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Math Help - Matrix transformation

  1. #1
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    Matrix transformation

    How do I start this question? Do I let P equal the 2x2 matrix abcd?
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  2. #2
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    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?
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  3. #3
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    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.
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  4. #4
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    If you don't know what an eigenvector is, then I suppose you could set

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

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

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

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

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

    for real k_{1}, k_{2}, then there is no solution. Does that make sense?
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  5. #5
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    Yes, I have already multiplied it out. It is quite lengthy to do.
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  6. #6
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    Ah. So I get

    \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?
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  7. #7
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    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.
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  8. #8
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    Well, that would only indicate that k_{1}=k_{2}=0 is a possibility. What about that denominator out front?
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  9. #9
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    (-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.
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  10. #10
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    I'm not sure you're seeing the right thing here. If c=d=0, then what must ad-bc=? And what is ad-bc relative to P?
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  11. #11
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    I see, P must be a null matrix.
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  12. #12
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    I don't exactly know what you mean by "null matrix". What is that?
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  13. #13
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    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.
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  14. #14
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    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.
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