# Diagonalizable

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Suppose A is diagonalizable with distinct eigenvalues...

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#### Solution Preview

I will use the notation a1...ak for the eigenvalues (rather than using lambda, which is difficult to write).

I will also denote just by m the minimal polynomial of A. We know that because A is diagonalisable:

m(t)=(t-a1)*...*(t-ak)

So P_j(t)=(t-a1)*...*(t-ak)/c_j where in the product the factor (t-aj) is missing, as it ...

#### Solution Summary

This is a proof regarding a diagonalizable matrix and orthogonal projections.

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