How do I start this question? Do I let P equal the 2x2 matrix abcd?

Printable View

- August 3rd 2010, 04:58 AMStuck ManMatrix transformation
How do I start this question? Do I let P equal the 2x2 matrix abcd?

- August 3rd 2010, 05:38 AMAckbeet
I would go about it by showing that A doesn't have 2 linearly independent eigenvectors. Then there's a theorem that states you can't diagonalize an n x n matrix unless you have n linearly independent eigenvectors. Make sense?

- August 3rd 2010, 05:50 AMStuck Man
I don't know what an eigenvector is. I think diagonalising is a method of finding an inverse matrix but I'm not familiar with that either. I'll look into these things.

- August 3rd 2010, 06:02 AMAckbeet
If you don't know what an eigenvector is, then I suppose you could set

Then you can compute the "inverse", if it exists, as

Multiply out and show that if you set it equal to

for real then there is no solution. Does that make sense? - August 3rd 2010, 06:33 AMStuck Man
Yes, I have already multiplied it out. It is quite lengthy to do.

- August 3rd 2010, 06:45 AMAckbeet
Ah. So I get

Is that what you get? If so, you can tell, almost by inspection, that you can't make this into the diagonal matrix specified above. Why is that? - August 3rd 2010, 07:00 AMStuck Man
Yes I got the same. If c and d equal 0 then k1 and k2 would be 0 since they only comprise of symbols multiplied by c or d.

- August 3rd 2010, 07:01 AMAckbeet
Well, that would only indicate that is a possibility. What about that denominator out front?

- August 3rd 2010, 07:23 AMStuck Man
(-c^2)/(ad-bc) = (d^2)/(ad-bc) = 0

Multiply all by ad-bc then (-c^2) = (d^2). Any valued squared is positive so c and d can only be 0. - August 3rd 2010, 08:32 AMAckbeet
I'm not sure you're seeing the right thing here. If , then what must ? And what is relative to ?

- August 3rd 2010, 09:26 AMStuck Man
I see, P must be a null matrix.

- August 3rd 2010, 09:33 AMAckbeet
I don't exactly know what you mean by "null matrix". What is that?

- August 3rd 2010, 09:45 AMStuck Man
I see what you mean now. The discriminant of P is 0 hence P is a singular matrix. A null matrix or zero matrix has every element 0.

- August 3rd 2010, 09:50 AMAckbeet
Well, the correct term is "determinant". But you have the idea. P is technically a null matrix as well. But what's important for your problem is that it is singular. So I'd say you're pretty much done, then.