A Unitarily Normal Problem

I couldn’t solve this linear algebra question as an undergraduate. Today, we revisit it and defeat it once and for all.

Problem 1. Let $V$ be a finite-dimensional (complex) vector space equipped with the inner product $\langle \cdot, \cdot \rangle$ . Let $T$ be an invertible linear operator. Suppose that for any $u,v \in V$ ,

$\displaystyle \frac{\langle T(u), T(v)\rangle}{\sqrt{\langle T(u), T(u) \rangle \langle T(v), T(v) \rangle}} = \frac{\langle u, v\rangle}{\sqrt{\langle u, u \rangle \langle v, v \rangle}}.$

Prove that $T$ is a scalar multiple of some unitary operator.

It is not hard to show that the converse holds (i.e. if $T$ is a scalar multiple of a unitary operator then the equation holds).

In the original exam paper, the examiner actually offered sub-steps to the question. The solution will therefore follow the suggested roadmap and furthermore, address the result more directly.

(Click for Solution)

Solution. We first observe that $(T^* \circ T)^* = T^* \circ T$ implies that $T^* \circ T$ is self-adjoint, therefore normal, and thus unitarily diagonalisable.

Thus, there exists an orthonormal basis $B := \{v_1,\dots,v_n\}$ for $V$ such that $(T^* \circ T)v_i = \lambda_i v_i$ , where each $\lambda_i \in \mathbb{C}$ .

We observe that

$\langle T(v_i),T(v) \rangle = \langle (T^* \circ T)(v_i),v \rangle = \langle \lambda_i v_i,v \rangle = \lambda_i \langle v_i,v \rangle.$

In particular,

${\| T(v_i) \|}^2 = \langle T(v_i),T(v_i) \rangle = \lambda_i \langle v_i,v_i \rangle = \lambda_i$

so that $\lambda_i > 0$ . Thus, substituting $u = v_i$ into the equation

$\displaystyle \frac{\langle T(u), T(v)\rangle}{\| T(u)\| \| T(v) \|} = \frac{\langle u, v\rangle}{\| u \| \| v \|}$

and simplifying,

$\displaystyle \frac{ \lambda_i \langle v_i,v \rangle}{ \sqrt{\lambda_i} \| T(v) \| } = \frac{\langle v_i, v\rangle}{\| v \|}.$

Simplifying for any $v \in V$ such that $\langle v_i, v\rangle \neq 0$ ,

$\displaystyle \| T(v) \| = \sqrt{\lambda_i} \| v\|.$

Since $\lambda_i$ was chosen arbitrarily, we must have $\lambda_1 = \cdots = \lambda_n = \lambda$ so that

$\displaystyle \| T(v) \| = \sqrt{\lambda} \| v\|\quad \iff \quad \left\|\left( \frac T{\sqrt{\lambda}} \right) v \right\| = \|v\|.$

It is not hard to show that this equation holds for any $v \in V$ . Thus, $S:=(1/\sqrt{\lambda})T$ is unitary, and $T = \sqrt{\lambda} S$ , as required.

—Joel Kindiak, 19 Oct 24, 0150H

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A Unitarily Normal Problem

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A Unitarily Normal Problem

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