Vtyjkyuk

for simplicity a specific case:

$f:=(a_i)_i in mathbbN in mathbbR[X]$

$A in mathbbR^4times 4$

In one of our assignments f is evaluated as $sum_i=0^deg(f) a_iA^i$.

How is this evaluation possible formally since $mathbbR$ is not a subring of $mathbbR^4 times 4$? Reference: https://en.wikipedia.org/wiki/Polynomial_ring#Polynomial_evaluation

greets

Ultimately, what matters is that the operations make sense after substitution.

If you take some structure that is not a ring, say the natural numbers $mathbbN$, and you try to compute $f(a)$ for $a$ in $mathbbN$, you probably won't find an integer.

This is why, in the article, the element $a$ has to be part of a ring.

Now, why does $R$ have to contain $K$? This is to allow the multiplication by the $a_i$ coefficients of the polynomial.

Take the ring $M_4(mathbbZ)$, matrices with integers coefficients, and $M$ a matrix in that set. Then, what is $pi M$? It is not part of our ring $M_4(mathbbZ)$.

In our case, you can multiply a matrix that has real coefficients by a real number, and get back a matrix with real coefficients (some call this an algebra).

You can also add two real matrices together, and get back a real matrix. Thus, all the operations make sense.

Another way to answer your question would be to say that $$tilde K = lambda I_4, lambda in mathbbR$$ is isomorphic to $(mathbbR, +, .) = K$. Indeed, the function $$lambda in mathbbR mapsto lambda I_4 in tilde K$$
is a ring isomorphism, mapping $K$ to $tilde K$. So, in a way, $M_4(mathbbR)$ contains $mathbbR$.

Another interesting property to think about is commutativity. Indeed, matrices do not commute in general. However, if you stick to one matrix, and transformations such as $(+, .)$ applied iteratively to the $a_iA^k$, this works just fine.