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Look at small cases

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[QUICK DESCRIPTION] A standard question of Gel'fond's, when listening to a talk, is: "What is the simplest nontrivial example?" This article shows how useful this principle can be.

[EXAMPLE | Counting and the Lagrange Identity]

This example is borrowed from [http://disquisitionesmathematicae.wordpress.com this blog post]. It is a prime example of how a proof for a general argument can be worked out by considering simple cases.<comment thread="387" /> The Lagrange identity for $ \mathbb{C}$ says that
[maths equation0] |\sum_{i=0}^{n} a_{i}b_{i}|^{2}=\sum_{i=0}^{n} |a_{i}|^{2}\sum_{i=0}^{n} |b_{i}|^{2}-\sum_{1 \leq{i}<j\leq{n}} |a_{i}\overline{b_{j}}-a_{j}\overline{b_{i}}|^{2} [/maths]

We will analyze the case for n=3. It will provide a beautiful outline for a general proof.

The left hand side of the equality [eqref equation0] will be equal to

[maths equation1] |a_{1}b_{1} +a_{2}b_{2}+a_{3}b_{3}|^{2}=(a_{1}b_{1} +a_{2}b_{2}+a_{3}b_{3})(\overline{a_{1}b_{1}}+\overline{a_{2}b_{2}}+\overline{a_{3}b_{3}})
[/maths]

It would be helpful indeed to consider the following 3 x 3 matrix for the sum of all the elements of this matrix is equal to the LHS of [eqref equation1].

[maths]\left(\begin{array}{ccc}a_{1}\overline{a_{1}}b_{1}\overline{b_{1}}&a_{1}\overline{a_{2}}b_{1}\overline{b_{2}}&a_{1}\overline{a_{3}}b_{1}\overline{b_{3}}\\ a_{2}\overline{a_{1}}b_{2}\overline{b_{1}}&a_{2}\overline{a_{2}}b_{2}\overline{b_{2}}&a_{2}\overline{a_{3}}b_{2}\overline{b_{3}}\\ a_{3}\overline{a_{1}}b_{3}\overline{b_{1}}&a_{3}\overline{a_{2}}b_{3}\overline{b_{2}}&a_{3}\overline{a_{3}}b_{3}\overline{b_{3}}\end{array}\right)[/maths]

In fact, the elements of the matrix are the terms on the RHS in [eqref equation1].

Each term of the matrix above can thus be represented as

[maths]A_{ij}=a_{i}\overline{a_{j}}b_{i}\overline{b_{j}}[/maths]

Now consider the first term on the RHS in [eqref equation0]. For n=3, it is

[maths equation2](|a_{1}|^{2}+|a_{2}|^{2}+|a_{3}|^{2})(|b_{1}|^{2}+|b_{2}|^{2}+|b_{3}|^{2})=(a_{1}\overline{a_{1}}+a_{2}\overline{a_{2}}+a_{3}\overline{a_{3}})(b_{1}\overline{b_{1}}+b_{2}\overline{b_{2}}+b_{3}\overline{b_{3}})[/maths]

Again, as above, it would be helpful to consider the following matrix.

[maths] \left(\begin{array}{ccc}a_{1}\overline{a_{1}}b_{1}\overline{b_{1}}&a_{1}\overline{a_{1}}b_{2}\overline{b_{2}}&a_{1}\overline{a_{1}}b_{3}\overline{b_{3}}\\ a_{2}\overline{a_{2}}b_{1}\overline{b_{1}}&a_{2}\overline{a_{2}}b_{2}\overline{b_{2}}&a_{2}\overline{a_{2}}b_{3}\overline{b_{3}}\\ a_{3}\overline{a_{3}}b_{1}\overline{b_{1}}&a_{3}\overline{a_{3}}b_{2}\overline{b_{2}}&a_{3}\overline{a_{3}}b_{3}\overline{b_{3}}\end{array}\right)[/maths]

The elements of this matrix are again the terms on RHS in [eqref equation2].

We can thus call each of them

[maths]B_{ij}=a_{i}\overline{a_{i}}b_{j}\overline{b_{j}}[/maths]

The last term in [eqref equation0] is

[maths equation3] -(|a_{1}\overline{b_{2}}-a_{2}\overline{b_{1}}|^{2}+|a_{1}\overline{b_{3}}-a_{3}\overline{b_{1}}|^{2}+|a_{2}\overline{b_{3}}-a_{3}\overline{b_{2}}|^{2})[/maths]

This time we use two matrices to organize the terms in [eqref equation3].

[maths]\left(\begin{array}{ccc}0&a_{1}\overline{a_{2}}b_{1}\overline{b_{2}}&a_{1}\overline{a_{3}}b_{1}\overline{b_{3}}\\ a_{2}\overline{a_{1}}b_{2}\overline{b_{1}}&0&a_{2}\overline{a_{3}}b_{2}\overline{b_{3}}\\ a_{3}\overline{a_{1}}b_{3}\overline{b_{1}}&a_{3}\overline{a_{2}}b_{3}\overline{b_{2}}&0\end{array}\right)[/maths]

and

[maths]
\left(\begin{array}{ccc}0&a_{1}\overline{a_{1}}b_{2}\overline{b_{2}}&a_{1}\overline{a_{1}}b_{3}\overline{b_{3}}\\ a_{2}\overline{a_{2}}b_{1}\overline{b_{1}}&0&a_{2}\overline{a_{2}}b_{3}\overline{b_{3}}\\ a_{3}\overline{a_{3}}b_{1}\overline{b_{1}}&a_{3}\overline{a_{3}}b_{2}\overline{b_{2}}&0\end{array}\right)[/maths]

And we thus define elements of the first and the second matrix as $ C_{ij}$ and $ D_{ij}$ respectively where

[maths]C_{ij}=a_{i}\overline{a_{j}}b_{i}\overline{b_{j}}\sigma_{ij}[/maths]

[maths]D_{ij}=a_{i}\overline{a_{i}}b_{j}\overline{b_{j}}\sigma_{ij}[/maths]

and where $ \sigma_{ij}$ is defined to be $0$ if i=j and is equal to unity otherwise.

Now let $T_{ij}=A_{ij}+D_{ij}-B_{ij}-C_{ij}$

Therefore, [maths]T_{ij}=A_{ij}(\delta_{ij}+\sigma_{ij})+D_{ij}-B_{ij}(\delta_{ij}+\sigma_{ij})-C_{ij}[/maths]
where we have used $ \delta_{ij}+\sigma_{ij}=1$ ($ \delta_{ij}$ is the Kronecker delta)

And so,

[maths]T_{ij}=a_{i}\overline{a_{j}}b_{i}\overline{b_{j}}\delta_{ij}+a_{i}\overline{a_{j}}b_{i}\overline{b_{j}}\sigma_{ij}+a_{i}\overline{a_{i}}b_{j}\overline{b_{j}}\sigma_{ij}[/maths]

[maths]-a_{i}\overline{a_{i}}b_{j}\overline{b_{j}}\delta_{ij}-a_{i}\overline{a_{i}}b_{j}\overline{b_{j}}\sigma_{ij}-a_{i}\overline{a_{j}}b_{i}\overline{b_{j}}\sigma_{ij}[/maths]

[maths] T_{ij}=a_{i}\overline{a_{i}}b_{i}\overline{b_{i}}-a_{i}\overline{a_{i}}b_{i}\overline{b_{i}}=0[/maths]

Since each $ T_{ij}$ is zero, the sum $\sum_{i,j=1}^{n}T_{ij}=0$

which automatically implies the Lagrange Identity in $\mathbb{C}$.

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