# 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.

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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`

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=[,]`

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 `[,]`

Now, take...

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