Search in book...
Toggle Font Controls
Create new playlist

Name your new playlist

Playlist description (optional)
Sign In

Email address

Password

Forgot Password?

or

Continue with Facebook

Continue with Google
Sign Up

Full Name

Email address

Confirm Email Address

Password

or

Continue with Facebook

Continue with Google

8.3 Spectral Decomposition Methods

Here, we present an alternative approach to the computation of the matrix exponential eAt $e^{A t}$ , one that does not require that eigenvectors (including generalized ones) of the n×n $n \times n$ matrix A be found first. Assume that the characteristic polynomial of A is written in the form

p (λ) = (- 1) n | A - λ I |,

$p (λ) = {(- 1)}^{n} | A - λ I |,$ (1)

with leading term +λn $+ λ^{n}$ . [Compare Eqs. (4) and (5) in Section 6.1.] If the (not necessarily distinct) eigenvalues of A are λ1,λ2,…,λn $λ_{1}, λ_{2}, \dots, λ_{n}$ , then

p (λ) = (λ - λ 1) (λ - λ 2) \dots (λ - λ n) .

$p (λ) = (λ - λ_{1}) (λ - λ_{2}) \dots (λ - λ_{n}) .$ (2)

The Cayley-Hamilton theorem (Section 6.3) says that any matrix A satisfies its own characteristic equation; that is,

p (A) = \prod i = 1 n (A - λ i I) = 0

$p (A) = \prod_{i = 1}^{n} (A - λ_{i} I) = 0$ (3)

(where I denotes the n×n $n \times n$ identity matrix). This crucial fact is the key to our method in this section.

The way we proceed to calculate the matrix exponential eAt $e^{A t}$ depends on whether or not the eigenvalues of A are distinct.

The Case of Distinct Eigenvalues

If the eigenvalues of A are distinct (whether real or complex), the reciprocal 1/p(λ) $1 / p (λ)$ has a partial fraction decomposition of the form

1 p ( λ ) = a 1 λ - λ 1 + a 2 λ - λ 2 + \dots + a n λ - λ n,

$\frac{1}{p (λ)} = \frac{a_{1}}{λ - λ_{1}} + \frac{a_{2}}{λ - λ_{2}} + \dots + \frac{a_{n}}{λ - λ_{n}},$ (4)

where the numerators a1,a2,…,an $a_{1}, a_{2}, \dots, a_{n}$ are constants. Indeed, these constants can be found by multiplying Eq. (4) by p(λ) $p (λ)$ to get

1 = \sum i = 1 n a i b i (λ) = a 1 b 1 (λ) + a 2 b 2 (λ) + \dots + a n b n (λ),

$1 = \sum_{i = 1}^{n} a_{i} b_{i} (λ) = a_{1} b_{1} (λ) + a_{2} b_{2} (λ) + \dots + a_{n} b_{n} (λ),$ (5)

where the polynomial

b i (λ) = p ( λ ) λ - λ i = \prod j \neq i (λ - λ j)

$b_{i} (λ) = \frac{p (λ)}{λ - λ_{i}} = \prod_{j \neq i} (λ - λ_{j})$ (6)

is obtained from the characteristic polynomial p(λ) $p (λ)$ upon deletion of the factor (λ−λi) $(λ - λ_{i})$ corresponding to the ith eigenvalue.

It follows immediately from (6) that bi(λi)≠0 $b_{i} (λ_{i}) \neq 0$ , whereas bj(λi)=0 $b_{j} (λ_{i}) = 0$ if i≠j $i \neq j$ [because in this case bj(λi) $b_{j} (λ_{i})$ includes the factor (λi−λi) $(λ_{i} - λ_{i})$ ]. Consequently, substitution of λ=λi $λ = λ_{i}$ into Eq. (5) gives the value

a i = 1 b i ( λ i ) = 1 \prod j \neq i ( λ i - λ j )

$a_{i} = \frac{1}{b_{i} (λ_{i})} = \frac{1}{\prod_{j \neq i} (λ_{i} - λ_{j})}$ (7)

for i=1,2,…,n $i = 1, 2, \dots, n$ . Given these coefficient values, we define the projection matrices P1,P2,…,Pn $P_{1}, P_{2}, \dots, P_{n}$ of the fixed n×n $n \times n$ matrix A by writing

P i = a i b i (A) = a i \prod j \neq i (A - λ j I) = \prod j \neq i ( A - λ j I ) \prod j \neq i ( λ i - λ j ) .

$P_{i} = a_{i} b_{i} (A) = a_{i} \prod_{j \neq i} (A - λ_{j} I) = \frac{\prod_{j \neq i} (A - λ_{j} I)}{\prod_{j \neq i} (λ_{i} - λ_{j})} .$ (8)

These matrices have very special properties that are summarized by the following proposition.

Remark

In terms of the Kronecker delta defined by

δ i j = {10 if i = j if i \neq j,

$δ_{i j} = {\begin{array}{r} 1 & if i = j \\ 0 & if i \neq j \end{array},$

the conditions in (10) and (11) say that

P i P j = δ i j P i .

$P_{i} P_{j} = δ_{i j} P_{i} .$ (13)

Also, note that (12) says that Pi $P_{i}$ is effectively an “eigenmatrix” associated with the eigenvalue λi $λ_{i}$ of the matrix A—that is, multiplication of Pi $P_{i}$ by the matrix A (on the right) gives a scalar multiple of Pi $P_{i}$ .

Proof

We get (9) when we substitute A for λ $λ$ in the algebraic identity (5) and recall the definition in (8). If i≠j $i \neq j$ , then the product

P i P j = a i b i (A) \cdot a j b j (A) = a i a j \cdot \prod r \neq i (A - λ r I) \cdot \prod s \neq j (A - λ s I)

$P_{i} P_{j} = a_{i} b_{i} (A) \cdot a_{j} b_{j} (A) = a_{i} a_{j} \cdot \prod_{r \neq i} (A - λ_{r} I) \cdot \prod_{s \neq j} (A - λ_{s} I)$

evidently has the product p(A)=Πr=1n(A−λrI) $p (A) = Π_{r = 1}^{n} (A - λ_{r} I)$ as a factor. Hence (10) follows from the Cayley-Hamilton result p(A)=0 $p (A) = 0$ in (3). Next, (11) follows when we multiply (9) by Pi $P_{i}$ :

P i = P i I = P i (P 1 + P 2 + \dots + P n) = \sum j = 1 n P i P j = P 2 i,

$P_{i} = P_{i} I = P_{i} (P_{1} + P_{2} + \dots + P_{n}) = \sum_{j = 1}^{n} P_{i} P_{j} = P_{i}^{2},$

because of (10). Finally, to verify (12), we note that

P i A - λ i P i P i A - λ i P i = = = = = P i \cdot (A - λ i I) a i \prod j \neq i (A - λ j I) \cdot (A - λ i I) a i \prod j = i n (A - λ j I) a i p (A) 0,

$\begin{array}{rcl} P_{i} A - λ_{i} P_{i} & = & P_{i} \cdot (A - λ_{i} I) \\ = & a_{i} \prod_{j \neq i} (A - λ_{j} I) \cdot (A - λ_{i} I) \\ = & a_{i} \prod_{j = i}^{n} (A - λ_{j} I) \\ = & a_{i} p (A) \\ P_{i} A - λ_{i} P_{i} & = & 0, \end{array}$

by the use first of (8) and then of (2).

