Norm of rotation matrix as element in $M_2(A)$

Question

Let $A$ be a complex unital Banach algebra and let $R_t=\begin{pmatrix} \cos\frac{\pi t}{2} & -\sin\frac{\pi t}{2} \\ \sin\frac{\pi t}{2} & \cos\frac{\pi t}{2} \end{pmatrix}$. If I consider $M_2(A)$ with the operator norm coming from $A^2$ equipped with the norm $||(a,b)||=\sqrt{||a||^2+||b||^2}$, what is the norm of $R_t$ as an element of $M_2(A)$?

I was tempted to say it is 1 but I realized that it's probably not so straightforward. An upper bound would do as well (for example, is there an upper bound on the norm of an arbitrary matrix in $M_2(A)$ in terms of the norms of the entries?)

This should be a rather simple problem but I can't figure it out.

Answer 1

It works out the same way as in the real case (and the operator norm is 1) -- we just have to be a little more careful than usual. The following holds whenever $A$ is a normed real vector space, and $A^2=A\oplus A$ is equipped with the derived 2-norm:

Derive an inner product from the norm on $A^2$ in the usual way through the polarization identity: $$\langle v,w\rangle = \frac12(\|v+w\|^2-\|v\|^2-\|w\|^2)$$

EDIT: Argh, this is not possible in general -- only if the original norm on $A$ satisfies the parallelogram identity. That pretty much kills the rest of the answer in the general case.

It has all the usual nice properties -- in particular we can see by direct calculation that $$ \tag{*} \left\langle\bigl[{}^p_0\bigr],\bigl[{}^0_q\bigr]\right\rangle = 0 $$ for all $p,q\in A$. Due to the symmetry of $\|{\cdot}\|$ in the two coordinates we also have $$ \tag{**} \left\langle\bigl[{}^p_q\bigr],\bigl[{}^r_s\bigr]\right\rangle = \left\langle\bigl[{}^q_p\bigr],\bigl[{}^s_r\bigr]\right\rangle$$

Let $v=(a,b)$ and $w=R_t v$. If for brevity we set $c=\cos\frac{\pi t}{2}$ and $s=\sin\frac{\pi t}{2}$ we get $$ w = c\bigl[{}^a_0\bigr] - s\bigl[{}^b_0\bigr] + s\bigl[{}^0_a\bigr] + c\bigl[{}^0_b\bigr] $$ Now expand $\|w\|^2 = \langle w,w\rangle$ by linearity on both sides, and similarly move the $s$ and $c$ factors out in front. Of the 16 this produces, 8 vanish immediately due to $(*)$ and 4 of the rest (the cross terms such as $-sc\langle\bigl[{}^a_0\bigr],\bigl[{}^b_0]\rangle$) cancel each other due to $(**)$. This canceling is where using an inner product really pays off; it allows us to pull the minus sign on one of the sines in the matrix out of the inner product so it can cancel the other sine, which wouldn't have been possible when working with norms all of the way.

What is left is $$ \begin{align} \|w\|^2 &= c^2\left\|(a,0)\right\|^2 + s^2\left\|(b,0)\right\|^2 + s^2\left\|(0,a)\right\|^2 + c^2\left\|(0,b)\right\|^2 \\ &= (c^2+s^2)\|a\|^2+(c^2+s^2)\|b\|^2 \\ &= \|a\|^2+\|b\|^2 = \|v\|^2 \end{align}$$ So the operator norm is indeed $1$.