Find the eigenvalues and eigenvectors of A geometrically. Where A= (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.
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:
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.
`A-LI = [[-1,1],[1,-1]]`
You want a vector `v` such that `[[-1,1],[1,-1]]v=[,]`
We can get this by trial and error:
`[[-1,1],[1,-1]] [,[z]] = [[-1+z],[1-z]] = [,]`
This will be true if z = 1
And we have:
the eigenvalue L=1 has eigenvector `[,]`
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