The following theorem expresses the matrix A in terms of its projection matrices.

Remark 1

If we square both sides in (14) and then use (13), we get

(λ 1 P 1 + λ 2 P 2 + \dots + λ n P n) \cdot (λ 1 P 1 + λ 2 P 2 + \dots + λ n P n) = \sum i, j = 1 n λ i λ j P i P j,

$(λ_{1} P_{1} + λ_{2} P_{2} + \dots + λ_{n} P_{n}) \cdot (λ_{1} P_{1} + λ_{2} P_{2} + \dots + λ_{n} P_{n}) = \sum_{i, j = 1}^{n} λ_{i} λ_{j} P_{i} P_{j},$

A 2 = \sum i = 1 n λ 2 i P i = λ 21 P 2 + λ 22 P 2 + \dots + λ 2 n P n .

$A^{2} = \sum_{i = 1}^{n} λ_{i}^{2} P_{i} = λ_{1}^{2} P_{2} + λ_{2}^{2} P_{2} + \dots + λ_{n}^{2} P_{n} .$

Continuing in this fashion, we deduce by repeated multiplication that

A k = \sum i = 1 n λ k i P i = λ k 1 P 1 + λ k 2 P 2 + \dots + λ k n P n

$A^{k} = \sum_{i = 1}^{n} λ_{i}^{k} P_{i} = λ_{1}^{k} P_{1} + λ_{2}^{k} P_{2} + \dots + λ_{n}^{k} P_{n}$ (15)

for any positive integer k. We will see in the proof of Theorem 2 that this single fact is all we need to express the matrix exponential eAt $e^{A t}$ simply and explicitly in terms of the eigenvalues and projection matrices of the matrix A.

Remark 2

The relation in (15) holds also if k=12 $k = \frac{1}{2}$ . For instance, if n=2 $n = 2$ and B=λ1−−√P1+λ2−−√P2 $B = \sqrt{λ_{1}} P_{1} + \sqrt{λ_{2}} P_{2}$ , then the basic properties of projection matrices (Proposition 1) and (14) yield

B 2 = = = (λ 1 - - \sqrt P 1 + λ 2 - - \sqrt P 2) (λ 1 - - \sqrt P 1 + λ 2 - - \sqrt P 2) λ 1 P 21 + λ 1 λ 2 - - - - \sqrt P 1 P 2 + λ 2 λ 1 - - - - \sqrt P 2 P 1 + λ 2 P 22 λ 1 P 1 + λ 2 P 2 = A .

$\begin{array}{rcl} B^{2} & = & (\sqrt{λ_{1}} P_{1} + \sqrt{λ_{2}} P_{2}) (\sqrt{λ_{1}} P_{1} + \sqrt{λ_{2}} P_{2}) \\ = & λ_{1} P_{1}^{2} + \sqrt{λ_{1} λ_{2}} P_{1} P_{2} + \sqrt{λ_{2} λ_{1}} P_{2} P_{1} + λ_{2} P_{2}^{2} \\ = & λ_{1} P_{1} + λ_{2} P_{2} = A . \end{array}$

It follows that if (for each i) λi−−√ $\sqrt{λ_{i}}$ denotes either square root of the eigenvalue λi $λ_{i}$ of A, then

A - - \sqrt = λ 1 - - \sqrt P 1 + λ 2 - - \sqrt P 2

$\sqrt{A} = \sqrt{λ_{1}} P_{1} + \sqrt{λ_{2}} P_{2}$ (16)

is a square root of the matrix A.

Example 1

If A is a 2×2 $2 \times 2$ matrix with distinct eigenvalues λ1 $λ_{1}$ and λ2 $λ_{2}$ , then Eqs. (7), (8), and (17) give first

a 1 = 1 λ 1 - λ 2 and a 2 = 1 λ 2 - λ 1,

$\begin{array}{c} a_{1} = \frac{1}{λ_{1} - λ_{2}} & and & a_{2} = \frac{1}{λ_{2} - λ_{1}}, \end{array}$ (18)

then

P 1 = A - λ 2 I λ 1 - λ 2 and P 2 = A - λ 1 I λ 2 - λ 1,

$\begin{array}{c} P_{1} = \frac{A - λ_{2} I}{λ_{1} - λ_{2}} & and & P_{2} = \frac{A - λ_{1} I}{λ_{2} - λ_{1}}, \end{array}$ (19)

and finally

e A t = e λ 1 t P 1 + e λ 2 t P 2 .

$e^{A t} = e^{λ_{1} t} P_{1} + e^{λ_{2} t} P_{2} .$ (20)

Thus the calculation of the matrix exponential eAt $e^{A t}$ using the projection matrices P1 $P_{1}$ and P2 $P_{2}$ reduces to a routine matter of numerical substitution.

Example 2

The matrix

A = [10318]

$A = [\begin{array}{r} 10 & 1 \\ 3 & 8 \end{array}]$

has the characteristic polynomial

p (λ) = (10 - λ) (8 - λ) - 3 = λ 2 - 18 λ + 77 = (λ - 7) (λ - 11),

$p (λ) = (10 - λ) (8 - λ) - 3 = λ^{2} - 18 λ + 77 = (λ - 7) (λ - 11),$

so its eigenvalues are λ1=7 $λ_{1} = 7$ and λ2=11 $λ_{2} = 11$ . Hence Eq. (19) gives

P 1 = - 1 4 [10 - 11 3 1 8 - 11] = - 1 4 [- 1 3 1 - 3]

$P_{1} = - \frac{1}{4} [\begin{array}{c} 10 - 11 & 1 \\ 3 & 8 - 11 \end{array}] = - \frac{1}{4} [\begin{array}{r} - 1 & 1 \\ 3 & - 3 \end{array}]$

and

P 2 = 1 4 [10 - 7 3 1 8 - 7] = 1 4 [3311],

$P_{2} = \frac{1}{4} [\begin{array}{c} 10 - 7 & 1 \\ 3 & 8 - 7 \end{array}] = \frac{1}{4} [\begin{array}{r} 3 & 1 \\ 3 & 1 \end{array}],$

so Eq. (20) gives

e A t = = - 1 4 e 7 t [- 1 3 1 - 3] + 1 4 e 11 t [3311] 1 4 [e 7 t + 3 e 11 t - 3 e 7 t + 3 e 11 t - e 7 t + e 11 t 3 e 7 t + e 11 t] .

