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The Theorem that the Sum of the Eigenvalues of a Matrix is Equal to its Trace

Theorem:
Let A and B be two complex-valued matrices of dimensions n×m and m×n, respectively. Then both AB and BA are defined and

tr(AB) = tr(BA)

Proof:
The element in the j-th row, j-th column of AB is given by

Σ_{k}a_{j,k}b_{k,j}

The trace of AB is then

Σ_{j}Σ_{k}a_{j,k}b_{k,j} where k runs from 1 to m and j runs from 1 to n.

On the other hand, the element in the k-th row, k-th column of BA is

Σ_{j}b_{k,j}b_{k,j}a_{j,k}

and the trace of BA is

Σ_{k}Σ_{j}b_{k,j}b_{k,j}a_{j,k}

But the order of summation can be reversed and also the order of the terms in the product can be reversed.
When this is done the result is:

Σ_{j}Σ_{k}a_{j,k}b_{k,j} where k runs from 1 to m and j runs from 1 to n.

This is exactly the same as the expression for the trace of AB.

The Trace of a Matrix and Its Eigenvalues

First a simple version of the proposition will be considered. It is not necessary to consider this case separately but it makes the
proof of the theorem easier to absorb. Let M be a complex-valued n×n matrix that is diagonalizable; i.e., there exists V such that

V^{-1}MV = Λ

where Λ is the diagonal matrix of the eigenvalues of M.

Now consider V^{-1}MV as (V^{-1}M)V. Then the previous theorem applies
so

Thus the sum of the eigenvalues of a diagonalizable matrix is equal to its trace. A matrix M is diagonalizable if all of
its eigenvalues are different; i.e., the multiplicity of every eigenvalue is 1.

When the multiplicities of some of a matrix's eigenvalues of greater than 1 it is not diagonalizable but instead for any matrix A there exists
an invertible matrix V such that

V^{-1}AV = J

where J is of the canonical Jordan form, which has the eigenvalues of the matrix on the principal diagonal
and elements of 1 or 0 mext to the principal diagonal on the right and zeroes everywhere else.

As in the simple case of diagonalizable matrices considered before

The canonical Jordan form matrix J has the eigenvalues of A on its principal diagonal so the trace of J is equal to the sum of the
eigenvalues of A.
Thus