Why can an algebraic integer (in $\mathbb{C}$ integer over $\mathbb{Z}$) be defined as an eigenvalue of some matrix with integer coefficients?
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What do you mean by eigenvalue of some polynomial? – Sungjin Kim Aug 06 '16 at 03:27
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1what is an eigenvalue of a polynomial? – Sarvesh Ravichandran Iyer Aug 06 '16 at 03:27
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Forgive me for such mistake :P I was absentminded while writing, edited anyways.. – NTlearner95 Aug 06 '16 at 03:28
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1For this standard result see e.g. S. Fallat, Algebraic integers and tensor products of matrices, or Robin Chapman's Notes on algebraic integers or many other expositions that can be found with obvious web searches. – Bill Dubuque Aug 06 '16 at 03:41
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1actually, the integer is best thought of as the companion matrix itself. See http://math.stackexchange.com/questions/1883573/finding-units-of-mathbbz-sqrt33/1883652#1883652 and http://math.stackexchange.com/questions/1881822/show-that-19-5-sqrt32-8-sqrt34-is-a-unit-in-mathbbz-sqrt32/1881834#1881834 – Will Jagy Aug 06 '16 at 03:46
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See also this answer. – Bill Dubuque Aug 06 '16 at 03:48
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An algebraic integer is generally defined as a root of a monic polynomial in $\mathbb{Z}[x]$. If $A$ is a matrix with integer entries, then $\det(A-\lambda I)$ is a monic polynomial in $\mathbb{Z}$. This shows that any eigenvalue of an integer matrix is an algebraic integer.
To get the converse, you need only show that any such polynomial is the characteristic polynomial of some such matrix. Do you know how to show that part?
G Tony Jacobs
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1Ohh okay, now I can do it. Its just opposite thing of getting rational canonical form. – NTlearner95 Aug 06 '16 at 03:54