Let [A] be the square matrix (n*n)
The eigenvalues (call a generic one lambda) are such that
[A][X]=lambda[X] where [X] is some n-vector. Let [I] be the n*n identity matrix. Then lambda [X]= lambda [I][X] and the eigenvalues verify the matrix equation
[A-lambda I][X]=0
The eigenvalues are the roots of the polynomial equation obtained by setting det([A-lambda I])=0. Solving this n-degree polynomial equation will give you the eigenvalues.
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