Chapter 7 · Section 7.8

7.8 Basic Theorems on Homogeneous Linear Systems

In Sections 7.6 and 7.7 we used several facts without proof: that $n$ independent solutions give every solution, that the Wronskian can be checked at a single point, that $e^{A t} = Φ (t) Φ (0)^{- 1}$ , and that eigenvectors of distinct eigenvalues are independent. This section proves them. The proofs are short and are written in small steps. They also work for systems whose coefficients depend on $t$ , which the eigenvalue method cannot handle, so the results here are more general than anything we have used so far. Section 7.3 did the same for $2 \times 2$ systems; Section 4.6 did it for second-order equations. You will recognize many arguments.

Throughout, $I$ is an interval of the real line (open, closed or half-open), and we consider

x^{'} = A (t) x (H) and x^{'} = A (t) x + f (t) (N),

where $A (t) = (a_{ij} (t))$ is an $n \times n$ matrix function and $f (t)$ a vector function, all of whose entries are continuous on $I$ . A solution on $I$ is a differentiable vector function that satisfies the equation at every $t$ in $I$ .

Existence and uniqueness

Everything rests on one theorem. We state it without proof; one proof applies the method of successive approximations (Section 8.3) to the system, and it works because linear right-hand sides satisfy a Lipschitz condition on every closed bounded subinterval of $I$ .

Theorem 7.8.1 (Existence and uniqueness)

Let the entries of $A (t)$ and $f (t)$ be continuous on the interval $I$ , let $t_{0}$ be a point of $I$ , and let $x_{0}$ be any constant vector. Then the initial-value problem

x^{'} = A (t) x + f (t), x (t_{0}) = x_{0}

has exactly one solution, and this solution is defined on the whole interval $I$ .

Two features deserve attention. First, the solution exists on all of $I$ , not just near $t_{0}$ . This is special to linear systems; a nonlinear equation such as $y^{'} = y^{2}$ can blow up in finite time even though its right side is smooth. Second, for constant $A$ and $f = 0$ we can take $I = (- \infty, \infty)$ , so every solution of $x^{'} = A x$ is defined for all $t$ .

The following consequence is used in almost every proof below.

Theorem 7.8.2 (A solution that vanishes once vanishes everywhere)

If $x$ is a solution of (H) on $I$ and $x (t_{0}) = 0$ for one point $t_{0}$ in $I$ , then $x (t) = 0$ for all $t$ in $I$ .

Proof. The zero function is a solution of (H) with value $0$ at $t_{0}$ . So is $x$ . By the uniqueness part of Theorem 7.8.1 the two solutions coincide. $■$

Geometrically, for $n = 2$ and constant $A$ this says what we saw in the phase portraits of Section 7.6: no trajectory other than the equilibrium itself ever reaches the origin. More generally, two different solutions can never take the same value at the same time.

Superposition

Theorem 7.8.3 (Superposition principle)

If $x_{1}, \dots, x_{k}$ are solutions of (H) on $I$ and $c_{1}, \dots, c_{k}$ are constants, then $x = c_{1} x_{1} + \dots + c_{k} x_{k}$ is also a solution of (H) on $I$ .

Proof.

Differentiate term by term: $x^{'} = c_{1} x_{1}^{'} + \dots + c_{k} x_{k}^{'}$ .
Use that each $x_{j}$ is a solution: $x_{j}^{'} = A x_{j}$ . So $x^{'} = c_{1} A x_{1} + \dots + c_{k} A x_{k}$ .
Matrix multiplication is distributive and constants can be moved through $A$ : $c_{1} A x_{1} + \dots + c_{k} A x_{k} = A (c_{1} x_{1} + \dots + c_{k} x_{k}) = A x$ . $■$

The theorem fails for (N): the sum of two solutions of (N) has forcing $2 f$ , not $f$ . That is why the homogeneous system comes first.

Linear independence and the Wronskian

Recall from Definition 7.5.8 that vector functions $x_{1}, \dots, x_{n}$ are linearly dependent on $I$ if some combination with constant coefficients, not all zero, vanishes at every $t$ in $I$ .

Definition 7.8.1 (Wronskian)

The Wronskian of $n$ vector functions $x_{1}, \dots, x_{n}$ , each with $n$ components, is the determinant of the $n \times n$ matrix having them as columns:

W (t) = W (x_{1}, \dots, x_{n}) (t) = det [x_{1} (t) x_{2} (t) \dots x_{n} (t)] .

