Given an operator T that has in some basis matrix M(T)= prove there is no eigenbasis for T.
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Originally Posted by tjkubo Given an operator T that has in some basis matrix M(T)= prove there is no eigenbasis for T. This matrix isn't diagonalizable since the eigenspace is less n.
In other words, find the eigenvalues and all eigenvectors corresponding to those eigenvalues. If, for an n by n matrix, there are not n independent eigenvectors, they cannot form a basis for the n dimensional space and so there is no "eigenbasis".
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