Editing Characteristic polynomial (section)

==Characteristic polynomial of a product of two matrices==

If <math>A</math> and <math>B</math> are two square <math>n \times n</math> matrices then characteristic polynomials of <math>AB</math> and <math>BA</math> coincide: 
<math display=block>p_{AB}(t)=p_{BA}(t).\,</math>

When <math>A</math> is [[Non-singular matrix|non-singular]] this result follows from the fact that <math>AB</math> and <math>BA</math> are [[Similar matrices|similar]]:
<math display=block>BA = A^{-1} (AB) A.</math>

For the case where both <math>A</math> and <math>B</math> are singular, the desired identity is an equality between polynomials in <math>t</math> and the coefficients of the matrices. Thus, to prove this equality, it suffices to prove that it is verified on a non-empty [[open subset]] (for the usual [[Topological space|topology]], or, more generally, for the [[Zariski topology]]) of the space of all the coefficients. As the non-singular matrices form such an open subset of the space of all matrices, this proves the result.

More generally, if <math>A</math> is a matrix of order <math>m \times n</math> and <math>B</math> is a matrix of order <math>n \times m,</math> then <math>AB</math> is <math>m \times m</math> and <math>BA</math> is <math>n \times n</math> matrix, and one has
<math display=block>p_{BA}(t) = t^{n-m} p_{AB}(t).\,</math>

To prove this, one may suppose <math>n > m,</math> by exchanging, if needed, <math>A</math> and <math>B.</math> Then, by bordering <math>A</math> on the bottom by <math>n - m</math> rows of zeros, and <math>B</math> on the right, by, <math>n - m</math> columns of zeros, one gets two <math>n \times n</math> matrices <math>A^{\prime}</math> and <math>B^{\prime}</math> such that <math>B^{\prime}A^{\prime} = BA</math> and <math>A^{\prime}B^{\prime}</math> is equal to <math>AB</math> bordered by <math>n - m</math> rows and columns of zeros. The result follows from the case of square matrices, by comparing the characteristic polynomials of <math>A^{\prime}B^{\prime}</math> and <math>AB.</math>