Download The Exponential Function. The function eA = An/n! is defined for all

The Exponential Function.
The function eA =
∞
X
An /n! is defined for all (square) matrices A. We first apply this
n−0
function to systems of linear differential equations. Consider the differential equation
dx
= Ax; x(0) = x0
dt
(1)
where x0 is a constant column vector. We can find a solution to this equation immediately
d
as follows. We have etA = AeAt . This can be verified directly by using the infinite series
dt
for etA . Therefore, setting x(t) = etA x0 , we have dx/dt = AetA x0 = Ax(t). Further, the
initial condition is x(0) = eO x0 = Ix0 = x0 . We don’t even have to refer to a uniqueness
theorem to prove that this is the unique solution. For if x(t) satisfies (1) then
d(e−tA x)
dx
= e−tA
− Ae−tA x = e−tA Ax − Ae−tA x = (e−tA A − Ae−tA )x = 0
dt
dt
Therefore e−tA x is constant. Since its value at t = 0 is x0 , we have e−tA x = x0 and so
x(t) = etA x0 . (Note: AetA = etA A, since this is the substitution of A for z in the equality
zetz = etz z. This substitution principle is used throughout.)1
At this point, we mention some simple properties of the exponential all of which follow
simply from its definition as a power series.
∗
1. (eA )∗ = eA .
−1
2. If A ∼ B then eA ∼ eB . In fact Q−1 eA Q = eQ AQ .
3. If A is a diagonal matrix with diagonal entries λ1 , . . . , λN , then eA is a diagonal matrix
with diagonal entries eλ1 , . . . , eλN .
4. As in 3, if A is an upper diagonal matrix.
For fixed matrix A, the function f (t) = etA maps the real numbers into the group of nonsingular matrices. This group is called the General Linear Group, and is denoted GLn . etA is
non-singular since etA e−tA = eO = I so e−tA is the inverse of etA . The non-singular matrices
form a group since they are closed under multiplication and taking inverses. The function
f (t) = etA is a group homomorphism mapping the additive group of real numbers into the
multiplicative group GLn :
F : R(+) → GLn : f (t + s) = f (t)f (s)
f (t) (or the image of f (t)) is called a one parameter subgroup of GLn . (The one parameter
is the variable t.)
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But beware of substituting two non-commuting matrices into an equality. Even the simple (z + w)2 =
z + 2zw + w2 fails unless z and w commute. Thus, we don’t have (A + B)2 = A2 + 2AB + B 2 unless
AB = BA. Similarly, we don’t expect eA+B = eA eB unless A and B commute.
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Some other groups of matrices (the classical groups) are:
1. SLn : The Special Linear Group: matrices of determinant 1.
2. Un : The Unitary Group: all unitary matrices U .
3. SUn : The Special Unitary Group: all unitary matrices U with det(U ) = 1.
4. On : The Orthogonal group.
5. SOn : The Special Orthogonal group: the orthogonal matrices whose determinant is 1.
Note that the adjective “special” indicates that we restrict ourselves to matrices of determinant 1. Also GLn and SLn come in two flavors: the real and complex variety.
We have seen that eA maps all matrices into GLn . We claim that this map, restricted to
some neighborhood of O is 1-1 and onto a neighborhood of I in GLn . To see this, consider
the power series for log(1 + z).
∞
X
log(1 + z) = z − z 2 /2 + z 3 /3 + · · · =
(−1)n+1 z n /n, |z| < 1
n=1
setting 1 + z = w, we have the following power series in w − 1:
log w = (w − 1) − (w − 1)2 /2 + (w − 1)3 /3 + · · · =
∞
X
(−1)n+1 (w − 1)n /n, |(w − 1)| < 1
n=1
Therefore, we can define log(A) =
∞
X
(−1)n+1 (A − I)n /n for all matrices A in the ball
n=1
kA − Ik < 1. We have the identity elog z = z for |z − 1| < 1, and log(ez ) = z for sufficiently
small2 z. We have the same for matrices: elog(A) = A for kA − Ik < 1. Similarly, log(eA ) = A
for A sufficiently small. This shows that eA for A near O and log A for A near I are inverses.
This analysis shows that the one parameter subgroups of GLn , for the various values of A,
cover some neighborhood of I in GLn .
How can we find one parameter subgroups of the other classical groups defined above? We
start with the following striking theorem:
det(eA ) = etrace(A)
(2)
To prove (2), we know that A is similar to an upper diagonal matrix B. If the eigenvalues of
A are λ1 , . . . , λN , then these are the diagonal elements of B. Therefore the diagonal elements
of eB are, as noted above, eλ1 , . . . , eλN . Thus,
det(A) = det(B) = eλ1 · · · eλN = eλ1 +···+λN = etrace(B) = etrace(A)
This proves the result.
2
For example, |z| < 1/2.
2
Theorem: If trace(A) = 0, then f (t) = etA is a one parameter subgroup of SLn . Conversely,
if f (t) = etA is a one parameter subgroup of SLn then trace(A) = 0.
Proof: If trace(A) = 0, then det(f (t)) = det(etA ) = etrace(tA) = et trace(A) = e0 = 1. So f (t)
is in SLn . Conversely, if f (t) = etA is in SLn , then det(etA ) = 1, so etrace(tA) = et trace(A) = 1
for all t. Take derivatives to get trace(A) et trace(A) = 0. So trace(A) = 0.
In a similar way, we can show
Theorem: If A is real and skew symmetric: A∗ = −A, then f (t) = etA is a one parameter
subgroup of SOn . Conversely, if f (t) = etA is a one parameter subgroup of On then A∗ = −A.
∗
Proof: Let A∗ = −A, and f (t) = etA . Then f (t)∗ = (etA )∗ = etA = e−tA = f (t)−1 . So f is
orthogonal. Since A∗ = −A, we also have trace(A) = 0, so det(f (t)) = 1 and f (t) ∈ SOn .
The converse result is proven similarly to the above proof, and is left to the reader.
Note that any homomorphism f (t) of R(+) into On automatically has its determinant 1 for
all values of t. This is because f (0) = I, so the determinant of f is 1 at t = 0. Since the
determinant of f is continuous, and since, as an orthogonal matrix, its determinant if 1 or
−1, it follows that det(f (t)) = 1 for all values of t.
∗
The situation for Un is similar, since in this case we have (eA )∗ = eA . So it is an easy matter
to show
Theorem: If A is skew hermitian: A∗ = −A, then f (t) = etA is a one parameter subgroup
of Un . Conversely, if f (t) = etA is a one parameter subgroup of Un then A∗ = −A.
The proof is similar to the above proof. But in this case, the mapping need not be into SUn ,
since the trace of a skew-hermitian matrix need not be 0. In general, if A∗ = −A, then taking
traces, we find trace(A) = −trace(A). Therefore, the trace of a skew-hermitian matrix is
pure imaginary. In fact, for a skew-hermitian matrix A, any diagonal element aii satisfies
aii + aii = 0. Thus, each diagonal element is pure imaginary. In order for f (t) = etA to be a
one parameter subgroup of SUn , we need A to be skew-hermitian and to have traceA = 0.
This follows from the above results.
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