If $T:V\to V$ is diagonalizable and $S \subset V$ is a $T$-invariant subspace, I have to show that $T|_{S}$ is diagonalizable, but without using the minimal polynomial.
I can not think how to do it without using the minimal polynomial.
Thanks
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I actually think it's easier without the minimal polynomial. This is hardly formal, but think of $S$ as corresponding to a part of the diagonal matrix. A part of a diagonal matrix is still a diagonal matrix. Making this into a formal statement is quite straight forward, in terms of kernels, basis vectors, transition matrices, or really whatever you choose. – GPerez Mar 29 '14 at 00:51
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i can do it: take ${\v_{1},..v_{p} }$ basis of $S$. after complete this basis with eigenvectors: $B={\v_{1},..v_{p}, u_{p+1},...,u_{n} }$ , and make matrix change basis with basis $D$ with all eigenvectors. And the block in matrix change is the matrix change for $T|_{S}$ ? – user89940 Mar 29 '14 at 01:09
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1Does this answer your question? Diagonalizable transformation restricted to an invariant subspace is diagonalizable – PatrickR Apr 28 '21 at 07:03
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@PatrickR yes look fine – user89940 Apr 28 '21 at 11:47