Ask a question

Please help me with this math problem?

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(Jt).
(d) Use exp(Jt) to solve the initial value problem x'=Jx, x(0)=x^0.

2 Answers by Expert Tutors

Tutors, sign in to answer this question.
Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
4.9 4.9 (174 lesson ratings) (174)
Let me continue the Andre's answer:
For eJt he obtained the following:
eJtn=0(1/n!){(λn, 0);(nλn-1, λn)}tnn=0{( (λntn)/n!, 0);( (nλn-1tn)/n!, (λntn)/n!)}=
=[Σn=1{( (λt)n/n!, 0); ( (λt)n-1t/(n-1)!, (λt)n/n!)}]+{(1,0);(0,1)}={ (eλt, 0); (teλt; eλt) }
So, eJt={ (eλt, 0); (teλt; eλt) }
The initial value problem x′=Jx; x(0)=x^0; has a solution in the form:
x=x^0*eJt; Since we determined eJt in the previous part, we can formally multiply matrix eJt by a vector x^0 to get a solution. I assume x^0 means (1,1) vector. Then the solution is:
Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
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)
J³=J²*J=(λ², 0, 2λ, λ²)*(λ, 0, 1, λ)=(λ³, 0, 3λ², λ³)
J4=J³*J=(λ³, 0, 3λ, λ³)*(λ, 0, 1, λ)=(λ4, 0, 4λ³, λ4)
(b) In part (a), we see a pattern emerging, so we claim
Jn=(λn, 0, nλn-1, λn) for any positive integer n.
The inductive argument goes like this:
1. The statement is true for n=1, because
J1=J=(λ, 0, 1, λ) by definition.
2. Let the statement be true for some arbitrary but fixed n=k:
Jk=(λk, 0, kλk-1, λk)
Then for n=k+1
Jk+1=Jk*J=(λk, 0, kλk-1, λk)* (λ, 0, 1, λ)=(λk+1, 0, (k+1)λk, λk+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 et about t=0:
exp(A)= I + A + 1/2! A² + 1/3! A³ + .... =Σ( (1/n!) An, n=0..infinity),
where I is the identity matrix and A0=I.
exp(Jt) = Σ( (1/n!) (Jt)n, n=0..infinity) = Σ( (1/n!) Jntn, n=0..infinity)
For our matrix J=(λ, 0, 1, λ), we showed in part (b) Jn=(λn, 0, nλn-1, λn), so that
exp(Jt) =Σ( (1/n!) (λn, 0, nλn-1, λn) tn, n=0..infinity)
We can distribute the summation and the tn into each entry and turn the series of matrices into a matrix of series. We get
exp(Jt) = (Σ (1/n!)λntn, 0, Σ (1/n!) nλn-1tn, Σ (1/n!) λntn)
Now use the definition of the Taylor series of the function eλt about t=0 and get
exp(Jt) = ( eλt, 0, t eλt, eλt).

(d) We have the initial value problem
x'=Jx, with x(0)=x0
I claim that the solution is
x= exp(Jt) x0.
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.
x' = (exp(Jt) x0)' =  d/dt (eλt, 0, t eλt, eλt) x0 =  ( λ eλt, 0, λt eλt + eλt, λ eλt) x0
J x = (λ, 0, 1, λ)*(exp(Jt) x0) = (λ, 0, 1, λ)*(eλt, 0, t eλt, eλt) x0
= ( λ eλt, 0, λt eλt+ eλt, λ eλt) x0
Both sides of x'=Jx are equal to ( λ eλt, 0, λt eλt+ eλt, λ eλt) x0, so
x= exp(Jt) x0
is a solution to the equation. By the uniqueness theorem it is the only solution. qed