**J**=(λ, 0, 1, λ), where λ is an arbitrary real number. (2x2 matrix, λ and 0 on the left, 1 and λ on the right.)

**J**^2,

**J**^3, and

**J**^4.

**J**^n=(λ^n, 0, nλ^(n-1), λ^n).

**J**t).

**J**t) to solve the initial value problem x'=

**J**x, x(0)=x^0.

Let **J**=(λ, 0, 1, λ), where λ is an arbitrary real number. (2x2 matrix, λ and 0 on the left, 1 and λ on the right.)

(a) Find **J**^2, **J**^3, and **J**^4.

(b) Use an inductive argument to show that **J**^n=(λ^n, 0, nλ^(n-1), λ^n).

(c) Determine exp(**J**t).

(d) Use exp(**J**t) to solve the initial value problem x'=**J**x, x(0)=x^0.

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Let me continue the Andre's answer:

For e^{J}^{t} he obtained the following:

e^{Jt}=Σ_{n=0}^{∞}(1/n!){(λ^{n}, 0);(nλ^{n-1}, λ^{n})}t^{n}=Σ_{n=0}^{∞}{( (λ^{n}t^{n})/n!, 0);( (nλ^{n-1}t^{n})/n!, (λ^{n}t^{n})/n!)}=

=[Σ_{n=1}^{∞}{( (λt)^{n}/n!, 0); ( (λt)^{n-1}t/(n-1)!, (λt)^{n}/n!)}]+{(1,0);(0,1)}={ (e^{λt}, 0); (te^{λt}; e^{λt}) }

So, e^{J}^{t}={ (e^{λt}, 0); (te^{λt}; e^{λt}) }

The initial value problem x′=**J**x; x(0)=x^0; has a solution in the form:

x=x^0*e^{J}^{t}; Since we determined e^{J}^{t} in the previous part, we can formally multiply matrix e^{J}^{t} by a vector x^0 to get a solution. I assume x^0 means (1,1) vector. Then the solution is:

x(t)=(e^{λt},(t+1)e^{λt})

I'm assuming you know how to multiply matrices and what a Taylor series is. If not, let me know.

(a) J² just means multiply matrix J by itself, J² = J*J. Similarly, J³=J*J*J etc.

Let's do J² for your matrix:

J²=J*J=(λ, 0, 1, λ)*(λ, 0, 1, λ)=(λ², 0, 2λ, λ²)

Similarly,

J³=J²*J=(λ², 0, 2λ, λ²)*(λ, 0, 1, λ)=(λ³, 0, 3λ², λ³)

J^{4}=J³*J=(λ³, 0, 3λ, λ³)*(λ, 0, 1, λ)=(λ^{4}, 0, 4λ³, λ^{4})

(b) In part (a), we see a pattern emerging, so we claim

J^{n}=(λ^{n}, 0, nλ^{n-1}, λ^{n}) for any positive integer n.

The inductive argument goes like this:

1. The statement is true for n=1, because

J^{1}=J=(λ, 0, 1, λ) by definition.

2. Let the statement be true for some arbitrary but fixed n=k:

J^{k}=(λ^{k}, 0, kλ^{k-1}, λ^{k})

Then for n=k+1

J^{k+1}=J^{k}*J=(λ^{k}, 0, kλ^{k-1}, λ^{k})* (λ, 0, 1, λ)=(λ^{k+1}, 0, (k+1)λ^{k}, λ^{k+1}).

Therefore, by the principle of mathematical induction, the statement is true for all n=1,2,3,...

(c) The matrix exponential exp(A) of a matrix A is defined as a Taylor series of matrices analogous to the Taylor series of the function e^{t} about t=0:

exp(A)= I + A + 1/2! A² + 1/3! A³ + .... =Σ( (1/n!) A^{n}, n=0..infinity),

where I is the identity matrix and A^{0}=I.

Therefore,

exp(Jt) = Σ( (1/n!) (Jt)^{n}, n=0..infinity) = Σ( (1/n!) J^{n}t^{n}, n=0..infinity)

For our matrix J=(λ, 0, 1, λ), we showed in part (b) J^{n}=(λ^{n}, 0, nλ^{n-1}, λ^{n}), so that

exp(Jt) =Σ( (1/n!) (λ^{n}, 0, nλ^{n-1}, λ^{n}) t^{n}, n=0..infinity)

We can distribute the summation and the t^{n} into each entry and turn the series of matrices into a matrix of series. We get

exp(Jt) = (Σ (1/n!)λ^{n}t^{n}, 0, Σ (1/n!) nλ^{n-1}t^{n}, Σ (1/n!) λ^{n}t^{n})

Now use the definition of the Taylor series of the *function* e^{λt} about t=0 and get

exp(Jt) = ( e^{λt}, 0, t e^{λt}, e^{λt}).

(d) We have the initial value problem

x'=Jx, with x(0)=x_{0}

I claim that the solution is

x= exp(Jt) x_{0}.

Actually, this statement is true for *all* matrices J, but the proof is especially simple for the triangular matrix J given in this problem, J=(λ, 0, 1, λ). I will show that x' and Jx are equal to the same thing.

First,

x' = (exp(Jt) x_{0})' = d/dt (e^{λt}, 0, t e^{λt}, e^{λt}) x_{0} = ( λ e^{λt}, 0, λt e^{λt} + e^{λt}, λ e^{λt}) x_{0}

Second,

J x = (λ, 0, 1, λ)*(exp(Jt) x_{0}) = (λ, 0, 1, λ)*(e^{λt}, 0, t e^{λt}, e^{λt}) x_{0}

= ( λ e^{λt}, 0, λt e^{λt}+ e^{λt}, λ e^{λt}) x_{0}

Both sides of x'=Jx are equal to ( λ e^{λt}, 0, λt e^{λt}+ e^{λt}, λ e^{λt}) x_{0}, so

x= exp(Jt) x_{0}

is a solution to the equation. By the uniqueness theorem it is the *only* solution. qed

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