1. If A and B are square matrices of the same size, then the spectrums of AB and BA are the same. More precisely, the generalized eigenvalues of AB and BA over the complex numbers occur with the same multiplicity. This is clear when A is invertible, because then AB and BA are similar. If A is not invertible, then we can approximate A by invertible matrices. The spectrum of AB is just the zeroes of det(xI-AB) (counted with multiplicities). The claim then follows easily. This generalizes the fact that the traces and determinants of AB and BA are the same.
2. If A and B are real square matrices of the same size and if there is a complex matrix U such that
3. This is more or less for my own benefit, to recall the following proof of the well-known fact that any normal matrix is diagonalizable over C. Key: Assume A is normal, i.e. A commutes with A*. Then if x is an eigenvector of A, x is also an eigenvector of A*. In fact if Ax = tx, then
when A commutes with A*. This allows us to proceed by induction on the dimension of the vector space: first pick any eigenvector x of A. Then its orthogonal complement is preserved by A, because x is also an eigenvector of A*. Hence we can reduce to one dimension lower, and by induction this completes the proof.
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Nice post Polam! In (1), you prove that AB and BA have the same characteristic polynomials by approximating A with a sequence of invertible matrices. I think there is another way of proving this without using limit. Simply observe that the characteristic polynomial of A is
\det(A-tI)=\sum_{i=1}^n(-1)^i\text{tr}(\bigwedge^{n-i}A)t^i
where \bigwedge^i A:\bigwedge^i C\to\bigwedge^i C is given by
\bigwedge^i A(z_1\wedge\cdots\wedge z_i)=Az_1\wedge\cdots\wedge Az_i
and \bigwedge^i(AB)=\bigwedge^i A\bigwedge^i B. So the equality of the characteristic polynomials of AB and BA boils down to the fact that tr(CD)=tr(DC).
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