Can You Simplify Finding Eigenvalues of an n x n Matrix?

coldstone
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Just wondering is there a way to get the characteristic equation of an n by n matrix without going through tedious calculations of solving multiple determinants of matrices?
 
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I think you're basically stuck using the determinant, but there are many ways to simplify a matrix so that calculating it's determinant is easier.
 
There are many ways
-Hamiltan-Cayley
Find powers of the matrix and find the polynomial they satisfy
-eigenvalues
The terms of the characteristic polynomial are home geneous combinations of the eigenvalues
p=(x-eigenvalue1)(x-eigenvalue2)(x-eigenvalue3)...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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