9
$\begingroup$

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:

Theorem 7.14

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.

$\endgroup$
  • 1
    $\begingroup$ Unless I am grossly misunderstanding you, the explanation of the breakdown is in the paragraph before the theorem. It works over $\mathbb{C}$ because in a complex field, $i$ can be an eigenvalue, as it is for the rotation. But it cannot be one over a real field. Does this help? I can expand to an answer if you'd like. $\endgroup$ – The Count Dec 28 '16 at 4:31
  • $\begingroup$ Yeah I would appreciate more explanation how the breakdown is more generally caused by the lack of complex eigenvalues $\endgroup$ – Yibo Yang Dec 28 '16 at 4:57
  • 2
    $\begingroup$ I mean, re: the eigenvalues... imagine how to show $Tv = 0$ for all $v$. Let's first show it for one single eigenvector. Ok, choose an eigenvector $v$ with eigenvector $\lambda$. Then $Tv = \lambda v$. Oh, but $0 = \langle Tv,v \rangle = \lambda \langle v,v \rangle$ then, so $\lambda = 0$. Then it all goes on by induction on the dimension of the underlying space from there. But you see the start of this ``attempt'' to show $T = 0$ begins by choosing an eigenvector and so necessarily relies on extending to a field which is algebraically closed (like the complex numbers). $\endgroup$ – tkr Dec 28 '16 at 5:12
  • $\begingroup$ Posted as per your request, @YiboYang. Let me know if anything is unclear. $\endgroup$ – The Count Dec 28 '16 at 17:18
  • $\begingroup$ I had a similar question to this and I thought I came up with a proof. Is this correct? My issue with my proof is I'm not sure if I can suppose that there exists such an eigenvector for all operators. Suppose $\langle Tv, v \rangle = 0$. If $v$ is an eigenvector, then $\langle \lambda v, v \rangle$ = 0. Then $\langle Tv, v \rangle - \langle \lambda v, v \rangle = 0$. Now, $v \neq 0 \implies T - \lambda I = 0$. And since $\lambda = 0, T = 0$. $\endgroup$ – jaslibra Apr 16 '17 at 21:44
6
$\begingroup$

One geometric answer is the somewhat surprising fact that rotation is essentially 2-dimensional. We're used to thinking, in 3-space, of "rotation about an axis," but it might be wiser to say "rotation in a plane". In 4-space, for instance, we have the rotation $$ \begin{bmatrix} 1 & 0 & &&\\ 0 & 1 & & &\\ & & c & -s \\ & & s & c \end{bmatrix} $$ where $s$ and $c$ are the sine and cosine of some angle. But this rotation fixes both the $x$- and the $y$-axis so it doesn't have 'an axis'; in fact, for every dimension, any rotation can be written as the product of such "rotations in a plane" (although this takes a little proving).

Now in the complex case, such a rotation in a plane is not so much a rotation as a uniform-scale by a complex constant $\gamma = c + is$ of modulus 1. The very notion of complex inner product says, for instance, that $1$ and $i$ are no longer perpendicular, for $$ <1, i> = 1 \cdot (-i) = -i \ne 0. $$

So while you used to "have room" to put $T(u)$ someplace orthogonal to $u$ (in the real case), you no longer do in the complex one, or at least there aren't enough different and unentangled places to do so, which is what Axler's proof shows: you can't make $u \pm w$ and $u \pm iw$ all map to things that'll be perpendicular to them.

$\endgroup$
3
$\begingroup$

I apologize for the long answer. I just threw as much exposition out for you as I could, since I just recently became comfortable with these ideas.

I am a big fan of "Done Right" and am in fact using it myself at the moment. The proof of the theorem is nice because it is detailed. Intuitively, I find it helpful to consider the complex plane, which represents all of $\mathbb{C}$, as an analogue to the real number line. Because numbers can have both real and imaginary parts, the complex plane is a two-dimensional real vector space over $\mathbb{R}$, but if we allow complex scalars, it is a one-dimensional vector space.

Consider writing any complex number $z$ in polar form, as $re^{i\theta}$, where $r=|z|$ and $\theta=\arg(z)$, the angle from the positive real axis. Now, if we only allow multiplication by real scalars, we can only move along the line through the origin containing $z$ since such a multiplication only changes the value of $r$. But if we allow multiplication by complex scalars, we can multiply by any $z'=r'e^{i\theta'}$ to get $zz'=rr'e^{i(\theta+\theta')}$, and it is clear that we can get to any other element of $\mathbb{C}$ via scalar multiplication by a complex scalar. So, in this sense, no element of $\mathbb{C}$ is orthogonal to any other element.

In $\mathbb{C}$, for example, let $a+ib$ be represented by $(a,b)$. Then we can get from $(1,0)$ to $(0,1)$, for example, by multiplying by $i$. Write it down: $(1,0)$ is $1+0i$, and multiplying by $i$ gives us $(0+i)$ which is $(0,1)$. It should be easy to see that, in general, for $v\in\mathbb{C}$, and $T:\mathbb{C}\to\mathbb{C}$, there exists $z\in \mathbb{C}$ such that $Tv=zv$ and taking the inner product of both sides with $v$ shows that $\langle Tv, v\rangle = \langle zv,v\rangle=z\langle v, v\rangle$, which equals zero for non-zero $z$ if and only if $\langle v, v\rangle=0$, which means $v=0$.

The reason this fails for real vector spaces is given above. Consider $\mathbb{R}^2$. Because the field is now real, we cannot get from $(1,0)$ to $(0,1)$ by scalar multiplication since $i$ is not in our field. We see clearly that if $T$ rotates in $\mathbb{R^2}$ by $\pi/2$, we can have $\langle Tv, v\rangle = \langle T(1,0),(1,0)\rangle=\langle (0,1),(1,0)\rangle=0$ taking the usual inner product, even though $v\neq 0$

I hope this is helpful. I ramble.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.