The exponential image of real matrices dedicated to the memory of my father,

Giovanni

Andrea Previtali

Como, 13 December 2002

Exponential power series on ℂ

As is well known exp as a complex function may be defined as a power series

exp(z) := ∑_{j=0}^{∞} \frac{z^{j}}{j!} .

Moreover an inverse function is defined in a neighborhood of 1:

log(1 − z) = − ∑_{j=1}^{∞} \frac{z^{j}}{j} ,

for any z ∈ ℂ, |z| < 1. Be careful, since exp has period 2πi, it can not be globally invertible!

… and on (K)_{n}

Given any field K, we would like to define an exponential map from (K)_{n}, the full matrix algebra over K of degree n, using the former power series.

Warm up: Assume char(K) = p > 0, then exp(Z), Z ∈ (K)_{n}, is defined iff Z^{p} = 0 (we can not divide by p! in K!) Hence Z must be nilpotent and

exp(Z) = ∑_{j=0}^{p−1} \frac{Z^{j}}{j!}

is a finite sum.

Conjugation in GL_{n}(K)

This is an example of a unipotent matrix, namely a matrix U such that U − I is a nilpotent matrix N. Conversely any such matrix admits a logarithm:

log(I + N) = − ∑_{j=1}^{p−1} \frac{(−N)^{j}}{j} .

Definition 1. Let GL_{n}(K) be the group of invertible matrices in (K)_{n}.

In particular GL_{1}(K) = K^{∗}. Set A^{T} = T^{−1}AT. Remark 2. For any A ∈ (K)_{n} and T ∈ GL_{n}(K), A = exp(B), for some B ∈ (K)_{n} iff A^{T} = exp(B^{T}).

Jordan Normal Form

Thus we may consider the equation exp(B) = A only when A is chosen up to conjugation. For example if K is algebraically closed we may assume: Theorem 3 (C. Jordan, 1870). Any A ∈ (K)_{n}, K an algebraically closed field, is conjugate in GL_{n}(K) to the diagonal sum of blocks λI + N, where λ ∈ K and N is the nilpotent matrix with all elements in position (i, i + 1) equal to 1 and zero elsewhere.

What about a non-algebraically closed field, e. g. ℝ? Fortunately there exists a so called rational canonical form, but we can do better and single out the two features of λI and N as above.

Jordan-Chevalley decomposition

Definition 4. We say that S ∈ (K)_{n} is semisimple, if S^{T} is diagonal for some matrix T ∈ GL_{n}(\overline{K}), where \overline{K} is the algebraic closure of K. Theorem 5 (Additive Jordan-Chevalley decomposition). For any A ∈ (K)_{n}:

1. there exist uniquely determined D semisimple and N nilpotent such that A = D + N and DN = ND;

2. there exist p, q ∈ K[x], such that p(0) = q(0), and D = p(A), N = q(A).

Definition 6. We call D and N the semisimple and nilpotent part of A.

Multiplicative version

Theorem 7 (Multiplicative Jordan-Chevalley decomposition). For any B ∈ GL_{n}(K):

1. there exist uniquely determined S semisimple and U unipotent such that B = SU = US;

2. S, U commute with any matrix commuting with A.

Definition 8. We call D and N the semisimple and unipotent part of A.

The additive form may be obtained using the euclidean algorithm in K[x]. The multiplicative version follows from the additive one observing that S := D is invertible and setting U = I + D^{−1}N.

Baker-Campbell-Hausdorff’s formula

Definition 9. Given A, B ∈ (K)_{n}, define [A, B] = AB − BA, as the bracket of A and B.

From now on we stick to fields of characteristic 0, e. g. ℝ and ℂ, the favorite fields for an analyst. Theorem 10 (Baker-Campbell-Hausdorff, 1907). For A, B ∈ (K)_{n}, exp(C) = exp(A) exp(B), where

C = A+B+\frac{[A, B]}{2}+\frac{[[A, B], A]}{12}+\frac{[[A, B], B]}{12}+C_{4},

C_{4} being the sum of terms obtained by bracketing A, B in some order more than three times.

Invertible Squares

Definition 11. We say that A ∈ (K)_{n} is divisible if there exists B ∈ (K)_{n} such that B^{m} = A, for any m ∈ ℕ. Remark 12. In particular

exp(A + B) = exp(A) exp(B),

whenever A and B commute. Thus I = exp(A − A) = exp(A) exp(−A) and exp(A) = exp(\frac{A}{m})^{m}, for any positive integer m.

