Q:

Two linear maps are "identical"

I'm trying to describe a situation, whose smallest example is the following (coefficients in the original problem are some horrible expressions):
We're given a linear map $\phi$ with:
$$
\begin{bmatrix}x \\y \\z \\w\end{bmatrix} \mapsto_\phi \begin{bmatrix} 2x + y \\3y + x \\2z + w \\3w + z\end{bmatrix}
$$
This map $\mathbb{R}^4 \to \mathbb{R}^4$ is just two copies of "the same" map combined by "direct product". 
What would be appropriate words for "the same" (something like isomorphic?)  and "combined" here?  I'm particularly interested in following reasoning: 

To compute eigenvectors of $\phi$ it is sufficient to study it's action on $\mathbb{R}^2$ as it's "the same" as action on the second $\mathbb{R}^2$

So stating that they have identical eigenvalues and therefore are isomorphic is the opposite of what I need here. 

A:

Your map is a product of maps: if you have $f:E\to F$ linear and $g:G\to H$ linear, then you have a map $f\times g:E\times G\to F\times H$ given by $(f\times g)(u,v)=(f(u),g(v))$. An eigenvalue of $f$ induces an eigenvalue for $f\times g$ (take an eigenvector $u_\lambda$ for $f$ associated to the eigenvalue $\lambda$, then $(u_\lambda,0)$ is an eigenvector for $f\times g$ associated to the same eigenvalue $\lambda$), and same for $g$. You are in the special case where $E=F=G=H=\mathbb{R}^2$, and $f=g$ the same linear map.