I have shown the easy case, when A is diagonalizable.

But I am stuck on the case when A is not.

So, I put A in its Jordan canonical form. Then say A = $SJS^{-1}$.

Then $e^A = Se^JS^{-1}$,

where $e^J$ is an upper triangular matrix with the Jordan blocks exponentiated.

Now for each Jordan block, I have that

$$e^{J_i} = e^{\lambda_i I + N}$$

where N is elementary nilpotent. But since scalar matrices commute with all matrices, it commutes with N.

Then $$e^{J_i} = e^{\lambda_i I + N} = e^{\lambda_i}e^{N}$$ where $e^{\lambda_i}$ is a diagonal matrix with non-zero diagonal, hence it is invertible / has non-zero determinant. What can I do with the $e^N$ factor? I know that it has a finite expansion, since it is nilpotent:

$$e^N = I + N + ... + \frac {N^k}{k!}$$

I'm not sure where to go from here...


  • 5
    $\begingroup$ If you want to avoid Jordan normal form altogether, you might try to prove directly that $\exp(A)\exp(-A) = I$ for any square complex matrix $A$ ( note that the powers of $A$ all commute with each other). $\endgroup$ Commented Oct 15, 2015 at 11:50
  • $\begingroup$ Very pretty, @GeoffRobinson -- thanks so much :-) $\endgroup$
    – User001
    Commented Oct 16, 2015 at 1:36

3 Answers 3


The simpler way to shows that $e^A$ is invertible is to note that, for commuting matrices $A,B$ we have, from the definition of the exponential, $e^{A+B}=e^Ae^B$. So, since $A$ and $-A$ commute, we have: $$ e^Ae^{-A}=e^{A-A}=e^O=I $$

Also, using Jacobi's formula (see here) we can find: $$ \det(e^A)=e^{Tr\;(A)} $$ that confirms the invertibility of $e^A$.


The eigenvalues $\lambda_i$ become $e^{\lambda_i}$ after matrix exponential and the exponential function has no zeros. Therefore any matrix over an algebraically closed field will have a matrix exponential full of non-zero eigenvalues and we don't even need to use Jordan.

To clarify, we just need to use definition of eigenvalue and the fact that degeneracy (non-invertibility) is the same as a one or more eigenvalues equal to 0. The eigenvalues are the roots to the characteristic polynomial which if matrix is under algebraically closed field must have as many of them as it's size. So there will always be $n$ eigenvalues regardless of diagonalization or jordan form and at least 1 of those needs to be 0 to make the matrix degenerate.

Important note: This only works when working over any field which the exponential function has no zeroes. I don't know if there may exist any fields where the exponential function can actually give 0.

  • $\begingroup$ A shorter variant on my argument. Nice! $\endgroup$
    – Ian
    Commented Oct 15, 2015 at 11:55
  • 1
    $\begingroup$ One has to say what one means by exponential function over a general field; there is a standard notion of this, and its definition precludes its image containing zero: en.wikipedia.org/wiki/Exponential_field (Nice answer, b.t.w., +1.) $\endgroup$ Commented Oct 15, 2015 at 19:13
  • $\begingroup$ Hi @mathreadler, how do you get $e^{\lambda_i}$ after matrix exponential, without using the Jordan form and without the assumption that A is diagonalizable (which would be trivial to see in this case)? Maybe I am overlooking something simple, but I don't really see it...thanks, $\endgroup$
    – User001
    Commented Oct 16, 2015 at 1:39
  • $\begingroup$ Hi James. I added some explanation. In short: the eigenvalues are the roots to the characteristic polynomial. If matrix is over algebraically closed field we will always have $n$ roots. Applying a function to a matrix also applies it to the eigenvalues and the exponential function never gives 0 no matter the input. $\endgroup$ Commented Oct 16, 2015 at 6:35

As $e^N$ has only $1$s on its diagonal (the $\sum_{k\ge 1} N^k/k!$ part is strictly upper diagonal), $e^{J_i}$ has only $e^{\lambda_i}$ on its diagonal. We therefore have $$ \det(e^{J_i}) = e^{k_i\lambda_i} $$ where $k_i$ is the size of $J_i$, this gives $$ \det(e^A) = \det S^{-1}\prod_i \det(e^{J_i}) \det S = e^{\sum_i k_i \lambda_i} = e^{\mathrm{tr}\, A} $$ As $0 \not\in \exp(\mathbf C)$, we have that $\det(e^A) = e^{\text{something}} \ne 0$.

  • $\begingroup$ Hi @Martini, how do you know that the factor $e^N$ has only 1s on its diagonal? I am guessing you are saying that because N is strictly upper triangular, then it has all zeroes on its diagonal, and when exponentiated, the zeros become 1s on the diagonal. But how does that exponentiation work? We know that exponentiation of a diagonal matrix is just the matrix with the diagonal entries exponentiated, but I don't know of such a rule for upper triangular matrices. Thanks, $\endgroup$
    – User001
    Commented Oct 16, 2015 at 1:43
  • $\begingroup$ I also tried expanding out the $e^N$ in power series...and still don't see why the diagonal has to be all 1s...thanks @martini $\endgroup$
    – User001
    Commented Oct 16, 2015 at 1:45

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