Find the eigenvalues and eigenvectors of A geometrically. Where A=[0110] (a reflection in the line y=x). I tried this and got the eigenvalues of 1 and -1 but i'm not sure how to explain it and how to find the eigenvectors for each eigenvalue.
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The algorithm for finding the eigenvalues is:
det (A - LI) = 0
where A is your matrix, I is the identity matrix, and the possibilities for L are the eigenvalues. Thus:
`"det"[[0-L,1],[1,0-L]]=0`
`L^2-1=0`
`L= +-1`
So, your eigenvalues are correct.
To find the eigenvectors, take each possibility for L, plug it into A - LI, and find the null space. That is, find a vector which, if you multiplied it by A-LI, would be 0.
So:
Take L=1
`A-LI = [[-1,1],[1,-1]]`
You want a vector `v` such that `[[-1,1],[1,-1]]v=[[0],[0]]`
We can get this by trial and error:
`[[-1,1],[1,-1]] [[1],[z]] = [[-1+z],[1-z]] = [[0],[0]]`
This will be true if z = 1
And we have:
the eigenvalue L=1 has eigenvector `[[1],[1]]`
Now, take...
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