Find a 2x2 Matrix A for Given Eigenspaces E_2 and E_4

snoggerT
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Find a 2\times 2 matrix A for which

E_4 = span [1,-1] and E_2 = span [-5, 6]

where E_(lambda) is the eigenspace associated with the eigenvalue (lambda)


relevant equations: Av=(lambda)v

The Attempt at a Solution



I've pretty much gotten most of the eigenspace/value problems down, but this one I'm clueless on how to work. Seems that you would have to work backwards , but I don't know how to do that.
 
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Suppose you have a matrix A and it is diagonalizable. That means you can write it as
A = C D C^{-1}
When A is given, how can you construct the matrices C and D?

Once you answer this question, you'll also see the answer to your problem.
 
You can just do it directly. You are told that
\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)\left(\begin{array}{c}1 \\ -1\end{array}\right)= 4\left(\begin{array}{c}1\\ -1\end{array}\right)= \left(\begin{array}{c}4\\ -4\end{array}\right)
which gives you two equations for a, b, c, d and
\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)\left(\begin{array}{c}-5 \\ 6\end{array}\right)= 2\left(\begin{array}{c}-5\\ 6\end{array}\right)= \left(\begin{array}{c}-10\\ 12\end{array}\right)
which gives you another two equations. Solve those 4 equations.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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