The first result holds for any vector functions, solutions or not.

Theorem 7.8.4 (A nonzero Wronskian implies independence)

If $W (x_{1}, \dots, x_{n}) (t_{0}) \neq = 0$ for at least one $t_{0}$ in $I$ , then $x_{1}, \dots, x_{n}$ are linearly independent on $I$ .

Proof. Suppose $c_{1} x_{1} (t) + \dots + c_{n} x_{n} (t) = 0$ for all $t$ in $I$ . In particular this holds at $t_{0}$ . In matrix form it says $Φ (t_{0}) c = 0$ , where $Φ (t_{0}) = [x_{1} (t_{0}) \dots x_{n} (t_{0})]$ and $c = (c_{1}, \dots, c_{n})^{T}$ . Since $det Φ (t_{0}) = W (t_{0}) \neq = 0$ , Theorem 7.5.2 gives $c = 0$ . So the only vanishing combination is the trivial one. $■$

The converse is false for general vector functions, as the next example shows. It becomes true for solutions of (H); that is Theorem 7.8.7 below.

Example 7.8.1

Let $f_{1} (t) = (1, t)^{T}$ and $f_{2} (t) = (t, t^{2})^{T}$ . Show that they are independent on every interval, although their Wronskian is identically zero. Conclude that they cannot both be solutions of one system $x^{'} = A (t) x$ with $A$ continuous on an interval.

Solution

Wronskian. $W (t) = 1 t t t^{2} = t^{2} - t^{2} = 0$ for all $t$ .
Independence. If $c_{1} f_{1} + c_{2} f_{2} = 0$ for all $t$ in an interval, the first component gives $c_{1} + c_{2} t = 0$ there. A linear polynomial that vanishes at more than one point is the zero polynomial, so $c_{1} = c_{2} = 0$ .
Conclusion. If they were solutions of such a system, Theorem 7.8.7 (proved below) would say: independent solutions have a Wronskian that is never zero. But here $W \equiv 0$ . So no continuous $A (t)$ can have both as solutions. Figure 7.8.1 shows what goes wrong: at each fixed $t$ the two vectors are parallel, although the multiplier $t$ between them changes with time.

Figure 7.8.1. The vectors

f_{1} (t) = (1, t)^{T}

(solid) and

f_{2} (t) = (t, t^{2})^{T}

(dashed) for four values of

t

. At every instant they lie on one line, so

W (t) = 0

; yet no single constant multiplier works for all

t

Liouville's formula

For solutions of (H) the Wronskian satisfies a simple first-order differential equation of its own. To derive it we need a rule for differentiating a determinant.

Theorem 7.8.5 (Derivative of a determinant)

If the entries of an $n \times n$ matrix $Φ (t)$ are differentiable, then $\frac{d}{d t} det Φ$ is the sum of $n$ determinants; the $i$ -th of them is $det Φ$ with row $i$ replaced by its derivative.

Proof. For $n = 2$ , by the product rule,

\frac{d}{d t} (ϕ_{11} ϕ_{22} - ϕ_{12} ϕ_{21}) = (ϕ_{11}^{'} ϕ_{22} - ϕ_{12}^{'} ϕ_{21}) + (ϕ_{11} ϕ_{22}^{'} - ϕ_{12} ϕ_{21}^{'}) = ϕ_{11}^{'} ϕ_{21} ϕ_{12}^{'} ϕ_{22} + ϕ_{11} ϕ_{21}^{'} ϕ_{12} ϕ_{22}^{'} .

For general $n$ the same thing happens: a determinant is a sum of terms $\pm ϕ_{1 j_{1}} ϕ_{2 j_{2}} \dots ϕ_{n j_{n}}$ , each containing exactly one entry from each row. Differentiating a product of $n$ factors by the product rule gives $n$ terms, in the $i$ -th of which only the factor from row $i$ is differentiated. Collecting, for each $i$ , the terms in which row $i$ was differentiated gives exactly the determinant with row $i$ replaced by its derivative. $■$

Theorem 7.8.6 (Liouville's formula; Abel's formula for systems)

Let $x_{1}, \dots, x_{n}$ be solutions of (H) on $I$ , let $W (t)$ be their Wronskian and let $t_{0}$ be in $I$ . Then for all $t$ in $I$

W (t) = W (t_{0}) exp (\int_{t_{0}}^{t} tr A (s) d s), tr A = a_{11} + a_{22} + \dots + a_{nn} .

