Norm on $\operatorname{End}(E) \otimes A$ Suppose $A$ is a finite dimensional $\mathbb C$-algebra with a submultiplicative norm (for all $a, b \in A$, $||ab|| \leq ||a|| ||b||)$. Let $E$ be a finite dimensional complex vector space with a hermitian inner product. $\operatorname{End}E$ is the endomorphism ring of $E$.

Is there a natural way to put a norm on $\operatorname{End}E \otimes A$?

This is a claim at the bottom of page 27 in the paper:
https://arxiv.org/abs/math/0702575
Also, according to the paper, for $H$ a hermitian endomorphism, we should have $\| H \otimes 1\|$ equal the largest eigenvalue of $H$. I suppose we could normalize the norm on $A$ so that $\|1_A\| = 1$. Under the operator norm on $\operatorname{End}E$, we do indeed have $\|H\|$ equal the absolute value of the largest eigenvalue of $H$.
Thank you.
 A: There is the operator norm on $End(E)$ and for $x \in End(E) \otimes A$ $$N(x) =\inf \{\sum \|b_i\|\|c_i\|,x = \sum b_i \otimes c_i\}$$
Let $x = \sum b_i \otimes c_i,y = \sum d_j \otimes e_j$ such that $N(x) =\sum \| b_i\|\|c_i\|,N(y)= \sum\|d_j\|\|e_j\|$ we have
$$N(xy) = N((\sum b_i \otimes c_i)(\sum d_j \otimes e_j))=
N(\sum b_id_j \otimes c_ie_j) \le \sum \|b_id_j\| \|c_ie_j\|\\ \le  \sum \|b_i\| \|d_j\| \|c_i\| \|e_j\| =( \sum \|b_i\| \|c_i\|)(\sum \|d_j\| \|e_j\|) = N(x)N(y) $$
A: $\operatorname{End}(E) \otimes A$ is simply the tensor product of two Banach algebras, equipped with the operator norm on $\operatorname{End}(E)$ and $\|\cdot\|$ on $A$ respectively.
Wikipedia defines the norm as
$$\pi(x) := \inf\left\{\sum_{i}\|A_i\|\|a_i\| : x = \sum_i A_i\otimes a_i\right\}$$
For Hermitian $H$ we have $\|H \otimes 1_A\| = \|H\|\|1_A\| = \|H\|$ which is equal to the largest eigenvalue of $H$.
Since $\operatorname{End}(E) \otimes A$ is a finite-dimensional vector space, all norms are equivalent to this one anyway.
