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Is every monic real polynomial the characteristic polynomial of some real square matrix? I strongly suspect so, but I couldn't find a proof.

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The matrix you are looking for is called the companion matrix of the polynomial. (The proof that the characteristic polynomial of the companion matrix of $f$ is $f$ is easy.)

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The canonical answer is the companion matrix as Rob Arthan said, and it works verbatim over any field or even commutative ring.

Yet, as you ask for the reals specifically let me offer a different answer.

For the complex numbers, it is possible to factor the polynomial as a product $\prod_{i=1}^n (X - \lambda_i)$, and then one can take the diagonal matrix with $\lambda_1 , \dots, \lambda_n$ on the diagonal.

For the real numbers, this does not work, but it is still possible to write the polynomial as $\prod_{i=1}^s (X^2 + \alpha_i X + \beta_i) \prod_{i=1}^r (X-\gamma_i)$ with $2s + r = n$ and reals $a_i,b_i,c_i$.

For the $\prod_{i=1}^r (X-c_i)$ you could do the same as before, that is put the $c_i$ on the diagonal. And if you could find a $2 \times 2$ matrix with characteristic polynomial $X^2 + a_i X + b_i$ then you could put those $2 \times 2$ blocks on the diagonal too.

Thus, your task is reduced to the $2 \times 2$ case. There you could recall that the characteristic polynomial of $$\pmatrix{a & b \\ c & d }$$ is $X^2 - (a+d)X + (ad - bc)$. Then what remains to do is finding a solution to $$-a -d = \alpha_i$$ $$(ad - bc) = \beta_i$$ There are varied ways to achieve this, a neat one is to decide to try $a=0$ and $c=1$, then $d = - \alpha_i$ and $b = - \beta_i$. Looking back we see this is the $2\times 2$ companion matrix.

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You can make partial attempts. Look at some special kind of matrices for which it is easy to find the characteristic polynomials. If eigenvalues are easy to find then the characteristic polynomial will be easy too. Inspect those polynomials and see what kind of polynomials are realizable as characteristic polynomials.

First see matrices $n\times n$ with $n$ eigen values real. What is the characteristic polynomial.

See $2\times2$ polynomials with eigenvalues conjugate complex numbers. In both cases characteristic polynomials are readily written down. When conceptual proof eludes it is instructive to experiment with as many special cases as possible.

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