## Determinant of an Exponential

**Point of post:** In this post I prove a neat theorem from matrix analysis three different ways.

There’s a neat little theorem which for some reason, I haven’t heard of until now. I’ll prove it in three different ways. The theorem

**Theorem:** Let be the set of all complex matrices and consider the topology induced by the norm

Then, given a matrix such that

converges, the following formula holds:

As stated we will prove it three ways.

**Proof 1:** By Schur’s Theorem we have that

where is the multiset of eigenvalues, is some unitary matrix, and is an arbitrary, inconsequential number. And so,

And so, for a fixed

which, just adding entrywise, gives

And thus,

but recalling that we’re treating this matrix as really being some -tuple, and we assumed that the limit exists we see that (also remembering that the matrices are “constants”) the above equals

where we’ve disregarded that each of the ‘s is evaluated at the limit, and technically is a “different” , but it turns out it won’t matter (reusing the asterisk for different values is a common occurrences in matrix analysis textbooks). But, evidently we see that the above evaluates to

And so

But,

and thus

But, the determinant of a triangular matrix is the product of it’s diagonal elements. So, finally

**Proof 2:** To cut the length of this already long post I will assume the reader is able to prove the following lemma

**Lemma:** Let then if is the multiset of eigenvalues of , then for any polynomial we have that is the multiset of eigenvalues for

**Proof: **Omitted.

Now, from this we see that if has the multiset of eigenvalues of then for a fixed we see that the multiset of eigenvalues for

is

and so (since the determinant is the product of the eigenvalues of matrix counting multiplicity)

Thus,

But, notice that since

is continuous (it’s a polynomial of the -tuple’s coordinates) we see that

and so by prior discussion

But, since each of the factors in the product converges we may split the limit across multiplication to finally get

as desired.

**Proof 3:** This is similar to **Proof 1** except it doesn’t need a theorem as powerful as Schur’s, but the theorem I’m going to use might actually be a consequence…so haha I don’t know if there’s a point. Oh well..

**Lemma:** Let have the topology described above, then the set of all diagonalizable matrices is dense in .

**Proof: **Omitted.

Now, let , then if is the multiset of eigenvalues for we have that

for some . Thus,

And, so more simply than before, for a fixed we have that

And so

But, as was said before (or similarly) the are “constant” and so by the way we are defining the topology we may pass the limit to the entries of the diagonal matrix. Namely, the above says

Thus,

(where we’ve implicitly used that ). It follows that

and

agree on a dense subset of , but since both are evidently continuous it is a simple topological consequence that they must agree on all of from where the conclusion follows.

thanks alex!

Comment by Yunjiang Jiang | March 31, 2011 |

Of course, friend!

Comment by drexel28 | March 31, 2011 |