Linear Algebra? (Complex eigenvalues/spaces)?

Question

Can someone give me a hand with question 2(a) here: http://conley.math.ualberta.ca/joso/courses/math225/HW3.pdf

The eigenvalues I've found are = 2+2i, and = 2-2i

I'm having the most problem with the eigenspace part

Answer 1

The eigenvalues are the roots of the characteristic polynomial. You should have see that before.

An eigenspace is the solution to :

(A-lamda*I)x = 0, for eigenvalue lamda, I is the identity, x is a vector.

None of this should be news to you. I just don't want to do your homework for you.

Answer 2

Remember what your ultimate goal in this problem was:
to find nonzero solutions to Ax = lambda*x
You've found your possible lambdas, now plug each of them into the equation to find x.

For the first eigenvalue, you'll get:
[[0,1],[-8,4]] [[x1],[x2]] = (2 + 2i) [[x1],[x2]]

Which will multiply out to give you two equations:
x2 = (2 + 2i) x1
-8x1 + 4x2 = (2 + 2i) x2

Since (2+2i) is an eigenvalue, we know that these two equations are linearly depentent (give you the same solution set). So let's use the simpler one, the first one.

You have one degree of freedom, since you have one equation and two unknowns. What you want is a single non-zero vector to be your basis. So you can choose a value for one of your variables, and solve for the other. I usually use 1 for the value of one of the variables: it makes the math much easier!

If you set x1 = 1 in this case, you get x2 = 2 + 2i
So a basis for your eigenspace is: {<1, 2+2i>}

I'm SURE you can do the other one!

Answer 3

a million. 4x - 3y = 8. -8 + 6y = 16. Simplify the 2d equation: upload 8 to both part, giving 6y = 24. Divide both part via 6, giving y = 4. Use this result contained in the first equation via substituting the y-fee: 4x - 3y = 8 will grow to be 4x - 3(4) = 8, or 4x - 12 = 8. upload 12 to both part, giving 4x = 20. Divide both part via 4, giving x = 5. the answer element is then (5,4). 2. enable R be the usual fee, and P the proper type fee. we are provided that P = R + 0.20. the cost of 11 gallons of customary, in money, is 11R. the cost of 16 gallons of proper type is 16P = 16(R + 0.20). the completed is 11R + 16(R + 0.20) = fifty 8.fifty 5, or 11R + 16R + 3.20 = fifty 8.fifty 5, or 27R + 3.20 = fifty 8.fifty 5, or 27R = fifty 5.35, or R = $2.05. P = R + 0.20 = $2.25, answer b.