Complex numbers + linear algebra = :S

[philnow]

Homework Statement

Find all complex numbers Z (if any) such that the matrix: (it is 2 by 2)

(2)(-1)
(4)(2)

multiplied by a vector V = ZV has a nonzero solution V.

part b)

for each Z that you found, find all vectors V such that (the same matrix)*V=ZV.

The Attempt at a Solution

I'm not sure that I even know what this question is asking... could anyone clarify this a little?

 

[philnow]

And please don't mistake my lack of effort for laziness, I genuinely don't understand the question.

 

[Hurkyl]

The question is "solve Av=zv".

(A is the matrix you were given)
(v and z are the indeterminate variables you're solving for)
(v is a vector variable)
(z is a complex variable)

 

[Dick]

In other words, as Hurkyl clarified, it's asking you for eigenvalues and eigenvectors of your matrix. Search on those keywords if you need examples.

 

[philnow]

So I'm looking for complex numbers such that, when multiplied by a vector, it gives the same results as the matrix multiplied by the same vector?

 

[Dick]

So I'm looking for complex numbers such that, when multiplied by a vector, it gives the same results as the matrix multiplied by the same vector?

Yes, the number times the vector should be the same as the matrix times the vector.

 

[philnow]

I googled eigenvectors/values but I don't see the relevance to this question :s

How would I find a set of complex numbers that multiplies the same way as a matrix of real numbers? Any hints please?

 

[Dick]

I'm surprised you didn't see the relevance. But look, let v be column vector (x,y). So z*v is column vector (z*x,z*y) What is Av written out as a column vector? Now equate the two column vectors so you have a system of two equations, write them down. What condition does z have to satisfy to get a solution where x and y are not both zero? It's a quadratic. It has two roots. They are complex. See if you can find them. Look back at examples of finding eigenvalues again. Because that is what you are doing.

 

[philnow]

That makes a lot of sense, thanks Dick :)