Some musings on Cauchy-Schwarz, Part II

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 product spaces. It is not quite true that every complex inner product is also a real inner product, because real inner products must take values in $\R$. However, the real part of a complex inner product is a real inner product on the associated real vector space, and the norm it gives is unchanged: if $V$ is a complex inner product space with inner product $\langle\cdot,\cdot\rangle$ and associated norm $||\cdot||$, then, for all $\vx \in V$, we have

$$\langle\vx,\vx\rangle=||\vx||^2\in \R$$

and so we also have

$$||\vx||^2=\Re(\langle\vx,\vx\rangle)\,.$$

We do get slightly different notions of orthogonal (perpendicular). Indeed, if $V$ is finite-dimensional, we will need twice as many elements in a real orthonormal basis than we have in a complex orthonormal basis. Also, as far as the real inner product on our space $V$ is concerned, $\vx$ is always orthogonal to $\i\vx \in V$, because $\langle \vx, \i\vx\rangle$ is purely imaginary. (It will be $\pm \i ||\vx||^2$ depending on your convention as to whether inner products are linear in the second variable or in the first variable.)

Note that 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, we have $\Re(\bar z w)= a_1 b_1 + a_2 b_2$ which is just the usual dot product $(a_1,a_2)\cdot(b_1,b_2)$. So if we identify $\C$ with $\R^2$ in the usual way, then the usual dot product (on $\R^2$) is the real part of the usual complex inner product on the (one-dimensional) complex vector space $\C$.

This gives us another way to see the relationship between our two standard formulae for the dot product in $\R^2$. If we write $z$ and $w$ in exponential form as $z=r \e^{\i s}$ and $w=R \e^{\i t}$, then we have $\bar z = r \e^{-\i s}$ and $\bar z w = r R \e^{\i(t-s)}$. Taking the real part we see that

$$(a_1,a_2)\cdot(b_1,b_2) = \Re(\bar z w) =  r R \cos(t-s)\,,$$

and, of course, $t-s$ is the angle between our two vectors, while $rR$ is the product of the norms.

I don't think that this is the usual way to see it, though!