Hence any exponential matrix is invertible and divisible, in particular it is a square. Example 13. exp(ℝ) = ℝ^{+} = (ℝ^{∗})^{2}

Claim

Exponential matrices coincide with invertible squares

Separation of Variables

We show that

B = exp(A)  (1)

admits a solution A iff the semisimple and unipotent parts S and U of B are exponential matrices. Indeed let A = D + N, then B = exp(D) exp(N), since DN = ND. But D semisimple forces exp(D) to be such and N nilpotent implies exp(N) to be unipotent, moreover exp(D) and exp(N) commute. Thus our equation is equivalent to

S = exp(D) and U = exp(N),  (2)

since the semisimple and unipotent part of any invertible matrix are uniquely determined.

The unipotent part

Now the logarithmic map shows that U = exp(N) has always a solution

N = − ∑_{i=1}^{∞} \frac{(I − U)^{i}}{i} ,

a finite sum since I − U is nilpotent.

… and the semisimple one

So we should only look at the semisimple parts. We are now in a position to prove Theorem 14. exp(ℂ)_{n} = GL_{n}(ℂ).

In fact up to conjugation we may assume that D is diagonal. Since B is invertible, the same holds for S. But any diagonal element of S, say s, is non-zero, hence there is d such that s = exp(d) and S = exp(D) is soluble.

The Real case

Unfortunately not every semisimple real matrix is diagonalizable in ℝ. We may assume that any solution D (if any) to S = exp(D) is the diagonal sum of blocks of size 1 or 2 according as the corresponding factor of the characteristic polynomial χ_{D}(x) of D has degree 1 or 2. If the block equals (d), then there exists d ∈ ℝ such that s = exp(d) iff s ∈ ℝ^{+}.

Block of Size 2

Otherwise we may assume our block is

d = ( \begin{matrix}a & b \\ −b & a \end{matrix} ) ,

for real a, b. Now exp(d) has the same shape as d and either has two non-real complex conjugates eigenvalues or is a scalar real matrix. Definition 15. Given S ∈ (K)_{n} denote with σ(S) the set of eigenvalues of S, its spectrum. This is the set of zeros of χ_{S}(x) = det(xI − S), the characteristic polynomial. We call m(λ), the highest power of (x − λ) dividing χ_{S}(x), the algebraic multiplicity of λ.

Even Multiplicity

For any real matrix B, σ(B) = \overline{σ(S)}, that is the spectrum of a real matrix is invariant under conjugation. In our situation σ(B) = σ(S) contains negative real elements only with even multiplicity. Conversely assume that S is a semisimple real matrix all of whose eigenvalues are complex conjugate and for λ ∈ σ(S) ∩ ℝ^{−}, m(λ) is even. Then S = exp(D) admits a solution in (ℝ)_{n}.

Squares

We now characterize invertible squares in (ℝ)_{n}. As before

B = A^{2}  (3)

admits a solution over ℝ for a given B ∈ GL_{n}(ℝ) iff

S = D^{2} and U = V^{2},

where S, D, and U, V are the semisimple and unipotent parts of B, A, respectively. Now the Binomial Theorem

\sqrt{1 + x} = ∑_{j=0}^{∞} \frac{1}{2}jx^{j}

evaluated in x = U − I, shows that V is always expressible as a finite sum in powers of U − I.

Main Characterization

Therefore our equation has solutions iff S = D^{2} has. As before we may assume that D is a diagonal sum of blocks of size 1 or 2.In the first case s = d^{2} admits real solutions iff s > 0. For size 2, we may

assume s = ( \begin{matrix}a & b \\ −b & a \end{matrix} ). Let (c + id)^{2} = a + ib,

then

s = ( \begin{matrix}c & d \\ −d & c \end{matrix} )^{2} .

Since squaring is a surjective map over ℂ, we see that any block may be obtained this way. We have proved the following: Theorem 16. Let B ∈ GL_{n}(ℝ)_{n}, then the following are equivalent:

1. B = exp(A), for some A ∈ (ℝ)_{n};

2. B = C^{2}, for some C ∈ GL_{n}(ℝ)_{n};

3. If λ ∈ σ(B) ∩ ℝ^{−}, then m(λ) ≡_{2} 0.

Bijective Maps

We conclude with a criterion for bijectivity of maps among nilpotent matrices. Lemma 17. Given a nilpotent algebra 𝒥, consider Δ and E two operators on 𝒥 such that

i) Δ is invertible and Δ(𝒥^{i}) ⊆ 𝒥^{i}, for any i;

ii) E(𝒥^{i}) ⊆ 𝒥^{i+1}, for any i.

Then Δ + E is an invertible operator on 𝒥.

Proof. From the assumptions (EΔ^{−1})^{n}(𝒥) ⊆ 𝒥^{n} = 0, for a suitable n, hence (1 + EΔ^{−1}) is invertible and the same holds for E + Δ . □