I am told a matrix A has characteristic polynomial: $(\lambda−1)^3(a\lambda+\lambda^2+b),$ and that $\text {tr}(A)=12,$ and $\det(A) =14.$
I am asked to find the eigenvalues.
Is the only to do this by simply using the fact that the sum of the eigenvalues is the trace, and the product is the determinant?
I had done so, and had gotten that the eigenvalues are 1, 2 and 7, (1 with multiplicity 3), which was correct.
I am asking because won't the number of eigenvalues have to be less than the number of rows and columns of the matrix? What if, if given different trace and determinant values, I somehow found eigenvalues which satisfied the trace and determinant given, but had a number of eigenvalues that exceeded the number of columns and rows of the matrix?
Also, I was not given the dimensions of the matrix A.
Best Answer
I hope I understand everything you said.