Teaching linear algebra this semester has made me face up to the fact that
for a linear operator $<semantics>T<annotation\; encoding="application/x-tex">T</annotation></semantics>$ on a *real* inner product space,
$$<semantics>\u27e8Tx,x\u27e9=0\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\forall x\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\iff \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{T}^{*}=-T<annotation\; encoding="application/x-tex">\; \backslash langle\; T\; x,\; x\; \backslash rangle\; =\; 0\; \backslash ,\backslash ,\; \backslash forall\; x\; \backslash ,\backslash ,\; \backslash iff\; \backslash ,\backslash ,\; T^\backslash ast\; =\; -T\; </annotation></semantics>$$
whereas for an operator on a *complex* inner product space,
$$<semantics>\u27e8Tx,x\u27e9=0\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\forall x\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\iff \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}T=0.<annotation\; encoding="application/x-tex">\; \backslash langle\; T\; x,\; x\; \backslash rangle\; =\; 0\; \backslash ,\backslash ,\; \backslash forall\; x\; \backslash ,\backslash ,\; \backslash iff\; \backslash ,\backslash ,\; T\; =\; 0.\; </annotation></semantics>$$
In other words, call an operator $<semantics>T<annotation\; encoding="application/x-tex">T</annotation></semantics>$ a **quarter-turn** if $<semantics>\u27e8Tx,x\u27e9=0<annotation\; encoding="application/x-tex">\backslash langle\; T\; x,\; x\; \backslash rangle\; =\; 0</annotation></semantics>$ for all $<semantics>x<annotation\; encoding="application/x-tex">x</annotation></semantics>$. 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 $<semantics>X<annotation\; encoding="application/x-tex">X</annotation></semantics>$,

$$<semantics>\u27e8x,y\u27e9=\frac{1}{4}(\Vert x+y{\Vert}^{2}-\Vert x-y{\Vert}^{2}).<annotation\; encoding="application/x-tex">\; \backslash langle\; x,\; y\; \backslash rangle\; =\; \backslash frac\{1\}\{4\}\; \backslash bigl(\; \backslash |\; x\; +\; y\; \backslash |^2\; -\; \backslash |\; x\; -\; y\; \backslash |^2\; \backslash bigr).\; </annotation></semantics>$$

All we used about $<semantics>\u27e8-,-\u27e9<annotation\; encoding="application/x-tex">\backslash langle\; -,\; -\; \backslash rangle</annotation></semantics>$ here is that it’s a symmetric
bilinear form (and that $<semantics>\Vert w{\Vert}^{2}=\u27e8w,w\u27e9<annotation\; encoding="application/x-tex">\backslash |w\backslash |^2\; =\; \backslash langle\; w,\; w\; \backslash rangle</annotation></semantics>$). In other words,
for a symmetric bilinear form $<semantics>\beta <annotation\; encoding="application/x-tex">\backslash beta</annotation></semantics>$ on $<semantics>X<annotation\; encoding="application/x-tex">X</annotation></semantics>$, writing $<semantics>Q(w)=\beta (w,w)<annotation\; encoding="application/x-tex">Q(w)\; =\; \backslash beta(w,\; w)</annotation></semantics>$,
we have

$$<semantics>\beta (x,y)=\frac{1}{4}(Q(x+y)-Q(x-y))<annotation\; encoding="application/x-tex">\; \backslash beta(x,\; y)\; =\; \backslash frac\{1\}\{4\}\; \backslash bigl(\; Q(x\; +\; y)\; -\; Q(x\; -\; y)\; \backslash bigr)\; </annotation></semantics>$$

for all $<semantics>x,y\in X<annotation\; encoding="application/x-tex">x,\; y\; \backslash in\; X</annotation></semantics>$.

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 $<semantics>\beta <annotation\; encoding="application/x-tex">\backslash beta</annotation></semantics>$ is.) For a
not-necessarily-symmetric bilinear form $<semantics>\beta <annotation\; encoding="application/x-tex">\backslash beta</annotation></semantics>$, all we can say is

$$<semantics>\frac{1}{2}(\beta (x,y)+\beta (y,x))=\frac{1}{4}(Q(x+y)-Q(x-y))<annotation\; encoding="application/x-tex">\; \backslash frac\{1\}\{2\}\; \backslash bigl(\; \backslash beta(x,\; y)\; +\; \backslash beta(y,\; x)\; \backslash bigr)\; =\; \backslash frac\{1\}\{4\}\; \backslash bigl(\; Q(x\; +\; y)\; -\; Q(x\; -\; y)\; \backslash bigr)\; </annotation></semantics>$$

or more simply put,

$$<semantics>\beta (x,y)+\beta (y,x)=\frac{1}{2}(Q(x+y)-Q(x-y)).<annotation\; encoding="application/x-tex">\; \backslash beta(x,\; y)\; +\; \backslash beta(y,\; x)\; =\; \backslash frac\{1\}\{2\}\; \backslash bigl(\; Q(x\; +\; y)\; -\; Q(x\; -\; y)\; \backslash bigr).\; </annotation></semantics>$$