$\begin{array}{rcl} e^{A t} & = & - \frac{1}{4} e^{7 t} [\begin{array}{r} - 1 & 1 \\ 3 & - 3 \end{array}] + \frac{1}{4} e^{11 t} [\begin{array}{r} 3 & 1 \\ 3 & 1 \end{array}] \\ = & \frac{1}{4} [\begin{array}{r} e^{7 t} + 3 e^{11 t} & - e^{7 t} + e^{11 t} \\ - 3 e^{7 t} + 3 e^{11 t} & 3 e^{7 t} + e^{11 t} \end{array}] . \end{array}$

Example 3

The 3×3 $3 \times 3$ matrix

A = ⎡ ⎣ ⎢ 3 - 4 16 06 - 15 02 - 5 ⎤ ⎦ ⎥

$A = [\begin{array}{r} 3 & 0 & 0 \\ - 4 & 6 & 2 \\ 16 & - 15 & - 5 \end{array}]$

has the characteristic polynomial

p (λ) p (λ) = = = = - | A - λ I | - (3 - λ) [(6 - λ) (- 5 - λ) + 30] - (3 - λ) [λ 2 - λ] λ (λ - 1) (λ - 3) .

$\begin{array}{rcl} p (λ) & = & - | A - λ I | \\ = & - (3 - λ) [(6 - λ) (- 5 - λ) + 30] \\ = & - (3 - λ) [λ^{2} - λ] \\ p (λ) & = & λ (λ - 1) (λ - 3) . \end{array}$

Hence the eigenvalues of A are λ1=0, λ2=1 $λ_{1} = 0, λ_{2} = 1$ , and λ3=3 $λ_{3} = 3$ . To find the matrix exponential

e A t = e λ 1 t P 1 + e λ 2 t P 2 + e λ 3 t P 3,

$e^{A t} = e^{λ_{1} t} P_{1} + e^{λ_{2} t} P_{2} + e^{λ_{3} t} P_{3},$ (21)

we need only calculate the three projections matrices P1, P2 $P_{1}, P_{2}$ , and P3 $P_{3}$ defined by Eq. (8). But first we need the coefficients

a 1 a 2 a 3 = = = 1 ( λ 1 - λ 2 ) ( λ 1 - λ 3 ) = 1 ( 0 - 1 ) ( 0 - 3 ) = 1 3, 1 ( λ 2 - λ 1 ) ( λ 2 - λ 3 ) = 1 ( 1 - 0 ) ( 1 - 3 ) = - 1 2, 1 ( λ 3 - λ 1 ) ( λ 3 - λ 2 ) = 1 ( 3 - 0 ) ( 3 - 1 ) = 1 6 and

$\begin{array}{rcl} a_{1} & = & \frac{1}{(λ_{1} - λ_{2}) (λ_{1} - λ_{3})} = \frac{1}{(0 - 1) (0 - 3)} = \frac{1}{3}, \\ a_{2} & = & \frac{1}{(λ_{2} - λ_{1}) (λ_{2} - λ_{3})} = \frac{1}{(1 - 0) (1 - 3)} = - \frac{1}{2}, & and \\ a_{3} & = & \frac{1}{(λ_{3} - λ_{1}) (λ_{3} - λ_{2})} = \frac{1}{(3 - 0) (3 - 1)} = \frac{1}{6} \end{array}$

that are given by Eq. (7). Then

P 1 P 2 = = = = a 1 (A - λ 2 I) (A - λ 3 I) 1 3 ⎡ ⎣ ⎢ 2 - 4 16 05 - 15 02 - 6 ⎤ ⎦ ⎥ ⎡ ⎣ ⎢ 0 - 4 16 03 - 15 02 - 8 ⎤ ⎦ ⎥ = ⎡ ⎣ ⎢ 04 - 12 0 - 5 15 026 ⎤ ⎦ ⎥, a 2 (A - λ 1 I) (A - λ 3 I) - 1 2 ⎡ ⎣ ⎢ 3 - 4 16 06 - 15 02 - 5 ⎤ ⎦ ⎥ ⎡ ⎣ ⎢ 0 - 4 16 03 - 15 02 - 8 ⎤ ⎦ ⎥ = ⎡ ⎣ ⎢ 0 - 4 10 06 - 15 02 - 5 ⎤ ⎦ ⎥,

$\begin{array}{rcl} P_{1} & = & a_{1} (A - λ_{2} I) (A - λ_{3} I) \\ = & \frac{1}{3} [\begin{array}{r} 2 & 0 & 0 \\ - 4 & 5 & 2 \\ 16 & - 15 & - 6 \end{array}] [\begin{array}{r} 0 & 0 & 0 \\ - 4 & 3 & 2 \\ 16 & - 15 & - 8 \end{array}] = [\begin{array}{r} 0 & 0 & 0 \\ 4 & - 5 & 2 \\ - 12 & 15 & 6 \end{array}], \\ P_{2} & = & a_{2} (A - λ_{1} I) (A - λ_{3} I) \\ = & - \frac{1}{2} [\begin{array}{r} 3 & 0 & 0 \\ - 4 & 6 & 2 \\ 16 & - 15 & - 5 \end{array}] [\begin{array}{r} 0 & 0 & 0 \\ - 4 & 3 & 2 \\ 16 & - 15 & - 8 \end{array}] = [\begin{array}{r} 0 & 0 & 0 \\ - 4 & 6 & 2 \\ 10 & - 15 & - 5 \end{array}], \end{array}$

and

P 3 = = a 3 (A - λ 1 I) (A - λ 2 I) 1 6 ⎡ ⎣ ⎢ 3 - 4 16 06 - 15 02 - 5 ⎤ ⎦ ⎥ ⎡ ⎣ ⎢ 2 - 4 16 05 - 15 02 - 6 ⎤ ⎦ ⎥ = ⎡ ⎣ ⎢ 102000000 ⎤ ⎦ ⎥ .

