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.
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.