Orthogonal and orthonormal for dot-product Consider the vectors $
u_1=\begin{pmatrix}
1\\ 
1\\ 
-1
\end{pmatrix},u_2=\begin{pmatrix}
2\\ 
2\\ 
4
\end{pmatrix},u_3=\begin{pmatrix}
1\\ 
-1\\ 
0
\end{pmatrix} $ in $\mathbb{R}^3$
and the map
$f:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined by $f(x)=u_1\left \langle u_1, x \right \rangle + u_2\left \langle u_2, x \right \rangle + u_3\left \langle u_3, x \right \rangle$
Show that the family of vectors $u=(u_1,u_2,u_3)$ is a orthogonal family with respect to the dot-product in $\mathbb{R}^3$ and that it is a basis. Is it a orthonormal basis?
I know that we say that 2 vectors are orthogonal if they are perpendicular to each other. i.e. the dot product of the two vectors is equal to zero. Would I just have to simply apply the inner product on my space pairwise and conclude that it is equal to 0?
I know that an orthonormal basis for an inner product space with finite dimension is a basis for the space whose vectors are orthonormal i.e they are all unit vectors and orthogonal to each other.
Even though I seem to get the concepts and the thinking behind I do not know how to apply it in this example.
 A: You're right that orthogonality means $\langle u_1,\,u_2\rangle=\langle u_2,\,u_3\rangle=\langle u_3,\,u_1\rangle=0$. As @MinusOne-Twelfth explained, the thogonality of $n$ nonzero vectors in $\Bbb R^n$ ensures they form a basis. For orthonormality, check whether $\langle u_1,\,u_1\rangle=\langle u_2,\,u_2\rangle=\langle u_3,\,u_3\rangle=1$; you should find this fails.
A: To show that the vectors are an orthogonal family, yes you just have to apply the dot product pairwise and show that each pair of vectors has a dot product of $0$.
To next argue that the vectors form a basis for $\Bbb{R}^3$, recall that a set of orthogonal vectors, none of which are the zero vector, is linearly independent. So $\{ u_1, u_2, u_3\}$ is a set of three linearly independent vectors in $\Bbb{R}^3$, and hence a basis for $\Bbb{R}^3$.
Finally, to check whether it is an orthonormal basis, you just have to check whether the norm of all the vectors is $1$ (equivalently whether $u_i\cdot u_i = 1$ for all $i=1,2,3$). Since $u_1\cdot u_1 = 3 \ne 1$, this is not the case. Hence the basis is not orthonormal.
