Explore BrainMass


Suppose A is diagonalizable with distinct eigenvalues...

See attached file for full problem description.


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:
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.