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

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

Σkaj,kbk,j

The trace of AB is then

ΣjΣkaj,kbk,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

Σjbk,jbk,jaj,k

and the trace of BA is

ΣkΣjbk,jbk,jaj,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:

ΣjΣkaj,kbk,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-1MV = Λ

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

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

tr(Λ) = tr(V-1MV) = tr((V-1M)V) = tr((V·V-1)M) = tr(I·M) = tr(M)

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-1AV = 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

tr(J) = tr(V-1AV) = tr(VV-1A) = tr(I·A) = tr(A)

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

tr(A) = Σλj


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