Eigenvalues of matrix exponential under multiplicity

Suppose $$A$$ is a possibly complex square $$n \times n$$ matrix. Then $$A$$ has, including multiplicity, $$n$$ eigenvalues $$\lambda_1, \dots, \lambda_n$$. The matrix exponential $$e^{A}$$, also an $$n \times n$$ matrix has eigenvalues $$\mu_1, \dots, \mu_n$$. It is clear, following the definition of the matrix exponential and its absolute convergence, that $$e^{\lambda_i}$$ are included among $$\mu_1, \dots, \mu_n$$.

However, is it also true that the multiplicities are the same? In other words, are $$(\mu_i)$$ and $$(e^{\lambda_i})$$ (slightly abusing notation here), equal as multisets? If so, how can we see that? I guess part of the work is to show that for each $$j$$ there is some $$i$$ such that $$\mu_j = e^{\lambda_i}$$.

• yes, that's true. It follows from the Jordan normal form of matrices and the fact that $A = QJQ^{-1}\implies e^A=Qe^JQ^{-1}$ Commented Sep 17, 2020 at 19:30
• @Exodd would you look at my answer and tell me if this is roughly what you had in mind? Commented Sep 17, 2020 at 19:41
• Note that if you only want the algebraic multiplicities then it "has to be true" that the multiplicities are the same, because 1) they're the same if the matrix is diagonalizable, and and 2) eigenvalues are a continuous function of the matrix, and every matrix can be approximated by diagonalizable matrices. Using JNF proves something stronger and less obvious which is that the geometric multiplicities are the same (and in fact even the Jordan block structure); here there isn't a continuity argument available. Commented Sep 17, 2020 at 19:53
• @QiaochuYuan, yes I was mostly interested in the algebraic multiplicity, but you are right that the JNF proof provides even more insight. The second point (2) is quite a useful intuition. Is it possible to make that type of argument rigorous? If so, I would be very interested in seeing a possible answer if you have the time. Let me know if you think what I wrote for the JNF proof looks OK to you as well. Commented Sep 17, 2020 at 20:44
• @Drew: yes, it can be made completely rigorous, I can write it up. Commented Sep 17, 2020 at 20:50

Quoting my comment:

Note that if you only want the algebraic multiplicities then it "has to be true" that the multiplicities are the same, because 1) they're the same if the matrix is diagonalizable, and and 2) eigenvalues are a continuous function of the matrix, and every matrix can be approximated by diagonalizable matrices.

This argument can be made fully rigorous as follows.

Step 1: Show that the algebraic multiplicities match for diagonalizable matrices, by computing that the exponential of a diagonal matrix $$\text{diag} \{ \lambda_i \}$$ is another diagonal matrix $$\text{diag} \{ \exp(\lambda_i) \}$$.

Step 2: Show that the characteristic polynomial $$\det(\lambda I - \exp(M))$$ of the exponential of a matrix $$M$$ is a continuous function of it (since it's a composition of two continuous functions). This says, concretely, that each coefficient of the characteristic polynomial is a continuous (and even smooth, and even analytic) function of the matrix entries of $$M$$.

Step 3: Show that a matrix $$M$$ can be approximated by a sequence of diagonalizable matrices $$D_i$$ whose eigenvalues converge to those of $$M$$. This can be done most explicitly by using the Jordan normal form of $$M$$ and perturbing its diagonal entries slightly so that they are all distinct.

Step 4: Approximate an arbitrary matrix $$M$$ by diagonalizable matrices $$D_i$$ as above and use the continuity of the characteristic polynomial to argue that, since the characteristic polynomials $$\det(\lambda I - \exp(D_i))$$ converge to $$\det(\lambda I - \exp(M))$$ and since, by construction, the eigenvalues of the $$D_i$$ converge to those of $$M$$, it follows that $$\det(\lambda I - \exp(M)) = \prod_{i=1}^n (\lambda - \exp(\lambda_i))$$ as desired.

