I have an homework to solve but I am behind with the theory. It doesn't look difficult, since it is about computations, just it would be good for me to find a short document which explains exactly what is introduced here. This is the text to which I refer:
"Consider on $\mathbb{R}^{3}$ the scalar product defined by assuming that the vectors $f_{1}=(1,0,0), f_{2}=(1,1,0), f_{3}=(1,1,1)$ are orthonormal:
(1) Compute the angles between the elements of the standard basis $e_{1}=(1,0,0), e_{2}=(0,1,0), e_{3}=(0,0,1)$ with respect to this scalar product. (2) Compute the orthogonal projections of the ei onto each other with respect to this scalar product.”
My interpretion of the text:
I have a 3D space and I am considering the vector space defined by the 3 orthonormal vectors. First question: what are orthonormal vectors? Also reading on the internet, I have understood they are orthogonal unit vectors. Cfr. https://en.wikipedia.org/wiki/Orthonormal_basis but of the vectors above only the first one is a unit vector.
The text asks for computing the angles between the elements of the standard basis with respect to the scalar product. Does this mean that I have to draw the 3 vectors above and the other 2 unit vectors and evaluating the degrees/radians of the angles between them? (of course, I would made any computation in analytical way; for now, I want to just understand what is asked by the text).
While the second task isn't very clear for me in this moment.
Could you give me some insights on the exercise(s), please? In addition, could you post here some examples, in order to let me understand me what is asked by the text, please?
The set $F = \{ \vec f_1, \vec f_2,\vec f_3 \}$ form a basis for $\mathbb R^3$ and what's more, we are told that with respect to some inner product $\langle \cdot, \cdot \rangle$ they are orthonormal, hence
$$\langle \vec f_i, \vec f_j \rangle = \delta_{ij} = \begin{cases} 1, i = j \\ 0, i \neq j \end{cases}$$
Thus given any two vectors $\vec a,\vec b \in \mathbb R^3$, write them with respect to the $F$ basis, $\vec a = a_1 \vec f_1 + a_2 \vec f_2 + a_3 \vec f_3$ and $\vec b = b_1 \vec f_1 + b_2 \vec f_2 + b_3 \vec f_3$. Then by linearity of inner products
$$\langle \vec a, \vec b \rangle = \sum_{i=1}^3 \sum_{j=1}^3 a_ib_j \langle \vec f_i, \vec f_j \rangle$$
As $\langle \vec f_i, \vec f_j \rangle = \delta_{ij}$ this sum reduces to
$$\langle \vec a, \vec b \rangle = \sum_{i=1}^3 a_ib_i = a_1b_1 + a_2b_2 + a_3b_3$$
Use this to solve the question asked by writing the $\vec e_i$ in the $F$ basis: $\vec e_1 = \vec f_1, \vec e_2 = \vec f_2 - \vec f_1, \vec e_3 = \vec f_3 - \vec f_2 - \vec f_1$
For example, the angle $\theta_{12}$ between $\vec e_1$ and $\vec e_2$ satisfies
$$\cos\theta_{12} \| \vec e_1 \| \ \| \vec e_2 \| = \langle \vec e_1, \vec e_2 \rangle$$
where $$\| \vec e_1 \| = \sqrt{\langle \vec e_1, \vec e_1 \rangle} = \sqrt{\langle \vec f_1, \vec f_1 \rangle} = \sqrt 1 = 1$$ and $$\| \vec e_2 \| = \sqrt{\langle e_2, e_2 \rangle} = \sqrt{\langle f_2 - f_1, f_2 - f_1 \rangle} = \sqrt 2$$
as $\langle \vec f_2 - \vec f_1, \vec f_2 - \vec f_1 \rangle = \langle \vec f_2, \vec f_2\rangle - \langle \vec f_2, \vec f_1\rangle - \langle \vec f_1, \vec f_2\rangle + \langle \vec f_1, \vec f_1\rangle = 1 - 0 - 0 + 1 = 2$
Also $$\langle e_1, e_2 \rangle = \langle f_1, f_2 - f_1 \rangle = \langle f_1, f_2 \rangle - \langle f_1, f_1 \rangle = 0 - 1 = -1$$ and thus we have
$$\cos\theta_{12} = \frac{\langle \vec e_1, \vec e_2 \rangle}{ \| \vec e_1 \| \ \| \vec e_2 \|} = \frac{-1}{\sqrt{2}}$$