Teaching linear algebra this semester has made me face up to the fact that
for a linear operator on a real inner product space,
whereas for an operator on a complex inner product space,
In other words, call an operator a quarter-turn if for all . Then the real quarter-turns correspond to the skew
symmetric matrices — but apart from the zero operator, there are no
complex quarter turns at all.
Where in my mental landscape should I place these facts?
The proofs of both facts are easy enough. Everyone who’s met an inner
product space knows the real polarization identity: in a real inner product
All we used about here is that it’s a symmetric
bilinear form (and that ). In other words,
for a symmetric bilinear form on , writing ,
for all .
The crucial point is that we really did need the symmetry. (For it’s clear
that the right-hand side is symmetric whether or not is.) For a
not-necessarily-symmetric bilinear form , all we can say is
or more simply put,
Now let be a linear operator on . There is a bilinear form
defined by . It’s not symmetric
unless is self-adjoint; nevertheless, the polarization identity just
stated tells us that
It follows that is a quarter-turn if and only if
for all . After some elementary rearrangement, this in turn is
for all , where is the adjoint of . But that just means
that . So, is a quarter-turn if and only if .
The complex case involves a more complicated polarization identity, but is
ultimately simpler. To be clear, when I say “complex inner product” I’m
talking about something that’s linear in the
first argument and conjugate linear in the second.
In a complex inner product space, the polarization formula is
This can be compared with the real version, which (in unusually heavy
notation) says that
And the crucial point in the complex case is that this time, we don’t
need any symmetry. In other words, for any bilinear form on
, writing , we have
So given a quarter-turn on , we can define a bilinear form
by , and it follows immediately from
this polarization identity that for all
— that is, .
So we’ve now shown that over ,
but over ,
Obviously everything I’ve said is very well-known to those who know it.
(For instance, most of it’s in Axler’s Linear Algebra Done Right.)
But how should I think about these results? How can I train my intuition
so that the real and complex results seem simultaneously obvious?
Whatever the intuitive picture, here’s a nice consequence, also in Axler’s book.
This pair of results immediately implies that whether we’re over or
, the only self-adjoint quarter-turn is zero. Now let be any
operator on a real or complex inner product space, and recall that is
said to be normal if it commutes with .
Equivalently, is normal if the operator is zero.
But is always self-adjoint, so is normal if and
only if is a quarter-turn.
Finally, a bit of routine messing around with inner products shows that
this is in turn equivalent to
So a real or complex operator is normal if and only if and
have the same length for all .