Now let $<semantics>T<annotation\; encoding="application/x-tex">T</annotation></semantics>$ be a linear operator on $<semantics>X<annotation\; encoding="application/x-tex">X</annotation></semantics>$. There is a bilinear form $<semantics>\beta <annotation\; encoding="application/x-tex">\backslash beta</annotation></semantics>$
defined by $<semantics>\beta (x,y)=\u27e8Tx,y\u27e9<annotation\; encoding="application/x-tex">\backslash beta(x,\; y)\; =\; \backslash langle\; T\; x,\; y\; \backslash rangle</annotation></semantics>$. It’s not symmetric
unless $<semantics>T<annotation\; encoding="application/x-tex">T</annotation></semantics>$ is self-adjoint; nevertheless, the polarization identity just
stated tells us that

$$<semantics>\u27e8Tx,y\u27e9+\u27e8Ty,x\u27e9=\frac{1}{2}(\u27e8T(x+y),x+y\u27e9-\u27e8T(x-y),x-y\u27e9).<annotation\; encoding="application/x-tex">\; \backslash langle\; T\; x,\; y\; \backslash rangle\; +\; \backslash langle\; T\; y,\; x\; \backslash rangle\; =\; \backslash frac\{1\}\{2\}\; \backslash bigl(\; \backslash langle\; T(x\; +\; y),\; x\; +\; y\; \backslash rangle\; -\; \backslash langle\; T(x\; -\; y),\; x\; -\; y\; \backslash rangle\; \backslash bigr).\; </annotation></semantics>$$

It follows that $<semantics>T<annotation\; encoding="application/x-tex">T</annotation></semantics>$ is a quarter-turn if and only if

$$<semantics>\u27e8Tx,y\u27e9+\u27e8Ty,x\u27e9=0<annotation\; encoding="application/x-tex">\; \backslash langle\; T\; x,\; y\; \backslash rangle\; +\; \backslash langle\; T\; y\; ,\; x\; \backslash rangle\; =\; 0\; </annotation></semantics>$$

for all $<semantics>x,y\in X<annotation\; encoding="application/x-tex">x,\; y\; \backslash in\; X</annotation></semantics>$. After some elementary rearrangement, this in turn is
equivalent to

$$<semantics>\u27e8(T+{T}^{*})x,y\u27e9=0<annotation\; encoding="application/x-tex">\; \backslash langle\; (T\; +\; T^\backslash ast)x,\; y\; \backslash rangle\; =\; 0\; </annotation></semantics>$$

for all $<semantics>x,y<annotation\; encoding="application/x-tex">x,\; y</annotation></semantics>$, where $<semantics>{T}^{*}<annotation\; encoding="application/x-tex">T^\backslash ast</annotation></semantics>$ is the adjoint of $<semantics>T<annotation\; encoding="application/x-tex">T</annotation></semantics>$. But that just means
that $<semantics>T+{T}^{*}=0<annotation\; encoding="application/x-tex">T\; +\; T^\backslash ast\; =\; 0</annotation></semantics>$. So, $<semantics>T<annotation\; encoding="application/x-tex">T</annotation></semantics>$ is a quarter-turn if and only if $<semantics>{T}^{*}=-T<annotation\; encoding="application/x-tex">T^\backslash ast\; =\; -T</annotation></semantics>$.

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

$$<semantics>\u27e8x,y\u27e9=\frac{1}{4}\sum _{p=0}^{3}{i}^{p}\Vert x+{i}^{p}y{\Vert}^{2}.<annotation\; encoding="application/x-tex">\; \backslash langle\; x\; ,\; y\; \backslash rangle\; =\; \backslash frac\{1\}\{4\}\; \backslash sum\_\{p\; =\; 0\}^3\; i^p\; \backslash |\; x\; +\; i^p\; y\; \backslash |^2.\; </annotation></semantics>$$

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

$$<semantics>\u27e8x,y\u27e9=\frac{1}{4}\sum _{p=0}^{1}(-1{)}^{p}\Vert x+(-1{)}^{p}y{\Vert}^{2}.<annotation\; encoding="application/x-tex">\; \backslash langle\; x\; ,\; y\; \backslash rangle\; =\; \backslash frac\{1\}\{4\}\; \backslash sum\_\{p\; =\; 0\}^1\; (-1)^p\; \backslash |\; x\; +\; (-1)^p\; y\; \backslash |^2.\; </annotation></semantics>$$

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 $<semantics>\beta <annotation\; encoding="application/x-tex">\backslash beta</annotation></semantics>$ on
$<semantics>X<annotation\; encoding="application/x-tex">X</annotation></semantics>$, writing $<semantics>Q(x)=\beta (x,x)<annotation\; encoding="application/x-tex">Q(x)\; =\; \backslash beta(x,\; x)</annotation></semantics>$, we have

$$<semantics>\beta (x,y)=\frac{1}{4}\sum _{p=0}^{3}{i}^{p}Q(x+{i}^{p}y).<annotation\; encoding="application/x-tex">\; \backslash beta(x,\; y)\; =\; \backslash frac\{1\}\{4\}\; \backslash sum\_\{p\; =\; 0\}^3\; i^p\; Q(x\; +\; i^p\; y).\; </annotation></semantics>$$

