of linear independent vectors would be a scalar multiple of these two vectors that form a Basis for $\mathbb{R}^2$ hence they have to be orthogonal. Right?
 A: 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.
A: Nope. $(1,1)$ and $(1,0)$ form a basis of $\mathbb R^2$.
A: 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.