The point of writing the argument in this slightly indirect way is to avoid having to prove that the roots of a polynomial are a continuous function of its coefficients; since the roots don't come in a prescribed order, especially over $$\mathbb{C}$$, it's a little tricky to make this rigorous (although it can be made rigorous and it is true), and so it's easier to just argue with the characteristic polynomial instead.

This density argument is a powerful technique and makes precise the intuition that statements like this "have to be true" because they're true for diagonalizable matrices. The same or similar density arguments can prove lots of other facts, for example that if $$A, B$$ are two square matrices then $$AB$$ and $$BA$$ have the same characteristic polynomial (note that they aren't necessarily similar).

This particular density argument shows more generally that if $$f$$ is a holomorphic function then the eigenvalues of $$f(M)$$ are $$f$$ applied to the eigenvalues of $$M$$, with matching algebraic multiplicities; this is a slight enhancement of the spectral mapping theorem for the holomorphic functional calculus.

Following the hint from Exodd, recall that $$A$$ is similar to a Jordan matrix $$J = \bigoplus_{k=1}^{p} J_{n_k}(\lambda_k).$$ Above, $$J_{n_k}(\lambda_k)$$ denotes an $$n_k \times n_k$$-Jordan block corresponding to eigenvalue $$\lambda_k$$. Here $$p$$ is the number of Jordan blocks.

By induction, one can check that for nonnegative integers $$t$$, the matrix $$A^t$$ is similar (under the same similarity transformation) to $$J^t = \bigoplus_{k=1}^{p} J_{n_k}^t(\lambda_k).$$ And consequently if $$T$$ denotes the similarity transformation: $$\exp(A) = \sum_{t\geq 0} \frac{TJ^tT^{-1}}{t!} = T \exp(J) T^{-1}.$$ Stated differently, $$e^{A}$$ is similar to $$e^J$$. Moreover, by direct calculation it is possible to check that: $$\exp(J) = \bigoplus_{k=1}^p \exp(J_{n_k}(\lambda_k)) = \bigoplus_{k=1}^p e^{\lambda_k} M_{n_k},$$ where above for positive integers $$n$$ we define $$M_{n}$$ by: $$M_{n} = \begin{bmatrix} 1 & \tfrac{1}{1!} & \cdots & \tfrac{1}{(n-2)!} & \tfrac{1}{(n-1)!} \\ 0 & 1 & \cdots & \tfrac{1}{(n-3)!} & \tfrac{1}{(n-2)!} \\ 0 & 0 & 1 & \cdots & \tfrac{1}{(n-3)!} \\ \vdots & & \ddots & \ddots & \vdots\\ 0 & \cdots & \cdots & 0 & 1 \\ \end{bmatrix}$$ Therefore, we see that $$e^A$$ is similar to a (block) upper triangular matrix with diagonal: $$(\underbrace{e^{\lambda_1}, \dots, e^{\lambda_1}}_{\mbox{n_1 times}}, \underbrace{e^{\lambda_2}, \dots, e^{\lambda_2}}_{\mbox{n_2 times}}, \ldots, \underbrace{e^{\lambda_p}, \dots, e^{\lambda_p}}_{\mbox{n_p times}})$$ And so the claim about multiplicity follows.

• Your last step is off: it's not true that the exponential of a Jordan block is literally another Jordan block of the same size. However, it is conjugate to another Jordan block of the same size. And in any case it's upper triangular with $\exp(\lambda_k)$ on the diagonals which is enough to conclude about the algebraic multiplicities. Commented Sep 17, 2020 at 20:50
• Oops. I didn't actually check this. I will try a computation by hand and edit accordingly. Commented Sep 18, 2020 at 6:10
• @QiaochuYuan Does this look better to you? (I also realized that my previous claim about similarity was overly strong, of course, can have multiple Jordan blocks with the same eigenvalues!) Commented Sep 18, 2020 at 16:38