Consequently $W$ is either identically zero on $I$ or never zero on $I$ .

Proof.

Rows of $Φ^{'}$ . Let $Φ = [x_{1} \dots x_{n}]$ . Since every column solves (H), $Φ^{'} = A Φ$ . Reading this row by row: row $i$ of $Φ^{'}$ equals $a_{i 1} \cdot (row 1 of Φ) + \dots + a_{in} \cdot (row n of Φ)$ .
One of the $n$ determinants. By Theorem 7.8.5, $W^{'}$ is a sum of $n$ determinants; in the $i$ -th, row $i$ of $Φ$ is replaced by the combination in step 1. A determinant is linear in each row, so this determinant splits into $n$ pieces: $a_{ik}$ times the determinant of $Φ$ with row $i$ replaced by row $k$ . For $k \neq = i$ that matrix has two equal rows, so its determinant is $0$ . Only $k = i$ survives, and it gives $a_{ii} det Φ = a_{ii} W$ .
Add up. Summing over $i$ , $W^{'} = (a_{11} + a_{22} + \dots + a_{nn}) W = tr A (t) W .$
Solve. This is a first-order linear equation for $W$ . Multiplying by the integrating factor $exp (- \int_{t_{0}}^{t} tr A d s)$ shows that $W (t) exp (- \int_{t_{0}}^{t} tr A d s)$ has derivative zero, hence equals its value $W (t_{0})$ at $t_{0}$ . This is the formula.
Zero or never zero. The exponential is never zero. So if $W (t_{0}) = 0$ , then $W \equiv 0$ on $I$ ; if $W (t_{0}) \neq = 0$ , then $W (t) \neq = 0$ for every $t$ in $I$ . $■$

Geometrically, for $n = 2$ the Wronskian is the signed area of the parallelogram spanned by $x_{1} (t)$ and $x_{2} (t)$ (for $n = 3$ , the volume of a parallelepiped). Liouville's formula says that this area changes at the relative rate $tr A$ . In particular it can shrink to zero only in the limit $t \to \pm \infty$ , never at a finite time (Figure 7.8.2).

Figure 7.8.2. Parallelograms spanned by the columns of

e^{A t}

at several times. Left:

A = (11 - 2 - 1)

tr A = 0

, so the area stays equal to

W (0) = 1

while the shape rotates and shears. Right:

A = (- 4 2 1 - 3)

(Example 7.6.2),

tr A = - 7

, so the area is

e^{- 7 t}

Connection with Section 4.6

The equation $y^{''} + p (t) y^{'} + q (t) y = 0$ is equivalent to the system $x^{'} = (0 - q 1 - p) x$ with $x = (y, y^{'})^{T}$ . For two solutions $y_{1}, y_{2}$ the Wronskian of the vector solutions $(y_{1}, y_{1}^{'})^{T}$ , $(y_{2}, y_{2}^{'})^{T}$ is exactly the Wronskian $y_{1} y_{2}^{'} - y_{1}^{'} y_{2}$ of Chapter 4, and the trace is $- p$ . So Liouville's formula reduces to Abel's formula $W (t) = W (t_{0}) exp (- \int_{t_{0}}^{t} p d s)$ (Theorem 4.6.5).

Example 7.8.2

Verify that $x_{1} = (t, 1)^{T}$ and $x_{2} = (t^{3}, 3 t^{2})^{T}$ solve $x^{'} = (0 - 3/ t^{2} 1 3/ t) x$ on $I = (0, \infty)$ , and check Liouville's formula with $t_{0} = 1$ .

Solution

Continuity. The entries $- 3/ t^{2}$ and $3/ t$ are continuous on $(0, \infty)$ (but not at $t = 0$ ), so the theorems apply on $I = (0, \infty)$ .
$x_{1}$ . $x_{1}^{'} = (1, 0)^{T}$ and $A x_{1} = (1, - 3 t / t^{2} + 3/ t)^{T} = (1, 0)^{T}$ .
$x_{2}$ . $x_{2}^{'} = (3 t^{2}, 6 t)^{T}$ and $A x_{2} = (3 t^{2}, - 3 t^{3} / t^{2} + 9 t^{2} / t)^{T} = (3 t^{2}, 6 t)^{T}$ .
Wronskian. $W (t) = t 1 t^{3} 3 t^{2} = 3 t^{3} - t^{3} = 2 t^{3}$ , so $W (1) = 2$ .
Liouville. $tr A = 3/ t$ and $\int_{1}^{t} 3/ s d s = 3 ln t$ , so the formula predicts $W (t) = 2 e^{3 l n t} = 2 t^{3}$ . It agrees. Note that $W \neq = 0$ on all of $(0, \infty)$ , as Theorem 7.8.6 requires; it vanishes only at $t = 0$ , where $A$ is not continuous.

