Unitary operators and a product of reflections

50 Views Asked by At

The book Advanced Linear Algebra by Steven Roman states that Reflections or Housholder transformations are self-adjoint and unitary. Moreover, Theorem 10.17 of this book states every unitary $\tau \in \mathcal{L}(V)$, where $V$ is an inner product space, is a product of reflections.

Now, consider $\mathbb{C}$ as a unitary space. Self-adjoint operators are real numbers and $i$ is a unitary operator. But, $i$ is not a product of reflections since an imaginary number cannot be a product of real numbers. What is the problem?

1

There are 1 best solutions below

1
On BEST ANSWER

This is akin to a typographical error. As you note the stated result is not true in complex vector spaces, at least if "Householder transformation"/"reflection" is defined as a map of the form $x \mapsto x - 2 \frac{\langle x, v \rangle}{\langle v, v \rangle} v$ for some nonzero $v \in V$ (see the displayed equation following the definition on the top of p. 244 of the third edition; while this is not expressly Roman's definition of "Householder transformation," he treats it as equivalent to his definition). [If one takes what Roman expressly uses as a definition of "Householder transformation"/"reflection" at the top of p. 244, one sees that the only "reflection" on a one-dimensional vector space is the additive inverse of the identity operator $I$, so that $I$ and $-I$ are the only products of "reflections" (thus defined) on a one-dimensional vector space, which does, as you point out, fail to include all unitaries on a one-dimensional unitary space.]

You can trace the problem in the proof of Roman's result to the mapping properties that are assumed of the transformations $x \mapsto x - 2 \frac{\langle x, u \rangle}{\langle u, u \rangle} u$ for nonzero $u \in V$. Roman asserts that these maps have the property that when $v$ and $w$ are distinct nonzero vectors of the same length, then $H_{v-w}$ maps $v$ to $w$ and vice versa. Indeed, this is part of his Theorem 10.16, p. 244 of the third edition (and note that he implicitly uses this mapping property in the first displayed equation in the proof of Theorem 10.17 on page 245). His proof of Theorem 10.16 begins "if $\|v\| = \|w\|$, then $(v - w) \perp (v + w)$ . . . ." and this is where the trouble arises. Of course if the inner product is symmetric (as it is in a real vector space) the conclusion $(v - w) \perp (v + w)$ (and the corresponding mapping property of Theorem 10.16) does follow from the hypothesis that $\|v\| = \|w\|$, while if the inner product is only conjugate symmetric (as it is in a complex vector space, or in particular in $V = \mathbb{C}^1$) it does not.

The failure of a Householder-style formula to give operators in the complex case with mapping properties analogous to those of the real case has been noted before on MSE, e.g. in the question What is the Householder matrix for complex vector space? and its answer.

I expect that Roman intended to state his result only for real vector spaces. As evidence of Roman's intent, if you look on page 292 of the third edition, Roman says "[r]ecall again that for a real inner product space, a reflection is defined as . . ." and then recites the definition from the top of p. 244, and on the top of page 293 he then expressly refers to what Theorem 10.16 says about the mapping properties of reflections "[f]or real inner product spaces . . . ." This suggests to me that Roman intended to define "reflection"/"Householder transformation" only for real vector spaces, and if that is the case, then maybe he also intended to state his Theorems 10.16 and 10.17 (which immediately follow his definition of those terms, and which concern those operators) for real vector spaces only.