Is the characteristic polynomial $X_A(t)$ of the matrix $A$ as a polynomial in the coefficients $a_{ij}$ of $A$ and the unknown $t$ irreducible in $\mathbb{Q}[a_{11},..,a_{nn},t]$? If so what is the proof for this?
Is the characteristic polynomial as a polynomial in the coefficients and x always irreducible?
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linear-algebra
abstract-algebra
irreducible-polynomials
1 Answers
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If for a given $n$, $X_A(t)$ was reducible in $\mathbb{Q}[a_{11},\ldots,a_{nn},t]$, then for any chosen rational values of $a_{11},\ldots,a_{nn}$, the new $X_A(t)$ would be reducible in $\mathbb{Q}[t]$.
Let f be any irreducible monic polynomial of degree $n$ in $Q[t]$, and let $A$ be the companion matrix for $f$. Then $X_A(t) = f$, which, by choice, is irreducible.
It follows that if $a_{11},\ldots,a_{nn}$ are unknowns, $X_A(t)$ is irreducible in $\mathbb{Q}[a_{11},\ldots,a_{nn},t]$.
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0I understand the first and second sentence. But your conclusion I do not understand. – 2017-01-06
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0it is a proof by contradiction, a matrix which has an irreducible characteristic polynomial exists for all $n$. therefore the polynomial which has the entries as unknowns must be irreducible. – 2017-01-06
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0Ok, Thanks! Now I got it! – 2017-01-06