$\begin{array}{rcl} P_{3} & = & a_{3} (A - λ_{1} I) (A - λ_{2} I) \\ = & \frac{1}{6} [\begin{array}{r} 3 & 0 & 0 \\ - 4 & 6 & 2 \\ 16 & - 15 & - 5 \end{array}] [\begin{array}{r} 2 & 0 & 0 \\ - 4 & 5 & 2 \\ 16 & - 15 & - 6 \end{array}] = [\begin{array}{r} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0 \end{array}] . \end{array}$

Finally, (17) gives the desired matrix exponential

e A t = = e 0 t ⎡ ⎣ ⎢ 04 - 12 0 - 5 15 026 ⎤ ⎦ ⎥ + e 1 t ⎡ ⎣ ⎢ 0 - 4 10 06 - 15 02 - 5 ⎤ ⎦ ⎥ + e 3 t ⎡ ⎣ ⎢ 102000000 ⎤ ⎦ ⎥ ⎡ ⎣ ⎢ e 3 t 4 - 4 e t - 12 + 10 e t + 2 e 3 t 0 - 5 + 6 e t 15 - 15 e t 0 - 2 + 2 e t 6 - 5 e t ⎤ ⎦ ⎥ .

$\begin{array}{rcl} e^{A t} & = & e^{0 t} [\begin{array}{r} 0 & 0 & 0 \\ 4 & - 5 & 2 \\ - 12 & 15 & 6 \end{array}] + e^{1 t} [\begin{array}{r} 0 & 0 & 0 \\ - 4 & 6 & 2 \\ 10 & - 15 & - 5 \end{array}] + e^{3 t} [\begin{array}{c} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0 \end{array}] \\ = & [\begin{array}{c} e^{3 t} & 0 & 0 \\ 4 - 4 e^{t} & - 5 + 6 e^{t} & - 2 + 2 e^{t} \\ - 12 + 10 e^{t} + 2 e^{3 t} & 15 - 15 e^{t} & 6 - 5 e^{t} \end{array}] . \end{array}$

As an application, the solution of the initial value problem

x' (t) = A x, x (0) = x 0 = [3711] T

$x ′ (t) = Ax, x (0) = x_{0} = {[\begin{array}{r} 3 & 7 & 11 \end{array}]}^{T}$

is (by Theorem 2 in Section 8.1) given by

x (t) = e A t x 0 = = ⎡ ⎣ ⎢ e 3 t 4 - 4 e t - 12 + 10 e t + 2 e 3 t 0 - 5 + 6 e t 15 - 15 e t 0 - 2 + 2 e t 6 - 5 e t ⎤ ⎦ ⎥ ⎡ ⎣ ⎢ 3711 ⎤ ⎦ ⎥ ⎡ ⎣ ⎢ 3 e 3 t - 45 + 52 e t 135 - 130 e t + 6 e 3 t ⎤ ⎦ ⎥ .

$\begin{array}{rcl} x (t) = e^{A t} x_{0} & = & [\begin{array}{c} e^{3 t} & 0 & 0 \\ 4 - 4 e^{t} & - 5 + 6 e^{t} & - 2 + 2 e^{t} \\ - 12 + 10 e^{t} + 2 e^{3 t} & 15 - 15 e^{t} & 6 - 5 e^{t} \end{array}] [\begin{array}{r} 3 \\ 7 \\ 11 \end{array}] \\ = & [\begin{array}{c} 3 e^{3 t} \\ - 45 + 52 e^{t} \\ 135 - 130 e^{t} + 6 e^{3 t} \end{array}] . \end{array}$

Second-Order Linear Systems

Spectral decompositions of matrices also yield convenient solutions of second-order matrix linear systems.

Example 4

Mass-spring system Consider the mass-and-spring system shown in Fig. 8.3.1, with m1=2, m2=1, k1=100 $m_{1} = 2, m_{2} = 1, k_{1} = 100$ , and k2=50 $k_{2} = 50$ . As in Example 1 of Section 7.4, the position vector x(t)=[x1(t)x2(t)]T $x (t) = {[\begin{array}{r} x_{1} (t) & x_{2} (t) \end{array}]}^{T}$ satisfies the second-order system x''=Ax $x ″ = Ax$ having coefficient matrix

A = [- 75 50 25 - 50] .

$A = [\begin{array}{r} - 75 & 25 \\ 50 & - 50 \end{array}] .$

FIGURE 8.3.1.

The mass-and-spring system of Example 4.

We find that A has characteristic polynomial p(λ)=λ2+125λ+2500=(λ+25)(λ+100) $p (λ) = λ^{2} + 125 λ + 2500 = (λ + 25) (λ + 100)$ and hence has eigenvalues λ1=−25 $λ_{1} = - 25$ and λ2=−100 $λ_{2} = - 100$ . By using (19), we calculate the projection matrices

P 1 P 2 = = A - λ 2 I λ 1 - λ 2 = 1 75 [25502550] = 1 3 [1212] and A - λ 1 I λ 2 - λ 1 = - 1 75 [- 50 50 25 - 25] = 1 3 [2 - 2 - 1 1]

$\begin{array}{rcl} P_{1} & = & \frac{A - λ_{2} I}{λ_{1} - λ_{2}} = \frac{1}{75} [\begin{array}{r} 25 & 25 \\ 50 & 50 \end{array}] = \frac{1}{3} [\begin{array}{r} 1 & 1 \\ 2 & 2 \end{array}] and \\ P_{2} & = & \frac{A - λ_{1} I}{λ_{2} - λ_{1}} = - \frac{1}{75} [\begin{array}{r} - 50 & 25 \\ 50 & - 25 \end{array}] = \frac{1}{3} [\begin{array}{r} 2 & - 1 \\ - 2 & 1 \end{array}] \end{array}$

of A. Starting with the spectral decomposition

A - - \sqrt = λ 1 - - \sqrt P 1 + λ 2 - - \sqrt P 2 = 5 i P 1 + 10 i P 2

$\sqrt{A} = \sqrt{λ_{1}} P_{1} + \sqrt{λ_{2}} P_{2} = 5 i P_{1} + 10 i P_{2}$

of A−−√ $\sqrt{A}$ , Theorem 2 yields

e A \sqrt t = e 5 i t P 1 + e 10 i t P 2 and e - A \sqrt t = e - 5 i t P 1 + e - 10 i t P 2 .

