#Linear Algebra - Eigenstuff
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eigenvectors corresponding to distinct eigenvalues are linearly independent
so U interset W = {0}
provided that both +1 and -1 are eigenvalues
if you take bases of the eigenspaces and put them together, you get a basis for V (because U direct sum W = V)
matrix is diagonalisable iff V has a basis of eigenvectors
@novel tulip
ive got it thank you
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