Inner products and Orthonormal Bases

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

-
Hint: By bilinearity (or sesquilinearity), defining an inner product on the basis (where the definition is forced upon us by the orthonormality condition) already defines it on the whole space. Now check that this forced expression actually gives an inner product. – t.b. May 12 '11 at 13:08
@Theo: I think you should post it as an answer. – Dennis Gulko May 12 '11 at 13:23
@Dennis: Actually, I was going to elaborate on this hint in case no-one else did, but as you asked for me to post it, I'm doing that now. – t.b. May 12 '11 at 13:33
Nir: I'm not invoking bilinearity in my answer below (if you read it closely), but it is the whole point! – t.b. May 12 '11 at 14:11
The simplest way is to fix an isomorphism $T\colon V\to F^n$, where $F$ is the ground field, that maps $B$ to the standard basis of $F$. Then define the inner product on $V$ by $\langle v,w\rangle_V = \langle T(v),T(w)\rangle_{F}$. Because $B$ is mapped to an orthonormal basis of $F^n$, this inner product makes $B$ into an orthonormal basis. – Arturo Magidin May 12 '11 at 16:35

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 $\lambda \in \mathbb{C}$:

1. $\overline{\langle y, x \rangle} = \langle x, y \rangle$
2. $\langle x + z, y \rangle = \langle x, y \rangle + \langle z, y \rangle$.
3. $\langle \lambda x, y \rangle = \lambda\,\langle x,y \rangle$.
4. $\langle x, x \rangle \geq 0$ and $\langle x, x \rangle = 0$.
5. $\langle x, x \rangle = 0$ if and only if $x = 0$.

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.

-
Thank you so much. sometimes simplicity is the answer. It was very helpful,and when I'll go through bilinearity or one of the terms you used in the previous answer, I'll sure go back to understand it more. – user6163 May 12 '11 at 16:27
@Nir: I let myself get a little carried away. Don't worry if you don't understand the other answer yet, but in some time I hope you'll see what I was trying to tell you. – t.b. May 12 '11 at 16:34
@Nir: One more thing: note that the formula I wrote down for the scalar product is the one you obtain if you're making Arturo's comment to your question explicit (by taking the isomorphism that sends $(v_{1},\ldots,v_{n})$ to the standard basis of $\mathbb{C}^{n}$). – t.b. May 13 '11 at 8:18

I was asked to post my hint as an answer. I'm going to elaborate a little, though.

The first thing to observe is that the inner product only depends on its values on a basis. Let me restrict to complex vector spaces (for me an inner product is linear in the first variable and anti-linear in the second one). So let $\langle\cdot,\cdot\rangle$ be an inner product on $\mathbb{C}^n$ and let $(v_1,\ldots,v_n)$ be a basis. Now if $x = x_{1} v_{1} + \cdots + x_{n}v_{n}$ and $y = y_{1} v_{1} + \cdots + y_{n} v_{n}$ are two arbitrary vectors, we get by using sesquilinearity that $$\langle x,y \rangle = \sum_{i,j = 1}^{n} \langle v_{i}, v_{j}\rangle\, x_{i} \, \bar{y}_{j}.$$ This shows that the values $a_{ij} = \langle v_{i},v_{j} \rangle$, $i,j = 1,\ldots,n$ determine the inner product entirely.

Let us consider the $n \times n$ matrix $A = (a_{ij})_{i,j =1}^{n}$ a bit more closely. First of all, notice that $$a_{ij} = \langle v_{i},v_{j} \rangle = \overline{\langle v_{j},v_{i} \rangle} = \overline{a_{ji}}$$ by the symmetry of the inner product. Writing this in matrix form, this is equivalent to $A^{\ast} = A$, where $A^{\ast}$ is the conjugate-transpose of $A$. In general, a matrix $A$ satisfying $A^{\ast} = A$ is called Hermitian. The second observation is that we can write the inner product as $$\langle x, y \rangle = (A \cdot y)^{\ast} \cdot x,$$ where $\cdot$ denotes the matrix product. Now the positive definiteness condition $\langle x, x \rangle \gt 0$ for all $x \neq 0$ can't be expressed as a simple condition on $A$ (but for instance, the combination of diagonal dominance and "Hermitian-ness" is sufficient - off-topic: is the ugly concoction "Hermitian-ness" used in the present case or even "Hermitianity"?! - be that as it may, the link on positive definiteness contains a number of good conditions).

The point I'm heading at is the following exercise you may or may not want to do (I've done one half of it in the above and left you the easier part):

Exercise: If $A = (a_{ij})_{i,j = 1}^{n}$ is an $n \times n$ Hermitian positive definite matrix, then the expression $$\langle x, y \rangle_{A} := (A \cdot y)^{\ast} \cdot x = \sum_{i,j = 1}^{n} a_{ij}\,x_{i}\,\overline{y_{j}}$$ defines an inner product on $\mathbb{C}^{n}$ and conversely, given an inner product, we get a Hermitian positive definite matrix $A$ by the procedure described above.

Finally, I'm addressing your actual question, so let $(v_{1},\ldots,v_{n})$ be a basis of $\mathbb{C}^{n}$. The condition that this basis should be orthonormal with respect to a stipulated inner product $\langle \cdot, \cdot\rangle$ states that the matrix $A = (a_{ij})_{i,j=1}^{n} = (\langle v_{i},v_{j} \rangle)_{i,j=1}^{n}$ must be the $n \times n$-identity matrix $I_{n}$ (why?). Now it is rather straightforward to check that the expression

$$\langle x, y \rangle_{I_{n}} = \sum_{i = 1}^{n} x_{i} \overline{y_{i}}$$

actually is an inner product on $\mathbb{C}^{n}$ for which $v_{1},\dots,v_{n}$ is an orthonormal basis (note that I'm expressing $x,y$ in terms of the basis $(v_{1},\ldots,v_{n})$!).

The case of $\mathbb{R}^{n}$ is very similar, simply omit all the complex conjugates, replace conjugate-transpose by ordinary transpose and Hermitian by symmetric.

-
I don't know of any standards, but if I were in your shoes, I'm for "Hermitian-ness" (unless something better-sounding comes up). I'll upvote in 12 hours... – J. M. May 12 '11 at 14:12
@J.M. Thanks for both things! (meta: this rep-cap is a bit annoying...) – t.b. May 12 '11 at 14:17
sesquilinearity is another word for bilinearity? – user6163 May 12 '11 at 14:34
@Nir: Sesquilinear means literally $1 \frac{1}{2}$-linear (in Latin). It means linear in the first variable and anti-linear in the second one (the missing half of linearity the anti-linearity in the second variable i.e., additivity $\langle x, y + z \rangle = \langle x,y \rangle + \langle x, z \rangle$ and anti-homogeneity $\langle x, \lambda y \rangle = \bar{\lambda} \, \langle x, y \rangle$ - note the bar! It is the appropriate replacement for bilinearity of inner products in the complex case (to have $\langle i\cdot v_{1}, i\cdot v_{1} \rangle = 1$ instead of $-1$ by bilinearity). – t.b. May 12 '11 at 14:42
@Nir: Can you be a bit more specific as to what is unclear? Let's discuss it, I'm here for a while and I'll try to answer your queries. – t.b. May 12 '11 at 14:51