$e^{\sqrt{A} t} = e^{5 i t} P_{1} + e^{10 i t} P_{2} and e^{- \sqrt{A} t} = e^{- 5 i t} P_{1} + e^{- 10 i t} P_{2} .$

Consequently, Eq. (22) in Theorem 3 gives the following general solution:

x (t) x (t) = = = e A \sqrt t c 1 + e - A \sqrt t c 2 = (e 5 i t P 1 + e 10 i t P 2) c 1 + (e - 5 i t P 1 + e - 10 i t P 2) c 2 P 1 (c 1 e 5 i t + c 2 e - 5 i t) + P 2 (c 1 e 10 i t + c 2 e - 10 i t) P 1 (a cos 5 t + b sin 5 t) + P 2 (a cos 10 t + b sin 10 t),

$\begin{array}{rcl} x (t) & = & e^{\sqrt{A} t} c_{1} + e^{- \sqrt{A} t} c_{2} = (e^{5 i t} P_{1} + e^{10 i t} P_{2}) c_{1} + (e^{- 5 i t} P_{1} + e^{- 10 i t} P_{2}) c_{2} \\ = & P_{1} (c_{1} e^{5 i t} + c_{2} e^{- 5 i t}) + P_{2} (c_{1} e^{10 i t} + c_{2} e^{- 10 i t}) \\ x (t) & = & P_{1} (a \cos 5 t + b \sin 5 t) + P_{2} (a \cos 10 t + b \sin 10 t), \end{array}$ (23)

where a=c1+c2 $a = c_{1} + c_{2}$ and b=i(c1−c2) $b = i (c_{1} - c_{2})$ . Now

p 1 a P 2 a = = 1 3 [1212] [a 1 a 2] = 1 3 (a 1 + a 2) [12] and P 1 b = 1 3 (b 1 + b 2) [12], 1 3 [2 - 2 - 1 1] [a 1 a 2] = 1 3 (2 a 1 - a 2) [1 - 1]

$\begin{array}{rcl} p_{1} a & = & \frac{1}{3} [\begin{array}{r} 1 & 1 \\ 2 & 2 \end{array}] [\begin{array}{c} a_{1} \\ a_{2} \end{array}] = \frac{1}{3} (a_{1} + a_{2}) [\begin{array}{c} 1 \\ 2 \end{array}] and P_{1} b = \frac{1}{3} (b_{1} + b_{2}) [\begin{array}{c} 1 \\ 2 \end{array}], \\ P_{2} a & = & \frac{1}{3} [\begin{array}{r} 2 & - 1 \\ - 2 & 1 \end{array}] [\begin{array}{c} a_{1} \\ a_{2} \end{array}] = \frac{1}{3} (2 a_{1} - a_{2}) [\begin{array}{r} 1 \\ - 1 \end{array}] \end{array}$

and

P 2 b = 1 3 (2 b 1 - b 2) [1 - 1] .

$P_{2} b = \frac{1}{3} (2 b_{1} - b_{2}) [\begin{array}{r} 1 \\ - 1 \end{array}] .$

Hence (23) finally takes the form

x (t) = (a cos 5 t + b sin 5 t) [12] + (c cos 10 t + d sin 10 t) [1 - 1] .

$x (t) = (a \cos 5 t + b \sin 5 t) [\begin{array}{r} 1 \\ 2 \end{array}] + (c \cos 10 t + d \sin 10 t) [\begin{array}{r} 1 \\ - 1 \end{array}] .$ (24)

This last equation expresses the motion x(t) of the mass-spring system in Fig. 8.3.1 as a linear combination of two natural modes of free oscillation. In the first mode, the two masses move with frequency ω1=5 $ω_{1} = 5$ in the same direction and with the amplitude of m2 $m_{2}$ being twice that of m1 $m_{1}$ . In the second mode, the two masses move with frequency ω2=10 $ω_{2} = 10$ in opposite directions and with equal amplitudes of motion.

The General Case

Now we want to take into account the possibility of multiple eigenvalues. If the n×n $n \times n$ matrix A has distinct eigenvalues λ1,λ2,…,λq $λ_{1}, λ_{2}, \dots, λ_{q}$ having multiplicities m1,m2,…,mq $m_{1}, m_{2}, \dots, m_{q}$ (respectively), then the reciprocal 1/p(λ) $1 / p (λ)$ has a partial-fraction decomposition of the form

1 p ( λ ) = a 1 ( λ ) ( λ - λ 1 ) m 1 + a 2 ( λ ) ( λ - λ 2 ) m 2 + \dots + a q ( λ ) ( λ - λ q ) m q,

$\frac{1}{p (λ)} = \frac{a_{1} (λ)}{{(λ - λ_{1})}^{m_{1}}} + \frac{a_{2} (λ)}{{(λ - λ_{2})}^{m_{2}}} + \dots + \frac{a_{q} (λ)}{{(λ - λ_{q})}^{m_{q}}},$ (25)

where (for each i=1,2,…,q $i = 1, 2, \dots, q$ ) the numerator ai(λ) $a_{i} (λ)$ is a polynomial in λ $λ$ of degree at most mi−1 $m_{i} - 1$ . Once these numerator polynomials have been found, we can define the projection matrices P1,P2,…,Pq $P_{1}, P_{2}, \dots, P_{q}$ of A by the formula

P i = a i (A) b i (A) = a i (A) Π j \neq i (A - λ j I) m j,

$P_{i} = a_{i} (A) b_{i} (A) = a_{i} (A) \underset{j \neq i}{Π} {(A - λ_{j} I)}^{m_{j}},$ (26)

analogous to Eq. (8), with the polynomial

b i (λ) = p ( λ ) ( λ - λ i ) m i = Π j \neq i (λ - λ j) m j

$b_{i} (λ) = \frac{p (λ)}{{(λ - λ_{i})}^{m_{i}}} = \underset{j \neq i}{Π} {(λ - λ_{j})}^{m_{j}}$ (27)

being obtained from the characteristic polynomial p(λ) $p (λ)$ upon deletion of the factor (λ−λi)mi ${(λ - λ_{i})}^{m_{i}}$ corresponding to the ith eigenvalue. It follows immediately from (27)that bi(λi)≠0 $b_{i} (λ_{i}) \neq 0$ , whereas bj(λi)=0 $b_{j} (λ_{i}) = 0$ if i≠j $i \neq j$ [in which case bj(λi) $b_{j} (λ_{i})$ includes the factor (λi−λi)mi ${(λ_{i} - λ_{i})}^{m_{i}}$ ]. The following properties of the projection matrices P1,P2,…,Pq $P_{1}, P_{2}, \dots, P_{q}$ are established in essentially the same way as in the proof of Proposition 1.

