Belonging to 1 and 1, respectively, to obtain the solutions

Theorem 3.13 Let A and B be n × n matrices that commute: AB = BA. Then _e(A+B)t _{= e}At_eBt_.
The proof is an exercise.



∈
Corollary 3.5.1 If A is a square matrix, then for all t R, the matrix e^At is nonsingular.



—
PROOF: The matrices A and A commute; therefore e^Ate^—At = e^(A—A)t = e^Zt, where Z is the zero matrix. Since e^Zt = I it follows that e^—At is the in- verse matrix of e^At. A matrix is nonsingular iff it is invertible; hence e^At is
nonsingular.


Theorem 3.14 Let A be a square matrix whose entries are constants. Then e^At is a fundamental matrix solution of →x^′ = A→x.



· · ·
The proof depends on the chain rule applied to matrix polynomials. If f (x) = c₀ + c₁x + c₂x² + c_kx^k is an ordinary polynomial, f (At) denotes the matrix


f (At) = c₀ I + c₁ A t + c₂ A² t² + · · · + c_k A^kt^k.
Lemma 3.5.2 Let f ^′(x) denote the derivative of f with respect to x of a polynomial f (x). If A is a square matrix with constant entries, then
d[ f (At)]/dt = A · f ^′(At).

PROOF. We only need to consider the special case where f (x) = c x^k is a monomial, because a general polynomial is a sum of monomials. f ^′(x) = c k x^k—1, so

A · f ^′(At) = A · (c k A^k—1 t^k—1) = c k A^k t^k—1.
On the other hand,
d[ f (At)]/dt = ^d (c A^k t^k) = c A^k (k t^k—1).
dt
Since these are equal, the proof is complete

PROOF OF THEOREM 3.14. By Corollary 3.5.1 e^At is nonsingular, so we only need to show that de^At /dt = Ae^At. Let F_m(At) denote the partial sum

_F ^m 1 _k

k!
_m(At) = _∑ (At) .
k=0
By Lemma 3.5.2, dF_m(At)/dt = A · F_m^′ (At). Since

F^′ At
^m 1 _k
At k—1

k!
_m( ) =
_∑ ( )
k=0

= AF_m—1(At),
it follows that lim_m→∞ F_m^′ (At) = lim_m→∞ F_m—1(At) = e^At. It is the definition of the sum of an infinite series that

lim F At

^∞ 1


At ^k


_eAt_.

Hence
_m(
m→∞
) = _∑ ( ) =

k!
k=0

_m ·
lim d[F (At)]/dt = A e^At, m→∞
and it might appear that the proof is complete. However, it still should be checked that

m→∞

dt

dt ^m→∞
lim ^d [F_m(At)] = ^d [ lim F_m(t)];


that is, that the limit of the derivatives is the derivative of the limit. The proof of this fact is omitted.

Calculation

The calculation of e^At, where A is a constant square matrix, involves devel- opments that are found in a full course in linear algebra. While all of the necessary ingredients are available in any linear algebra text, it is not un- usual for 14-week linear algebra courses to omit some of this material. In this text, the procedure for calculating e^At, where A is an arbitrary constant matrix with real or complex entries, will be described in detail, but we will depend on linear algebra texts for proofs.

×
There is a simple algorithm to calculate e^At when A is a 2 2 matrix. Although the algorithm does not extend to n > 2, it is a good warm up for the more general case.
Suppose that B is a 2 × 2 matrix with the special property that
tr B = 0.
The characteristic equation of B is then s² + d = 0, where d = det B. Here is the key formula:
B² = —dI (3.29)
The proof is just a reference to the Cayley-Hamilton theorem!

— —
It follows that B^2k = ( d)^k I, and B^2k+1 = B B^2k = ( d)^kB. Now that we know the powers of B we can figure out the exponential:

_eBt
^∞ 1

2k 2k

^∞ 1

2k+1 2k+1

k=0

k=0
^{= ∑} (2k)! ^{B t + ∑} (2k + 1)! ^{B t}

k=0
^∞ (—d)^k _2k
^∞ (—d)^k
2k+1

= _∑
k=0
t
(2k)!
^{I + ∑} (2k + 1)! ^t
B (3.30)

| | —
We can simplify this formula if we let d = ω². If d < 0, so that d =
ω² then (3.30) becomes

_eBt
^∞ (ωt)^2k
1 ^∞
_(ωt)2k+1

= _∑
k=0
(2k)! ^{I +} ω ^∑ (2k + 1)! ^B

k=0
1

= cosh(ωt)I +
sinh(ωt)B (3.31)
ω

As an exercise, you should prove that when d > 0,
e^Bt = cos(ωt)I + ¹ sin(ωt)B (3.32)
ω

Now for a matrix A with tr A = s /= 0, put B = A — ^s I, and note that

2

2

2

0. Furthermore, because the scalar ma-
tr B = tr A — tr ^s I = s — ^s (1 + 1) =

2

4 6
trix ^s I commutes with any matrix, by Theorem 3.13 we have e^At = e^st/2e^Bt.

4 3
Example 3.5.3 Find e^At, where (a) A = ^{1 2}
, (b) A = ^{2 —2}
, and

0 2
(c) A = ^{2 1} .

SOLUTION. In each case, we will put B = A — ( ¹ tr A)I.

1. 
   
   2
   
   4 1
   Since tr A = 4 we get B = A — 2I = ^{—1 2} . Thus d = det B = —9,
   

and ω = ^√| — 9| = 3. Hence

=

(

)

+

(

)

=

4

3

3

1
e^Bt cosh 3t I ¹ sinh 3t B
cosh(3t) — ¹ sinh(3t) ² sinh(3t) _,

3
and
₃ sinh(3t) cosh(3t) + ₃ sinh(3t)

_eAt
_e2t _B
e2t cosh(3t) — ¹ sinh(3t) ² sinh(3t)

=

=

4

3

3

1

₃ sinh(3t) cosh(3t) + ₃ sinh(3t)

=

3

3

3
¹ _e5t ₊ ² _e—t ¹ _(e5t _{— e}—t ₎ .

3

3

3
² _(e5t _{— e}—t ₎ ² _e5t ₊ ¹ _e—t

2. 
   
   4 2
   Since tr A = 8 we get B = A — 4I = ^{—2 —2} , and d = det B = 4.
   

Thus
_{eBt = cos(2t)I +} ¹ _{sin(2t)B =} cos(2t) — sin(2t) — sin(2t) _,

2
and
2 sin(2t) cos(2t) + sin(2t)

e = e e = e ^{— —}

.
_{At 4t Bt 4t} cos(2t) sin(2t) sin(2t)
2 sin(2t) cos(2t) + sin(2t)

3. 
   
   0 0
   Here tr A = 4 so B = A — 2I = ^{0 1} . Because d = 0, we can’t
   

plug into (3.31) or (3.32). However, we see that B² = 0 so


0 1

_{0 e}2t
e^Bt = I + tB = ^{1 t} It follows that e^At = e^2te^Bt = ^e2t ^te2t