Explaining maths

 Recall our claim from Part I Cauchy-Schwarz in complex inner product spaces can be deduced from Cauchy-Schwarz in real inner product spaces. This is essentially because every complex inner product space can also be regarded as a real inner product space with the same norm, but where the real inner product is the real part of the complex inner product. The last bit of the deduction uses a fairly standard "rotation trick": for any complex number $z$, we have  $$|z|=\max\{\Re(\alpha z):\alpha \in \C, |\alpha|=1\,\}\,.\qquad(*)$$ Let's check $(*)$ first. Let $z\in\C$, and set $$A=\{\Re(\alpha z):\alpha \in \C, |\alpha|=1\,\} \subseteq \R\,.$$  Clearly $A$ is non-empty.  Let $\alpha \in \C$ with $|\alpha|=1$. Then certainly we have $$\Re(\alpha z) \leq |\alpha z| = |z|\,.$$ Thus we have $$\sup A \leq |z|\,.$$ We now need to prove that equality holds, and that the supremum is actually a maximum. For this we just need to show that $|z| \in A$. This is trivial if $z=0$, b...

In this part, we take as our starting point the (easy) results discussed so far for real inner product spaces of dimensions less than or equal to $2$. That is, that for a given dimension $n$ (and we only need $n\leq 2$ below), up to "real inner product space isomorphism", all $n$-dimensional inner product spaces over $\R$ are the same as $\R^n$ with the usual dot product. In particular, the Cauchy-Schwarz inequality holds in all real inner product spaces of dimension $\leq 2$, with equality if and only if the two vectors involved are linearly dependent. Now let $V$ be any  inner product space over $\R$, with inner product $\langle \cdot,\cdot \rangle_V$ and associated norm $\|\cdot\|_V$. (Here $V$ can be infinite-dimensional, and need not be separable.) Let $\vx, \vy \in V$, let $W$ be the linear subspace of $V$ spanned by $\vx$ and $\vy$, i.e., $\lin\{\vx,\vy\}$. Clearly $W$ is then itself a real inner product space of dimension at most $2$. Thus the Cauchy-Schwarz inequalit...

 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 com...

In this part we will look at two-dimensional real inner product spaces, which are all essentially identical to $\R^2$ with the standard dot product. They are also all essentially identical to $\C$ when $\C$ is regarded as a  real  inner product space, using the  real  inner product (as discussed in Part II). $$\langle z,w \rangle = \Re(\bar z w)= \Re(z) \Re(w) + \Im(z) \Im(w)\,,$$ and the Cauchy-Schwarz inequality for all two-dimensional real inner product spaces is immediate from this, because $|\Re(\bar z w)| \leq |\bar z w|$. (Of course, Cauchy-Schwarz in $\R^2$ is easy to prove anyway!) We already discussed the connection between $\R^2$ and $\C$ in Part II. Recall the following from that part (also discussed in intermediate parts). The (linear in the second variable) standard complex inner product on $\C$ is given by $\langle z,w \rangle = \bar z w$, with the associated norm given by modulus. Writing $z=a_1+\i a_2$ and $w=b_1+\i b_2$, with $a_1,a_2,b_1,b_2$ real,...

In this very short part we look briefly at one-dimensional complex inner product spaces, essentially repeating some of the material from previous posts. As we noted in Part II, the standard (linear in the second variable) complex inner product on $\C$ is given by $\langle z,w \rangle_\C = \bar z w$, with the associated norm given by the usual modulus, $\|z\|_\C=|z|$. Since we have (by standard properties of $\C$) that $$|\langle z,w \rangle_\C|=|\bar z w| = |z| |w| = \|z\|_\C \|w\|_\C\,,$$ the Cauchy-Schwarz inequality holds for $\C$, and in fact equality always holds here. We can now repeat the reasoning from Part III to see that all one-dimensional complex inner product spaces are essentially the same as $\C$ with its standard inner product (given above). Unsurprisingly, this means that  equality  always holds in the Cauchy-Schwarz inequality in one-dimensional complex inner product spaces. In part IVb we will look at two-dimensional real inner product spaces, which are all ...

In this part we look at the very trivial case of one-dimensional inner product spaces over $\R$. As an exercise you can check any details omitted below, and identify the famous named theorems we are discussing the one-dimensional versions of! Almost everything discussed below is just the trivial one-dimensional case of standard facts about finite-dimensional, real inner product spaces. More advanced readers may wish to skip or merely skim Parts III and IV and move on to Part V. The standard inner product on $\R$ is just the usual product in $\R$, so, for $x,y \in \R$, $$\langle x,y\rangle_\R=xy\,.$$ This is, of course, the one-dimensional version of the dot product! You can check that this really is a (real) inner product. (This is mostly just the standard rules of arithmetic in $\R$, along with properties of modulus.) Unsurprisingly, the associated norm is given by $$||x||_\R=\sqrt{\langle x,x \rangle}_\R = \sqrt{x^2} = |x|\,,$$ i.e., the usual modulus of $x$. In this special case, th...

In this part we'll have an initial look at some of the connections between real inner product spaces and complex inner product spaces. In a previous post, I discussed the fact that every vector space over $\C$ can be regarded as a vector space over $\R$ with respect to the same addition, and the restricted scalar multiplication obtained by using only real scalars (where we regard $\R$ as a subset of $\C$, as usual). If we have a finite-dimensional vector space over $\C$, then the dimension doubles when we regard it as a vector space over $\R$. In particular, $\C$ is a $1$-dimensional complex vector space, but is a $2$-dimensional real vector space. This is no surprise, since we often identify $\C$ with $\R^2$ (and may well define $\C$ that way as a set). Similarly, every normed space over $\C$ can also be regarded as a normed apace over $\R$, with the same addition and norm, and restricting attention to multiplication by only real scalars. Now, however, we are talking about inner p...