In the distinct-eigenvalue case we saw in part (iv) of Proposition 1 that Pi(A−λiI)=0 $P_{i} (A - λ_{i} I) = 0$ for each i. This may no longer be so if mi>1 $m_{i} > 1$ , so we define

N i = P i (A - λ i I)

$N_{i} = P_{i} (A - λ_{i} I)$ (31)

for each i=1,2,…,q $i = 1, 2, \dots, q$ . Part (ii) of the following proposition asserts that these matrices are nilpotent. Note that Ni=0 $N_{i} = 0$ if mi=1 $m_{i} = 1$ (so that λi $λ_{i}$ is a nonrepeated eigenvalue).

The following theorem expresses the matrix A in terms of its projection and nilpotent matrices in the general case (allowing multiple eigenvalues).

As in the distinct-eigenvalue case of Theorem 2, the spectral decomposition of Theorem 4 can be used to calculate the matrix exponential eAt $e^{A t}$ in the general case.

Proof

First, we apply the fact that ∑i=1qPi=I $\sum_{i = 1}^{q} P_{i} = I$ and write

e A t = I e A t = \sum i = 1 q P i e A t .

$e^{A t} = I e^{A t} = \sum_{i = 1}^{q} P_{i} e^{A t} .$ (36)

Then, the typical term in (36) is

P i e A t P i e A t = = = = = = = = P i e (λ i I + A - λ i I) t = e λ i t I P i e (A - λ i I) t e λ i t P i (I + \sum k = 1 \infty (A - λ i I) k t k k !) e λ i t (P i + \sum k = 1 \infty P i (A - λ i I) k t k k !) e λ i t (P i + \sum k = 1 \infty P k i (A - λ i I) k t k k !) (because P k i = P i) e λ i t (P i + \sum k = 1 \infty N k i t k k !) (because N i = P i (A - λ i I)) e λ i t (P i + \sum k = 1 m i - 1 N k i t k k !) (because N m i i = 0) e λ i t (P i + \sum k = 1 m i - 1 P i (A - λ i I) k t k k !), so e λ i t P i (I + \sum k = 1 m i - 1 (A - λ i I) k t k k !) .

$\begin{array}{rcl} P_{i} e^{A t} & = & P_{i} e^{(λ_{i} I + A - λ_{i} I) t} = e^{λ_{i} t} I P_{i} e^{(A - λ_{i} I) t} \\ = & e^{λ_{i} t} P_{i} (I + \sum_{k = 1}^{\infty} {(A - λ_{i} I)}^{k} \frac{t^{k}}{k!}) \\ = & e^{λ_{i} t} (P_{i} + \sum_{k = 1}^{\infty} P_{i} {(A - λ_{i} I)}^{k} \frac{t^{k}}{k!}) \\ = & e^{λ_{i} t} (P_{i} + \sum_{k = 1}^{\infty} P_{i}^{k} {(A - λ_{i} I)}^{k} \frac{t^{k}}{k!}) (because P_{i}^{k} = P_{i}) \\ = & e^{λ_{i} t} (P_{i} + \sum_{k = 1}^{\infty} N_{i}^{k} \frac{t^{k}}{k!}) (because N_{i} = P_{i} (A - λ_{i} I)) \\ = & e^{λ_{i} t} (P_{i} + \sum_{k = 1}^{m_{i} - 1} N_{i}^{k} \frac{t^{k}}{k!}) (because N_{i}^{m_{i}} = 0) \\ = & e^{λ_{i} t} (P_{i} + \sum_{k = 1}^{m_{i} - 1} P_{i} {(A - λ_{i} I)}^{k} \frac{t^{k}}{k!}), so \\ P_{i} e^{A t} & = & e^{λ_{i} t} P_{i} (I + \sum_{k = 1}^{m_{i} - 1} {(A - λ_{i} I)}^{k} \frac{t^{k}}{k!}) . \end{array}$

Finally, we get (35) upon substitution of this last expression for Pi eAt $P_{i} e^{A t}$ in (36).

The complicated formula in (35) looks much simpler in low-dimensional special cases.

Example 5

If A is a 2×2 $2 \times 2$ matrix with a single eigenvalue λ1 $λ_{1}$ of multiplicity 2 and corresponding projection matrix P1 $P_{1}$ , then (35) reduces to eAt=eλ1tP1[I+(A−λ1I)t] $e^{A t} = e^{λ_{1} t} P_{1} [I + (A - λ_{1} I) t]$ . In this case, p(λ)=(λ−λ1)2 $p (λ) = {(λ - λ_{1})}^{2}$ so Eqs. (25)–(27) give a1(λ)=b1(λ)=1 $a_{1} (λ) = b_{1} (λ) = 1$ , and hence P1=I $P_{1} = I$ . It therefore follows that

e A t = e λ 1 t [I + (A - λ 1 I) t] .

$e^{A t} = e^{λ_{1} t} [I + (A - λ_{1} I) t] .$

Example 6

If A is a 3×3 $3 \times 3$ matrix with an eigenvalue λ1 $λ_{1}$ of multiplicity 1, an eigenvalue λ2 $λ_{2}$ of multiplicity 2, and corresponding projection matrices P1 $P_{1}$ and P2 $P_{2}$ , then (35) reduces to

e A t = e λ 1 t P 1 + e λ 2 t P 2 [I + (A - λ 2 I) t] .

$e^{A t} = e^{λ_{1} t} P_{1} + e^{λ_{2} t} P_{2} [I + (A - λ_{2} I) t] .$

Example 7

If A is a 5×5 $5 \times 5$ matrix with an eigenvalue λ1 $λ_{1}$ of multiplicity 2 and an eigenvalue λ2 $λ_{2}$ of multiplicity 3, then (35) reduces to

e A t = e λ 1 t P 1 [I + (A - λ 1 I) t] + e λ 2 t P 2 [I + (A - λ 2 I) t + 1 2 (A - λ 2 I) 2 t 2] .

$e^{A t} = e^{λ_{1} t} P_{1} [I + (A - λ_{1} I) t] + e^{λ_{2} t} P_{2} [I + (A - λ_{2} I) t + \frac{1}{2} {(A - λ_{2} I)}^{2} t^{2}] .$

In this case, the principal labor may consist of finding the characteristic polynomial and its partial-fractions decomposition (25), so that Eqs. (26)–(27) can be used to find the projection matrices P1 $P_{1}$ and P2 $P_{2}$ .

