Explain in detail please! Let A be a 2 times 2 matrix and consider the different

Question

Explain in detail please!

Let A be a 2 times 2 matrix and consider the differential equation where y = y(t) is a vector function of t in the form y(t) = [y1(t), y2(t)]T. Suppose that the matrix A has characteristic polynomial with a root lambda of multiplicity two. Suppose also that there are linearly independent vectors v1 and v2 in R2 such that y = elambdatv2 + telambdatv1 is a solution to the differential equation. Prove that v1 and v2 must satisfy the equations. Let Show that CA(lambda) = (lambda - 3)2. Find the eigenspace of A for lambda = 3, and show that it is one dimensional. Find a vector v2 such that Av2 = 3v2 + v1. Use the method of part (a) so give one solution to the differential equation y' = Ay for A as in part (b). Check your result.

Explanation / Answer

(a)

Suppose y = e^(Lt) v2 + te^(Lt) v1 is a solution to y'=Ay

But y' = e^(Lt) ( Lv2 + (1+Lt) v1 ) = Ay = e^(Lt)A(v2 + tv1)

=>

Lv2 + (1+Lt) v1 = A(v2+tv1) (since e^(Lt) > 0)

(Lv2-Av2+v1) + t(Lv1-Av1) = 0

Since a polynom of the form P(t)=a+tb can't have infinite roots, unless a=b=0 then :

Lv2-Av2+v1 = 0 and Lv1-Av1=0 => Av1=Lv1 and Av2=Lv2+v1

(b)

C(L) = det(A-LI)=(6-L)(-L)+9 = L^2-6L+9=(L-3)^2

Solve AX=3X with X=(a b)^T gives :

6a+9b = 3a => 3a+9b=0 => -a = 3b

-a = 3b

So it's one dimensional and a non-zero eigenvector v1 = (-3,1)^T

Av2 = 3v2+(-3,1)^T gives the system :

6a+9b = 3a-3 => a+3b=-1

-a = 3b+1      => a+3b=-1

So we can take the vector v2 = (2,1)

(c)

In (a) we have computed v1, v2 such that :

Av1=Lv1

Av2 = 3v2+v1

Since v1,v2 are linearly independant then the solution is

y(t) = e^(Lt) v2 + te^(Lt) v1

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