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For any field $k$, let $M_n(k)$ denote the ring of $n\times n$ matrices over $k$. Out of curiosity, do we know a way to relate polynomials (especially irreducible polynomials) in $k[x]$ with polynomials in $M_n(k)[x]$?

What about roots of polynomials in $M_n(k)[x]$? If a diagonalizable matrix $A$ is a root of a polynomial $f$ in $M_n(k)[x]$, then it's eigenvalues are roots of polynomials in $k[x]$ derived from $f$. Can we say more?

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    What is the process of deriving polynomials in $k[x]$ from those in $M_n(k)[x]$? I'm unfamiliar with such processes. – 2017-02-15
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    Also there are problems like the following. All the matrices of the form $$\pmatrix{0&a\cr-a^{-1}&0\cr}$$ are zeros of the polynomial $x^2+1$. So that quadratic has (potentially infinitely) many zeros in $M_2(k)$. Consequently the same quadratic has infinitely many different factorizations into a product of linear polynomials. – 2017-02-15
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    @JyrkiLahtonen I thought that, given for instance a root $A$ of $f \in M_n(k)$, if $A = P^{-1}DP$ where $D$ is diagonal, then $D$ is also a root of $f$. For every $X \in M_n(k)$, $f(X)$ is also a matrix in $M_n(k)$, so $f(D) = 0$ would give me $n^2$ polynomial equations over $k$. Also, by $x^2 + 1$ you mean $x^2 + \text{Id}$ where $\text{Id}$ is the identity matrix, right? Then I guess so, the bad behavior of $M_n(k)$ as a ring really messes up the behavior of $M_n(k)[x]$. – 2017-02-15

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Your first question is not reasonable.

It does not make sense to talk about irreducible polynomials over $M_n(k)$, the reason being that $M_n(k)$ is not factorial and so is $M_n(k)[x]$. Even worst, it does not make sense to talk about divisibility in $M_n(k)[x]$, since $M_n(k)$ is not an integral domain, which implies that $M_n(k)[x]$ is not an integral domain too.