Example 8

The matrix

A = ⎡ ⎣ ⎢ 422 - 2 0 - 2 113 ⎤ ⎦ ⎥

$A = [\begin{array}{r} 4 & - 2 & 1 \\ 2 & 0 & 1 \\ 2 & - 2 & 3 \end{array}]$ (37)

has the characteristic polynomial

p (λ) = - | A - λ I | = λ 3 - 7 λ 2 + 16 λ - 12 = (λ - 3) (λ - 2) 2

$p (λ) = - | A - λ I | = λ^{3} - 7 λ^{2} + 16 λ - 12 = (λ - 3) {(λ - 2)}^{2}$

and thus has eigenvalues λ1=3 $λ_{1} = 3$ of multiplicity 1 and λ2=2 $λ_{2} = 2$ of multiplicity 2. The partial-fractions decomposition

1 ( λ - 3 ) ( λ - 2 ) 2 = 1 λ - 3 + 1 - λ ( λ - 2 ) 2

$\frac{1}{(λ - 3) {(λ - 2)}^{2}} = \frac{1}{λ - 3} + \frac{1 - λ}{{(λ - 2)}^{2}}$

shows that a1(λ)=1 $a_{1} (λ) = 1$ and a2=1−λ $a_{2} = 1 - λ$ . Now b1(λ)=(λ−2)2 $b_{1} (λ) = {(λ - 2)}^{2}$ and b2(λ)=λ−3 $b_{2} (λ) = λ - 3$ , so (26)gives

P 1 = a 1 (A) b 1 (A) = (A - 2 I) 2 = ⎡ ⎣ ⎢ 222 - 2 - 2 - 2 111 ⎤ ⎦ ⎥ 2 = ⎡ ⎣ ⎢ 222 - 2 - 2 - 2 111 ⎤ ⎦ ⎥

$P_{1} = a_{1} (A) b_{1} (A) = {(A - 2 I)}^{2} = {[\begin{array}{r} 2 & - 2 & 1 \\ 2 & - 2 & 1 \\ 2 & - 2 & 1 \end{array}]}^{2} = [\begin{array}{r} 2 & - 2 & 1 \\ 2 & - 2 & 1 \\ 2 & - 2 & 1 \end{array}]$

and

P 2 = = a 2 (A) b 2 (A) = (I - A) (A - 3 I) ⎡ ⎣ ⎢ - 3 - 2 - 2 212 - 1 - 1 - 2 ⎤ ⎦ ⎥ ⎡ ⎣ ⎢ 122 - 2 - 3 - 2 110 ⎤ ⎦ ⎥ = ⎡ ⎣ ⎢ - 1 - 2 - 2 232 - 1 - 1 0 ⎤ ⎦ ⎥ .

$\begin{array}{rcl} P_{2} & = & a_{2} (A) b_{2} (A) = (I - A) (A - 3 I) \\ = & [\begin{array}{r} - 3 & 2 & - 1 \\ - 2 & 1 & - 1 \\ - 2 & 2 & - 2 \end{array}] [\begin{array}{r} 1 & - 2 & 1 \\ 2 & - 3 & 1 \\ 2 & - 2 & 0 \end{array}] = [\begin{array}{r} - 1 & 2 & - 1 \\ - 2 & 3 & - 1 \\ - 2 & 2 & 0 \end{array}] \end{array} .$

The result of Example 6 therefore gives

e A t e A t = = = e λ 1 t P 1 + e λ 2 t P 2 [I + (A - λ 2 I) t] e 3 t ⎡ ⎣ ⎢ 222 - 2 - 2 - 2 111 ⎤ ⎦ ⎥ + e 2 t ⎡ ⎣ ⎢ - 1 - 2 - 2 232 - 1 - 1 0 ⎤ ⎦ ⎥ ⎛ ⎝ ⎜ ⎡ ⎣ ⎢ 100010001 ⎤ ⎦ ⎥ + ⎡ ⎣ ⎢ 222 - 2 - 2 - 2 111 ⎤ ⎦ ⎥ t ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ 2 e 3 t - e 2 t 2 e 3 t - 2 e 2 t 2 e 3 t - 2 e 2 t - 2 e 3 t + 2 e 2 t - 2 e 3 t + 3 e 2 t - 2 e 3 t + 2 e 2 t e 3 t - e 2 t e 3 t - e 2 t e 3 t ⎤ ⎦ ⎥ .

$\begin{array}{rcl} e^{A t} & = & e^{λ_{1} t} P_{1} + e^{λ_{2} t} P_{2} [I + (A - λ_{2} I) t] \\ = & e^{3 t} [\begin{array}{r} 2 & - 2 & 1 \\ 2 & - 2 & 1 \\ 2 & - 2 & 1 \end{array}] \\ + e^{2 t} [\begin{array}{r} - 1 & 2 & - 1 \\ - 2 & 3 & - 1 \\ - 2 & 2 & 0 \end{array}] ([\begin{array}{r} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] + [\begin{array}{r} 2 & - 2 & 1 \\ 2 & - 2 & 1 \\ 2 & - 2 & 1 \end{array}] t) \\ e^{A t} & = & [\begin{array}{c} 2 e^{3 t} - e^{2 t} & - 2 e^{3 t} + 2 e^{2 t} & e^{3 t} - e^{2 t} \\ 2 e^{3 t} - 2 e^{2 t} & - 2 e^{3 t} + 3 e^{2 t} & e^{3 t} - e^{2 t} \\ 2 e^{3 t} - 2 e^{2 t} & - 2 e^{3 t} + 2 e^{2 t} & e^{3 t} \end{array}] . \end{array}$

For instance, suppose we want to solve the initial value problem

x' (t) = A x + f, x (0) = x 0,

$x ′ (t) = Ax + f, x (0) = x_{0},$ (38)

where A is the matrix in (37), x0=[293743]T, $x_{0} = {[\begin{array}{r} 29 & 37 & 43 \end{array}]}^{T},$ and f=[6000]T. $f = {[\begin{array}{r} 60 & 0 & 0 \end{array}]}^{T} .$ In this case, the variation of parameters formula in Eq. (28) of Section 8.2 gives

