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发帖数: 5298 | 1 If x is A's eigenvector, i.e., Ax=ax
Then Bx saitisfies, A(Bx)=B(Ax)=a(Bx).
If x is nondegenerate, then Bx=bx, i.e., x is B's eigenvector.
If x is degenerate, then Bx=\sum c_i x_i where Ax_i=a x_i.
We can find linearly independent combinations of x_i, say X_i, that satisfies
B X_i= b_i X_i.
In rephrase, this means we can diaognalize B in the subspace
of eigenvalue a of A. |
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