Let $V$ be a vector space with a finite Dimension above $\mathbb{C}$ or $\mathbb{R}$.
Let $B=(v_{1},v_{2},…,v_{n})$ be a basis of V.
How can I prove that there is an Inner product $\langle,\rangle$ which in respect to it, B is an orthonormal basis.
Thank you
Best Answer
Meta: I'm adding a second answer only performing the necessary calculations and not mentioning any sophisticated words. Nir, I'm sorry for having aimed a bit too high in the hope of telling you something useful.
Let $x = x_{1} v_{1} + \cdots + x_{n} v_{n}$ and $y = y_{1} v_{1} + \cdots + y_{n} v_{n}$ be any two vectors of $\mathbb{C}^{n}$. Define the expression $$\langle x, y \rangle = \sum_{i = 1}^{n} x_{i} \, \overline{y_i}.$$ I claim that this defines a scalar product such that $(v_{1},\ldots,v_{n})$ is orthonormal.
To see that $\langle \cdot, \cdot \rangle$ is an inner product, we need to check that for all $x,y,z \in \mathbb{C}^{n}$ and $\lambda \in \mathbb{C}$:
I'm going to do 1., 2. and 4., and leave 3. and 5. to you. Let's do 1. now: $$ \overline{\langle y, x \rangle} = \overline{\sum_{i = 1}^{n} y_{i} \, \overline{x_{i}}} = \sum_{i = 1}^{n} \overline{y_{i} \, \overline{x_{i}}} = \sum_{i = 1}^{n} \overline{y_{i}} \, x_{i} = \sum_{i = 1}^{n} x_{i} \, \overline{y_{i}} = \langle x, y \rangle.$$ Let's do 2.: $$ \langle x + z, y \rangle = \sum_{i =1}^{n} (x_{i} + z_{i})\,\overline{y_{i}} = \sum_{i=1}^{n} \left( x_{i}\,\overline{y_{i}} + z_{i}\,\overline{y_{i}} \right) = \sum_{i=1}^{n} x_{i}\,\overline{y_{i}} + \sum_{i = 1}^{n} z_{i} \, \overline{y_{i}} = \langle x,y \rangle + \langle z, y \rangle.$$ Let's do 4.: $$ \langle x, x \rangle = \sum_{i = 1}^{n} x_{i} \, \overline{x_{i}} = \sum_{i=1}^{n} |x_{i}|^2 \geq 0$$ since $x_{i} \, \overline{x_{i}} = |x_{i}|^2 \geq 0$.
Then we need to check that $(v_{1},\ldots,v_{n})$ is an orthonormal basis. I'll do two cases: $$\langle v_1, v_1 \rangle = 1 \cdot \overline{1} + 0 \cdot \overline{0}+ \cdots + 0 \cdot \overline{0} = 1$$ $$\langle v_1, v_{2}\rangle = 1 \cdot \overline{0} + 0 \cdot \overline{1} + 0 \cdot \overline{0} + \cdots + 0 \cdot \overline{0} = 0.$$ This should be enough for you to check that $\langle v_{k}, v_{k} \rangle = 1$ and $\langle v_{k}, v_{l} \rangle = 0$ if $k \neq l$ (actually $k \lt l$ suffices in view of condition 1. above.