Some musings on Cauchy-Schwarz, Part IVc

 To conclude Part IV, we prove the easy fact that all two-dimensional real inner product spaces are "essentially the same" as $\R^2$ with the usual dot product and the associated (Euclidean) norm $\|\vx\|_2 = \sqrt{\vx\cdot\vx}$.

Of course what we need is a linear isomorphism which preserves norms and inner products. The argument is almost identical to that in Part III (and, unsurprisingly, generalises via the usual Gram-Schmidt process to higher dimensions).

As an exercise you should check for yourselves the details of any unproven claims below.

Now let $V$ be any two-dimensional inner product space over $\R$, with inner product $\langle \cdot,\cdot \rangle_V$ and associated norm $\|\cdot\|_V$.

Let $\{\vv_1,\vv_2\}$ be a basis of $V$. Then $\vv_1\neq 0$, so we can normalize by setting

$$\Ve_1=\frac{1}{\|\vv_1\|_V} \vv_1\,.$$

Then $\{\Ve_1,\vv_2\}$ is easily seen to still be a basis of $V$.

Next we write $\vv_2=\va+\vb$ with $\va=\langle \Ve_1,\vv_2\rangle\Ve_1$ being the component of $\vv_2$ along $\Ve_1$ and $\vb=\vv_2-\va$ being the component of $\vv_2$ perpendicular to $\Ve_1$. Explicitly (as in the Gram-Schmidt process)

$$\vb = \vv_2 - \langle \Ve_1,\vv_2\rangle\Ve_1\,.$$

We have $\vb \neq \boldsymbol{0}$ (because $\{\Ve_1,\vv_2\}$ is a linearly independent set), and so we may normalise by setting $$\Ve_2=\frac{1}{\|\vb\|_V} \vb.$$ Then it is easy to see that $\{\Ve_1,\Ve_2\}$ is an orthonormal basis of $V$. 

Define a map $T:\R^2\to V$ by $T((x,y))=x \Ve_1 + y \Ve_2$. Then $T$ is easily seen to be a bijection. Moreover, $T$ is a linear isomorphism, and for all $\vx,\vy \in \R^2$, we have (via Pythagoras etc.)

$$\|T(\vx)\|_V=\|\vx\|_2$$

and

$$\langle T(\vx),T(\vy)\rangle_V=\vx \cdot \vy\,.$$

In other words, the linear isomorphism $T$ preserves norms and inner products. (It is an isomorphism of real inner product spaces.)

We can use $T$ and its inverse $T^{-1}$ to transfer any inner product space properties we have back and forth between $\R^2$ and $V$, and it is in this sense that all two-dimensional real inner product spaces are "essentially the same" as $\R^2$ with the usual dot product. In particular, the Cauchy-Schwarz inequality holds for all two-dimensional real inner product spaces because it holds for the dot product in $\R^2$.

In Part Va we see how the Cauchy-Schwarz inequality for general real inner product spaces follows easily from this (for the reasons we indicated before). Then in Part Vb we recall how to deduce the version for complex inner product spaces from the version for real inner product spaces, and give a quick summary of everything we have discussed so far.