I was at the store today when I had the following question. Suppose that $A$ is a real-valued $n \times m$ matrix. If we measure its operator norm as linear operator from $\mathbb{R}^{m} \longrightarrow \mathbb{R}^{n}$, does it co-incide with its operator norm as a $\mathbb{C}^{m} \longrightarrow \mathbb{C}^{n}$ operator? $ \newcommand{\abs}[1]{| #1 |} \newcommand{\ind}{\mathbf{1}} \newcommand{\norm}[1]{\lVert #1 \rVert} \newcommand{\ip}[2]{\langle #1, #2 \rangle} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\T}{\mathsf{T}} \newcommand{\Proj}{\mathcal{P}} \newcommand{\Tr}{\mathrm{Tr}} \newcommand{\A}{\mathcal{A}} $

The answer turns out to be yes, which is actually somewhat surprising (although maybe not that surprising) to me. I first thought the answer would be no, that allowing complex inputs could allow you to drive up the norm. Turns out this is not the case. Interestingly, though, the vectors which achieve the operator norm are much richer in the complex case.

Let's set up some notation. We'll fix a $n \times m$ matrix $A$ with real-valued coefficients. Let $\norm{A}_{\R}$ denote the real-valued operator norm, and $\norm{A}_{\C}$ denote the complex-valued operator norm. That is, $$ \norm{A}_{\R} = \sup_{x \in \R^{m} : \norm{x}_2 = 1} \norm{Ax}_{2} \:, \:\: \norm{A}_{\C} = \sup_{x \in \C^{m} : \norm{x}_2 = 1} \norm{Ax}_{2} \:. $$

Lemma: $\norm{A}_{\R} = \norm{A}_{\C}$.

Proof: Put $M = A^\T A$. We'll study the variational form $$ \norm{A}^2_{\C} = \sup_{x \in \C^{m}, \norm{x}_2 = 1} x^* M x \:. $$ Write $x \in \C^{m}$ as $x = a + j b$, where $a,b \in \R^{m}$. Then, $$ \begin{align*} x^* M x &= (a + jb)^* M (a + jb) = (a - jb)^\T M (a + jb) \\ &= a^\T M a + j a^\T M b - jb^\T M a + b^\T M b \\ &= a^\T M a + b^\T M b \:. \end{align*} $$ We have reduced the operator norm to the following real-valued optimization problem $$ \sup_{a,b \in \R^{m}} f(a, b) = a^\T M a + b^\T M b : \norm{a}_2^2 + \norm{b}_2^2 = 1 \:. $$ Now we just use the usual Lagrange multiplier argument. Define $L(a,b,\lambda)$ as $$ L(a,b,\lambda) = a^\T M a + b^\T M b - \lambda(a^\T a + b^\T b - 1) \:. $$ The necessary conditions for optimality are $$ \begin{align*} 0 &= M a - \lambda a \:, \\ 0 &= M b - \lambda b \:, \\ 1 &= a^\T a + b^\T b \:. \end{align*} $$ Hence, $a,b$ have to be eigenvectors of $M$ which correspond to the same eigenvalue. Substituting the necessary conditions back into the objective function, $$ f(a, b) = \lambda a^\T a + \lambda a^\T b = \lambda \:, $$ and so clearly we should pick $\lambda$ to correspond to the maximum eigenvalue of $M$. Hence $\lambda = \norm{A}_{\R}^2$. Furthermore, it can be verified that for any $\alpha \in [0, 1]$ and unit-norm $u, v$ such that $M u = \lambda u$ and $M v = \lambda v$, the vector $$ x = \sqrt{\alpha} u + j \sqrt{1-\alpha} v $$ is unit norm and satisfies $x^* M x = \lambda$. $\square$

The last point in the proof is interesting. Suppose for simplicity that $A$ satisfies $\sigma_1(A) > \sigma_2(A)$. Then there are only two unit vectors which achieve the supremum for $\norm{A}_{\R}$. On the other hand, in the $\norm{A}_{\C}$ case, there is a whole continnum of vectors achieving the supremum.

On a final note, this paper by Holtz and Karow proves this result for general $L^p$ spaces.