e - A t x (t) e - A t x (t) = = = = x 0 + \int t 0 e - A s f d s ⎡ ⎣ ⎢ 293743 ⎤ ⎦ ⎥ + \int t 0 ⎡ ⎣ ⎢ 2 e - 3 s - e - 2 s 2 e - 3 s - 2 e - 2 s 2 e - 3 s - 2 e - 2 s - 2 e - 3 s + 2 e - 2 s - 2 e - 3 s + 3 e - 2 s - 2 e - 3 s + 2 e - 2 s e - 3 s - e - 2 s e - 3 s - e - 2 s e - 3 s ⎤ ⎦ ⎥ ⎡ ⎣ ⎢ 6000 ⎤ ⎦ ⎥ d s ⎡ ⎣ ⎢ 293743 ⎤ ⎦ ⎥ + \int t 0 ⎡ ⎣ ⎢ 120 e - 3 s - 60 e - 2 s 120 e - 3 s - 120 e - 2 s 120 e - 3 s - 120 e - 2 s ⎤ ⎦ ⎥ d s ⎡ ⎣ ⎢ 293743 ⎤ ⎦ ⎥ + ⎡ ⎣ ⎢ 10 - 40 e - 3 t + 30 e - 2 t - 20 - 40 e - 3 t + 60 e - 2 t - 20 - 40 e - 3 t + 60 e - 2 t ⎤ ⎦ ⎥ = ⎡ ⎣ ⎢ 39 - 40 e - 3 t + 30 e - 2 t 17 - 40 e - 3 t + 60 e - 2 t 23 - 40 e - 3 t + 60 e - 2 t ⎤ ⎦ ⎥ .

$\begin{array}{rcl} e^{- A t} x (t) & = & x_{0} + \int_{0}^{t} e^{- A s} f d s \\ = & [\begin{array}{c} 29 \\ 37 \\ 43 \end{array}] \\ + \int_{0}^{t} [\begin{array}{c} 2 e^{- 3 s} - e^{- 2 s} & - 2 e^{- 3 s} + 2 e^{- 2 s} & e^{- 3 s} - e^{- 2 s} \\ 2 e^{- 3 s} - 2 e^{- 2 s} & - 2 e^{- 3 s} + 3 e^{- 2 s} & e^{- 3 s} - e^{- 2 s} \\ 2 e^{- 3 s} - 2 e^{- 2 s} & - 2 e^{- 3 s} + 2 e^{- 2 s} & e^{- 3 s} \end{array}] [\begin{array}{c} 60 \\ 0 \\ 0 \end{array}] d s \\ = & [\begin{array}{c} 29 \\ 37 \\ 43 \end{array}] + \int_{0}^{t} [\begin{array}{c} 120 e^{- 3 s} - 60 e^{- 2 s} \\ 120 e^{- 3 s} - 120 e^{- 2 s} \\ 120 e^{- 3 s} - 120 e^{- 2 s} \end{array}] d s \\ e^{- A t} x (t) & = & [\begin{array}{c} 29 \\ 37 \\ 43 \end{array}] + [\begin{array}{r} 10 - 40 e^{- 3 t} + 30 e^{- 2 t} \\ - 20 - 40 e^{- 3 t} + 60 e^{- 2 t} \\ - 20 - 40 e^{- 3 t} + 60 e^{- 2 t} \end{array}] = [\begin{array}{c} 39 - 40 e^{- 3 t} + 30 e^{- 2 t} \\ 17 - 40 e^{- 3 t} + 60 e^{- 2 t} \\ 23 - 40 e^{- 3 t} + 60 e^{- 2 t} \end{array}] . \end{array}$

Finally, multiplication by eAt $e^{A t}$ yields the solution

x (t) = = ⎡ ⎣ ⎢ 2 e 3 t - e 2 t 2 e 3 t - 2 e 2 t 2 e 3 t - 2 e 2 t - 2 e 3 t + 2 e 2 t - 2 e 3 t + 3 e 2 t - 2 e 3 t + 2 e 2 t e 3 t - e 2 t e 3 t - e 2 t e 3 t ⎤ ⎦ ⎥ ⎡ ⎣ ⎢ 39 - 40 e - 3 t + 30 e - 2 t 17 - 40 e - 3 t + 60 e - 2 t 23 - 40 e - 3 t + 60 e - 2 t ⎤ ⎦ ⎥ ⎡ ⎣ ⎢ - 10 + 67 e 3 t - 28 e 2 t 20 + 67 e 3 t - 50 e 2 t 20 + 67 e 3 t - 44 e 2 t ⎤ ⎦ ⎥

$\begin{array}{rcl} x (t) & = & [\begin{array}{c} 2 e^{3 t} - e^{2 t} & - 2 e^{3 t} + 2 e^{2 t} & e^{3 t} - e^{2 t} \\ 2 e^{3 t} - 2 e^{2 t} & - 2 e^{3 t} + 3 e^{2 t} & e^{3 t} - e^{2 t} \\ 2 e^{3 t} - 2 e^{2 t} & - 2 e^{3 t} + 2 e^{2 t} & e^{3 t} \end{array}] [\begin{array}{c} 39 - 40 e^{- 3 t} + 30 e^{- 2 t} \\ 17 - 40 e^{- 3 t} + 60 e^{- 2 t} \\ 23 - 40 e^{- 3 t} + 60 e^{- 2 t} \end{array}] \\ = & [\begin{array}{r} - 10 + 67 e^{3 t} - 28 e^{2 t} \\ 20 + 67 e^{3 t} - 50 e^{2 t} \\ 20 + 67 e^{3 t} - 44 e^{2 t} \end{array}] \end{array}$

of the initial value problem in (38).

8.3 Problems

1–20. Use projection matrices to find a fundamental matrix solution x(t)=eAt $x (t) = e^{A t}$ of each of the linear systems x'=Ax $x ′ = Ax$ given in Problems 1 through 20 of Section 7.3 .
21–30. Use projection matrices to find a fundamental matrix solution of each of the linear systems given in Problems 1 through 10 of Section 7.6 .
31–40. Use projection matrices, as in Example 8 of this section, to find the matrix exponentials and particular solutions desired in Problems 21 through 30 (respectively) of Section 8.2 .

In each of Problems 41 through 46, use the spectral decomposition methods of this section to find a fundamental matrix solution x(t)=eAt $x (t) = e^{A t}$ for the linear system x'=Ax $x ′ = Ax$ given in the problem.

Use projection matrices as in Example 4 of this section to solve Problems 47 through 50. You can simplify matters by taking c1=[10]T $c_{1} = {[\begin{array}{r} 1 & 0 \end{array}]}^{T}$ and c2=[01]T $c_{2} = {[\begin{array}{r} 0 & 1 \end{array}]}^{T}$ to calculate a typical particular solution x(t)=eA√tc1+e−A√tc2 $x (t) = e^{\sqrt{A} t} c_{1} + e^{- \sqrt{A} t} c_{2}$ that exhibits the fundamental frequencies and modes of oscillation.

..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.

Table of Contents for 8.3 Spectral Decomposition Methods

Create new playlist

Sign In

Sign Up

Table of Contents for
8.3 Spectral Decomposition Methods