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.
Best Answer
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.