Theorem 7.8.7 (Independence of solutions)

Let $x_{1}, \dots, x_{n}$ be solutions of (H) on $I$ . The following are equivalent:

$x_{1}, \dots, x_{n}$ are linearly independent on $I$ ;
$W (t) \neq = 0$ for every $t$ in $I$ ;
$W (t_{0}) \neq = 0$ for at least one $t_{0}$ in $I$ .

Proof. Statements 2 and 3 are equivalent by the "zero or never zero" part of Theorem 7.8.6, and 3 implies 1 by Theorem 7.8.4. It remains to show that 1 implies 3. We show the contrapositive: if 3 fails, then 1 fails.

Suppose $W (t_{0}) = 0$ at some (hence every) point $t_{0}$ . The columns of $Φ (t_{0})$ are then dependent (Theorem 7.5.3): there is $c \neq = 0$ with $c_{1} x_{1} (t_{0}) + \dots + c_{n} x_{n} (t_{0}) = 0$ .
Define $x (t) = c_{1} x_{1} (t) + \dots + c_{n} x_{n} (t)$ . By superposition (Theorem 7.8.3) it is a solution of (H), and $x (t_{0}) = 0$ .
By Theorem 7.8.2, $x (t) = 0$ for all $t$ in $I$ . That is a vanishing combination with constants not all zero, so the solutions are dependent on $I$ . $■$

This is the result we used in Sections 7.6 and 7.7: to show that $n$ solutions are independent, it suffices to evaluate their Wronskian at one convenient point, usually $t = 0$ .

Fundamental sets and the general solution

Definition 7.8.2 (Fundamental set, fundamental matrix)

A set of $n$ linearly independent solutions $x_{1}, \dots, x_{n}$ of (H) on $I$ is a fundamental set. The matrix $Φ (t) = [x_{1} (t) \dots x_{n} (t)]$ having them as columns is a fundamental matrix. By Theorem 7.8.7 a fundamental matrix satisfies

Φ^{'} (t) = A (t) Φ (t) and det Φ (t) \neq = 0 for every t in I .

Theorem 7.8.8 (Fundamental sets exist)

The system (H) has a fundamental set on $I$ .

Proof. Fix $t_{0}$ in $I$ and let $e_{1}, \dots, e_{n}$ be the columns of the identity matrix. By Theorem 7.8.1 there is, for each $j$ , a solution $x_{j}$ on $I$ with $x_{j} (t_{0}) = e_{j}$ . Their Wronskian at $t_{0}$ is $det I = 1 \neq = 0$ , so by Theorem 7.8.7 they are independent. $■$

Theorem 7.8.9 (General solution of (H))

Let $x_{1}, \dots, x_{n}$ be a fundamental set of (H) on $I$ , with fundamental matrix $Φ$ . Then every solution of (H) on $I$ can be written as

x (t) = c_{1} x_{1} (t) + \dots + c_{n} x_{n} (t) = Φ (t) c

for exactly one constant vector $c$ . In particular, the solution with $x (t_{0}) = x_{0}$ is $x (t) = Φ (t) Φ (t_{0})^{- 1} x_{0}$ .

Proof.

Choose the constants. Let $y$ be any solution and $t_{0}$ a point of $I$ . Since $det Φ (t_{0}) \neq = 0$ , the linear system $Φ (t_{0}) c = y (t_{0})$ has exactly one solution $c = Φ (t_{0})^{- 1} y (t_{0})$ (Theorem 7.5.2).
Compare. The function $z = Φ c - y$ is a solution of (H) by superposition, and $z (t_{0}) = 0$ by the choice of $c$ .
Conclude. By Theorem 7.8.2, $z \equiv 0$ , i.e. $y = Φ c$ on all of $I$ . Uniqueness of $c$ : if $Φ c = Φ d$ , evaluate at $t_{0}$ and multiply by $Φ (t_{0})^{- 1}$ . $■$

In the language of linear algebra, Theorems 7.8.3, 7.8.8 and 7.8.9 together say that the solutions of (H) form a vector space of dimension exactly $n$ , and a fundamental set is a basis of it. This is why the general solution of an $n \times n$ system always contains exactly $n$ arbitrary constants.

