TITLE: Why does the fact that "$Tv$ is orthogonal to $v$ for all $v$ implies T is the zero operator" break down for real inner product spaces?
QUESTION [10 upvotes]: Here's the awkwardly named theorem 7.14 (for which I can't think of a good name either) appearing in Axler's Linear Algebra Done Right, 3rd edition, p 210:

The proof is algebraic, and I can't glean from it any intuition about why this theorem breaks down over $\mathbf{R}$, as the author claimed. What's so special about $\mathbf{C}$ that allows an inner product to be written in the above form, which seems impossible for the inner product over $\mathbf{R}$?
I would also appreciate alternative (and more "intuitive", or perhaps  elementary) proofs for this result.

REPLY [2 votes]: How the proof fails. The cited proof relies on the equality $\langle T(u\pm iw), u\pm iw\rangle=0$. On a complex inner product space, this equality is just a special case of the given assumption that $\langle Tv,v\rangle=0$ for every vector $v\in V$. However, on a real inner product space, the equality no longer follows from the assumption, because $u\pm iw$ is not necessarily a vector in $V$ (note that $iw$ is not a legitimate scalar multiplication over $\mathbb R$).
Why the problem statement fails. The is because on a real inner product space, the identity $\langle Tv,v\rangle\equiv0$ is always satisfied by every skew-adjoint operator $T$. To illustrate concretely, consider a nonzero real skew-symmetric matrix $K$. Although there exist complex vectors $v$ (such as eigenvectors corresponding to nonzero eigenvalues of $K$) such that $\langle Kv,v\rangle=v^\ast Kv\ne0$, the inner product is always zero if $v$ is only allowed to be real.