So given a quarter-turn $<semantics>T<annotation\; encoding="application/x-tex">T</annotation></semantics>$ on $<semantics>X<annotation\; encoding="application/x-tex">X</annotation></semantics>$, we can define a bilinear form $<semantics>\beta <annotation\; encoding="application/x-tex">\backslash beta</annotation></semantics>$
by $<semantics>\beta (x,y)=\u27e8Tx,y\u27e9<annotation\; encoding="application/x-tex">\backslash beta(x,\; y)\; =\; \backslash langle\; T\; x,\; y\; \backslash rangle</annotation></semantics>$, and it follows immediately from
this polarization identity that $<semantics>\u27e8Tx,y\u27e9=0<annotation\; encoding="application/x-tex">\backslash langle\; T\; x,\; y\; \backslash rangle\; =\; 0</annotation></semantics>$ for all $<semantics>x,y<annotation\; encoding="application/x-tex">x,\; y</annotation></semantics>$
— that is, $<semantics>T=0<annotation\; encoding="application/x-tex">T\; =\; 0</annotation></semantics>$.

So we’ve now shown that over $<semantics>\mathbb{R}<annotation\; encoding="application/x-tex">\backslash mathbb\{R\}</annotation></semantics>$,

$$<semantics>T\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{is a quarter-turn}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\iff {T}^{*}=-T<annotation\; encoding="application/x-tex">\; T\; \backslash ,\backslash ,\backslash text\{is\; a\; quarter-turn\}\backslash ,\backslash ,\; \backslash iff\; T^\backslash ast\; =\; -\; T\; </annotation></semantics>$$

but over $<semantics>\u2102<annotation\; encoding="application/x-tex">\backslash mathbb\{C\}</annotation></semantics>$,

$$<semantics>T\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{is a quarter-turn}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\iff T=0.<annotation\; encoding="application/x-tex">\; T\; \backslash ,\backslash ,\backslash text\{is\; a\; quarter-turn\}\backslash ,\backslash ,\; \backslash iff\; T\; =\; 0.\; </annotation></semantics>$$

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 $<semantics>\mathbb{R}<annotation\; encoding="application/x-tex">\backslash mathbb\{R\}</annotation></semantics>$ or
$<semantics>\u2102<annotation\; encoding="application/x-tex">\backslash mathbb\{C\}</annotation></semantics>$, the only *self-adjoint* quarter-turn is zero. Now let $<semantics>T<annotation\; encoding="application/x-tex">T</annotation></semantics>$ be any
operator on a real or complex inner product space, and recall that $<semantics>T<annotation\; encoding="application/x-tex">T</annotation></semantics>$ is
said to be **normal** if it commutes with $<semantics>{T}^{*}<annotation\; encoding="application/x-tex">T^\backslash ast</annotation></semantics>$.

Equivalently, $<semantics>T<annotation\; encoding="application/x-tex">T</annotation></semantics>$ is normal if the operator $<semantics>{T}^{*}T-T{T}^{*}<annotation\; encoding="application/x-tex">T^\backslash ast\; T\; -\; T\; T^\backslash ast</annotation></semantics>$ is zero.

But $<semantics>{T}^{*}T-T{T}^{*}<annotation\; encoding="application/x-tex">T^\backslash ast\; T\; -\; T\; T^\backslash ast</annotation></semantics>$ is always self-adjoint, so $<semantics>T<annotation\; encoding="application/x-tex">T</annotation></semantics>$ is normal if and
only if $<semantics>{T}^{*}T-T{T}^{*}<annotation\; encoding="application/x-tex">T^\backslash ast\; T\; -\; T\; T^\backslash ast</annotation></semantics>$ is a quarter-turn.

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

$$<semantics>\Vert {T}^{*}x\Vert =\Vert Tx\Vert \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{for all}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\in X.<annotation\; encoding="application/x-tex">\; \backslash |\; T^\backslash ast\; x\; \backslash |\; =\; \backslash |\; T\; x\; \backslash |\; \backslash ,\backslash ,\backslash text\{for\; all\}\backslash ,\backslash ,\; x\; \backslash in\; X.\; </annotation></semantics>$$

So a real or complex operator $<semantics>T<annotation\; encoding="application/x-tex">T</annotation></semantics>$ is normal if and only if $<semantics>{T}^{*}x<annotation\; encoding="application/x-tex">T^\backslash ast\; x</annotation></semantics>$ and
$<semantics>Tx<annotation\; encoding="application/x-tex">T\; x</annotation></semantics>$ have the same length for all $<semantics>x<annotation\; encoding="application/x-tex">x</annotation></semantics>$.