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

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

- Aug 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?

- Aug 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.

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

$\displaystyle P=\begin{bmatrix}a &b\\ c &d\end{bmatrix}.$

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

$\displaystyle \displaystyle{P^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d &-b\\ -c &a\end{bmatrix}}.$

Multiply out $\displaystyle P^{-1}AP,$ and show that if you set it equal to

$\displaystyle \begin{bmatrix}k_{1} &0\\ 0 &k_{2}\end{bmatrix},$

for real $\displaystyle k_{1}, k_{2},$ then there is no solution. Does that make sense? - Aug 3rd 2010, 06:33 AMStuck Man
Yes, I have already multiplied it out. It is quite lengthy to do.

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

$\displaystyle \displaystyle{P^{-1}AP=\begin{bmatrix}\frac{-bc+(a+c)d}{ad-bc} &\frac{d^{2}}{ad-bc}\\ -\frac{c^{2}}{ad-bc} &\frac{ad-c(b+d)}{ad-bc}\end{bmatrix}=\frac{1}{ad-bc}\begin{bmatrix}-bc+(a+c)d &d^{2}\\ -c^{2} &ad-c(b+d)\end{bmatrix}.}$

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? - Aug 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.

- Aug 3rd 2010, 07:01 AMAckbeet
Well, that would only indicate that $\displaystyle k_{1}=k_{2}=0$ is a possibility. What about that denominator out front?

- Aug 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. - Aug 3rd 2010, 08:32 AMAckbeet
I'm not sure you're seeing the right thing here. If $\displaystyle c=d=0$, then what must $\displaystyle ad-bc=$? And what is $\displaystyle ad-bc$ relative to $\displaystyle P$?

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

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

- Aug 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.

- Aug 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.