A fundamental matrix is not unique; the next theorem tells us how any two are related, and justifies the formula for $e^{A t}$ used in Section 7.7.

Theorem 7.8.10 (Properties of fundamental matrices)

If $Φ$ is a fundamental matrix and $C$ is a constant nonsingular $n \times n$ matrix, then $Φ C$ is a fundamental matrix.
If $Φ$ and $Ψ$ are fundamental matrices of the same system (H), then $Ψ = Φ C$ for some constant nonsingular matrix $C$ .
If $A$ is constant and $Φ$ is any fundamental matrix of $x^{'} = A x$ , then $e^{A t} = Φ (t) Φ (0)^{- 1}$ .

Proof.

Part 1. $(Φ C)^{'} = Φ^{'} C = A Φ C = A (Φ C)$ , so each column of $Φ C$ is a solution; and $det (Φ C) = det Φ \cdot det C \neq = 0$ , so they are independent.
Part 2. Each column $ψ_{j}$ of $Ψ$ is a solution, so by Theorem 7.8.9 $ψ_{j} = Φ c_{j}$ for a constant vector $c_{j}$ . Putting these side by side, $Ψ = Φ C$ with $C = [c_{1} \dots c_{n}]$ . Taking determinants, $det Ψ = det Φ \cdot det C$ ; the left side is nonzero, so $det C \neq = 0$ .
Part 3. By Part 1 (with $C = Φ (0)^{- 1}$ ), $Φ (t) Φ (0)^{- 1}$ is a fundamental matrix, and it equals $I$ at $t = 0$ . In Section 7.7 we saw that the columns of $e^{A t}$ solve $x^{'} = A x$ and that $e^{A \cdot 0} = I$ . So column $j$ of each matrix is the solution with initial value $e_{j}$ at $t = 0$ . By uniqueness these columns coincide. $■$

Careful

In Part 1 the constant matrix must multiply from the right. $C Φ$ is usually not a fundamental matrix, because $(C Φ)^{'} = C A Φ$ , which equals $A (C Φ)$ only if $C A = A C$ (see Exercise 5).

Example 7.8.3

For the compartment system of Example 7.6.2, $A = (- 4 2 1 - 3)$ , compute $e^{A t}$ from the fundamental matrix found there, and use it to recover the solution with $x (0) = (6, 0)^{T}$ .

Solution

Fundamental matrix. $Φ (t) = (e^{- 2 t} 2 e^{- 2 t} e^{- 5 t} - e^{- 5 t})$ , so $Φ (0) = (12 1 - 1)$ , with determinant $- 3$ .
Inverse. $Φ (0)^{- 1} = \frac{1}{- 3} (- 1 - 2 - 1 1) = \frac{1}{3} (12 1 - 1)$ .
Product. $e^{A t} = Φ (t) Φ (0)^{- 1} = \frac{1}{3} (e^{- 2 t} + 2 e^{- 5 t} 2 e^{- 2 t} - 2 e^{- 5 t} e^{- 2 t} - e^{- 5 t} 2 e^{- 2 t} + e^{- 5 t}) .$ Check: at $t = 0$ this is $\frac{1}{3} (3003) = I$ .
Solution. $e^{A t} (6, 0)^{T}$ is $6$ times the first column of $e^{A t}$ : $(2 e^{- 2 t} + 4 e^{- 5 t}, 4 e^{- 2 t} - 4 e^{- 5 t})^{T}$ , as in Example 7.6.2.

Example 7.8.4

Check Liouville's formula for the three solutions found in Example 7.7.2.

Solution

Value at $0$ . At $t = 0$ the solutions are $(0, 1, 0)^{T}$ , $a = (1, - 1, 1)^{T}$ and $b = (- 1, 0, 0)^{T}$ . Expanding along the first column, $W (0) = 010 1 - 1 1 - 1 00 = - 1 \cdot 11 - 1 0 = - 1.$
Prediction. $tr A = - 3 - 3 + 1 = - 5$ (which also equals the sum of the eigenvalues $- 3 + (- 1 + 2 i) + (- 1 - 2 i)$ ). For a constant matrix Liouville's formula reads $W (t) = W (0) e^{(tr A) t} = - e^{- 5 t}$ .
Direct computation. Factor $e^{- 3 t}$ from the first column and $e^{- t}$ from the other two; expanding the remaining determinant along its first column leaves $- (cos^{2} 2 t + sin^{2} 2 t) = - 1$ . So $W (t) = - e^{- 3 t} e^{- t} e^{- t} = - e^{- 5 t}$ , as predicted.

