### Quarter-Turns

#### Posted by Tom Leinster

Teaching linear algebra this semester has made me face up to the fact that
for a linear operator $T$ on a *real* inner product space,
$\langle T x, x \rangle = 0 \,\, \forall x \,\,
\iff
\,\,
T^\ast = -T$
whereas for an operator on a *complex* inner product space,
$\langle T x, x \rangle = 0 \,\, \forall x \,\,
\iff
\,\,
T = 0.$
In other words, call an operator $T$ a **quarter-turn** if $\langle T x, x \rangle =
0$ for all $x$. 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 space $X$,

$\langle x, y \rangle = \frac{1}{4} \bigl( \| x + y \|^2 - \| x - y \|^2 \bigr).$

All we used about $\langle -, - \rangle$ here is that it’s a symmetric bilinear form (and that $\|w\|^2 = \langle w, w \rangle$). In other words, for a symmetric bilinear form $\beta$ on $X$, writing $Q(w) = \beta(w, w)$, we have

$\beta(x, y) = \frac{1}{4} \bigl( Q(x + y) - Q(x - y) \bigr)$

for all $x, y \in X$.

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 $\beta$ is.) For a
not-necessarily-symmetric bilinear form $\beta$, all we can say is

$\frac{1}{2} \bigl( \beta(x, y) + \beta(y, x) \bigr) = \frac{1}{4} \bigl( Q(x + y) - Q(x - y) \bigr)$

or more simply put,

$\beta(x, y) + \beta(y, x) = \frac{1}{2} \bigl( Q(x + y) - Q(x - y) \bigr).$

Now let $T$ be a linear operator on $X$. There is a bilinear form $\beta$ defined by $\beta(x, y) = \langle T x, y \rangle$. It’s not symmetric unless $T$ is self-adjoint; nevertheless, the polarization identity just stated tells us that

$\langle T x, y \rangle + \langle T y, x \rangle = \frac{1}{2} \bigl( \langle T(x + y), x + y \rangle - \langle T(x - y), x - y \rangle \bigr).$

It follows that $T$ is a quarter-turn if and only if

$\langle T x, y \rangle + \langle T y , x \rangle = 0$

for all $x, y \in X$. After some elementary rearrangement, this in turn is equivalent to

$\langle (T + T^\ast)x, y \rangle = 0$

for all $x, y$, where $T^\ast$ is the adjoint of $T$. But that just means that $T + T^\ast = 0$. So, $T$ is a quarter-turn if and only if $T^\ast = -T$.

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

$\langle x , y \rangle = \frac{1}{4} \sum_{p = 0}^3 i^p \| x + i^p y \|^2.$

This can be compared with the real version, which (in unusually heavy notation) says that

$\langle x , y \rangle = \frac{1}{4} \sum_{p = 0}^1 (-1)^p \| x + (-1)^p y \|^2.$

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 $\beta$ on
$X$, writing $Q(x) = \beta(x, x)$, we have

$\beta(x, y) = \frac{1}{4} \sum_{p = 0}^3 i^p Q(x + i^p y).$

So given a quarter-turn $T$ on $X$, we can define a bilinear form $\beta$ by $\beta(x, y) = \langle T x, y \rangle$, and it follows immediately from this polarization identity that $\langle T x, y \rangle = 0$ for all $x, y$ — that is, $T = 0$.

So we’ve now shown that over $\mathbb{R}$,

$T \,\,\text{is a quarter-turn}\,\, \iff T^\ast = - T$

but over $\mathbb{C}$,

$T \,\,\text{is a quarter-turn}\,\, \iff T = 0.$

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 $\mathbb{R}$ or
$\mathbb{C}$, the only *self-adjoint* quarter-turn is zero. Now let $T$ be any
operator on a real or complex inner product space, and recall that $T$ is
said to be **normal** if it commutes with $T^\ast$.

Equivalently, $T$ is normal if the operator $T^\ast T - T T^\ast$ is zero.

But $T^\ast T - T T^\ast$ is always self-adjoint, so $T$ is normal if and only if $T^\ast T - T T^\ast$ is a quarter-turn.

Finally, a bit of routine messing around with inner products shows that this is in turn equivalent to

$\| T^\ast x \| = \| T x \| \,\,\text{for all}\,\, x \in X.$

So a real or complex operator $T$ is normal if and only if $T^\ast x$ and $T x$ have the same length for all $x$.

## Re: Quarter-Turns

The first thing that jumps out at me is that $\mathbb{C}$ has a

scalarquarter-turn already. Such is famously absent over the Reals.