The nonhomogeneous system

Theorem 7.8.11 (General solution of (N))

Let $Φ$ be a fundamental matrix of (H) on $I$ and let $x_{p}$ be any one solution of (N) on $I$ . Then every solution of (N) on $I$ has the form

x (t) = Φ (t) c + x_{p} (t)

for a unique constant vector $c$ . One particular solution is

x_{p} (t) = Φ (t) \int_{t_{0}}^{t} Φ (s)^{- 1} f (s) d s,

which is the solution with $x_{p} (t_{0}) = 0$ .

Proof.

Differences solve (H). If $x$ is any solution of (N), then $(x - x_{p})^{'} = A x + f - A x_{p} - f = A (x - x_{p})$ . By Theorem 7.8.9, $x - x_{p} = Φ c$ for a unique $c$ .
The formula. Put $u (t) = \int_{t_{0}}^{t} Φ (s)^{- 1} f (s) d s$ . The integrand is continuous because $det Φ \neq = 0$ on $I$ and the entries of $Φ^{- 1}$ are quotients of continuous functions by $det Φ$ . By the fundamental theorem of calculus, $u^{'} = Φ^{- 1} f$ .
Check. By the product rule, $(Φ u)^{'} = Φ^{'} u + Φ u^{'} = A Φ u + Φ Φ^{- 1} f = A (Φ u) + f$ . So $x_{p} = Φ u$ solves (N), and $u (t_{0}) = 0$ gives $x_{p} (t_{0}) = 0$ . $■$

Combining with Theorem 7.8.9, the solution of (N) with $x (t_{0}) = x_{0}$ is

x (t) = Φ (t) Φ (t_{0})^{- 1} x_{0} + Φ (t) \int_{t_{0}}^{t} Φ (s)^{- 1} f (s) d s,

and for constant $A$ (using $Φ (t) = e^{A t}$ and $e^{A t} e^{- A s} = e^{A (t - s)}$ , valid because $A t$ and $- A s$ commute) this becomes $x (t) = e^{A (t - t_{0})} x_{0} + \int_{t_{0}}^{t} e^{A (t - s)} f (s) d s$ , the exact analogue of the scalar formula of Section 2.3.

Eigenvectors of distinct eigenvalues

Finally we prove the general form of Theorem 7.5.4, on which Theorems 7.6.2 and 7.7.1 rest.

Theorem 7.8.12 (Distinct eigenvalues, independent eigenvectors)

Let $v_{1}, \dots, v_{k}$ be eigenvectors of a matrix $A$ belonging to eigenvalues $λ_{1}, \dots, λ_{k}$ that are all different. Then $v_{1}, \dots, v_{k}$ are linearly independent.

Proof.

Assume the opposite. Suppose some nontrivial combination of the $v_{j}$ is $0$ . Among all such combinations pick one with the fewest nonzero coefficients. After renumbering, it reads $c_{1} v_{1} + \dots + c_{m} v_{m} = 0$ with all $c_{j} \neq = 0$ . Since eigenvectors are nonzero, $m \geq 2$ .
Apply $A - λ_{m} I$ . Because $(A - λ_{m} I) v_{j} = (λ_{j} - λ_{m}) v_{j}$ , we get $c_{1} (λ_{1} - λ_{m}) v_{1} + \dots + c_{m - 1} (λ_{m - 1} - λ_{m}) v_{m - 1} = 0 .$ The $v_{m}$ term has disappeared.
Contradiction. The coefficients $c_{j} (λ_{j} - λ_{m})$ , $j < m$ , are nonzero because $c_{j} \neq = 0$ and the eigenvalues are distinct. So this is a nontrivial vanishing combination with only $m - 1$ nonzero coefficients, fewer than the minimum we chose. This is impossible, so the eigenvectors are independent. $■$

Exercises

Show that $x_{1} = (1/ t, - 1/ t^{2})^{T}$ and $x_{2} = (t^{2}, 2 t)^{T}$ are solutions of $x^{'} = (0 2/ t^{2} 10) x$ on $(0, \infty)$ . Compute their Wronskian and explain the result with Liouville's formula.
Answer
$W (t) = 2 + 1 = 3$ for all $t$ ; constant because $tr A = 0$ . The two solutions form a fundamental set on $(0, \infty)$ .
The coefficients of a system satisfy $tr A (t) = t - 1/ t$ on $(0, \infty)$ , and two solutions have Wronskian $W (1) = 3$ . Find $W (t)$ .
Answer
$W (t) = 3 exp (\int_{1}^{t} (s - 1/ s) d s) = \frac{3}{t} e^{(t^{2} - 1) /2}$
Let $x_{1} = (t, 1)^{T}$ and $x_{2} = (t^{2}, 2 t)^{T}$ . Can they both be solutions of a system $x^{'} = A (t) x$ with $A$ continuous on $(- 1, 1)$ ? On $(0, \infty)$ ? In the second case find $A (t)$ .
Answer
$W (t) = t^{2}$ vanishes at $t = 0$ but not elsewhere, which contradicts Theorem 7.8.6 on $(- 1, 1)$ : impossible there. On $(0, \infty)$ it is possible: $A = Φ^{'} Φ^{- 1} = (0 - 2/ t^{2} 1 2/ t)$ .
Use $e^{A t} = Φ (t) Φ (0)^{- 1}$ to compute $e^{A t}$ for $A = (1320)$ (Example 7.6.1).
Answer
$e^{A t} = \frac{1}{5} (3 e^{3 t} + 2 e^{- 2 t} 3 e^{3 t} - 3 e^{- 2 t} 2 e^{3 t} - 2 e^{- 2 t} 2 e^{3 t} + 3 e^{- 2 t})$
Let $A = (1002)$ , $Φ = (e^{t} 0 0 e^{2 t})$ and $C = (0110)$ . Show that $Φ C$ is a fundamental matrix of $x^{'} = A x$ but $C Φ$ is not.
Answer
$Φ C = (0 e^{2 t} e^{t} 0)$ : its columns are the same two solutions, swapped. $C Φ = (0 e^{t} e^{2 t} 0)$ : the column $(0, e^{t})^{T}$ fails $x_{2}^{'} = 2 x_{2}$ .
Prove: if two solutions $x$ and $y$ of (H) satisfy $x (t_{1}) = y (t_{1})$ at one point $t_{1}$ of $I$ , then $x = y$ on all of $I$ .
Answer
$x - y$ solves (H) by superposition and vanishes at $t_{1}$ ; apply Theorem 7.8.2.
Three solutions of $x^{'} = A x$ with $A = - 2 01 531 1 - 4 - 4$ have $W (0) = 2$ . Find $W (t)$ without solving the system. Are they a fundamental set?
Answer
$tr A = - 3$ , so $W (t) = 2 e^{- 3 t}$ ; it is never zero, so yes.
Solve $x^{'} = (- 4 2 1 - 3) x + (0 3 e^{- 2 t})$ , $x (0) = 0$ , with the formula of Theorem 7.8.11 (compare Example 7.7.8).
Answer
$x = e^{- 2 t} (t - \frac{1}{3}, 2 t + \frac{1}{3})^{T} + \frac{1}{3} e^{- 5 t} (1, - 1)^{T}$
Prove: if $x_{1}, \dots, x_{n}$ are solutions of (H) and the vectors $x_{1} (t_{0}), \dots, x_{n} (t_{0})$ are linearly dependent for one $t_{0}$ in $I$ , then the functions are linearly dependent on $I$ .
Answer
Then $W (t_{0}) = 0$ (Theorem 7.5.3), and Theorem 7.8.7 gives dependence on $I$ . Directly: with $Φ (t_{0}) c = 0$ , $c \neq = 0$ , the solution $Φ c$ vanishes at $t_{0}$ , hence everywhere.
Write $y^{''} + p (t) y^{'} + q (t) y = 0$ ( $p, q$ continuous on $I$ ) as a system for $x = (y, y^{'})^{T}$ and deduce Abel's formula for the Wronskian $y_{1} y_{2}^{'} - y_{1}^{'} y_{2}$ of two solutions from Liouville's formula.
Answer
$A = (0 - q 1 - p)$ , $tr A = - p$ , so $y_{1} y_{2}^{'} - y_{1}^{'} y_{2} = W (t_{0}) exp (- \int_{t_{0}}^{t} p (s) d s)$ .