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Let $E$ be a finite dimensional vector space over a field $F$ and let $T:E\to E$ be a linear transformation. Let $W\subseteq E$ be a subspace such that $T(W)\subseteq W$. Suppose $T$ is diagonalizable. Is $T$ restricted to $W$ also diagonalizable?

my attempts : yes $ T $ is restricted to W is also diagonlisable because T has distinct eigenvalue

pliz help me and tell me the solution

thanks in advance

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  • $\begingroup$ Diagonalisable matrices need not have distinct eigenvalues. $\endgroup$ – Lord Shark the Unknown Mar 18 '18 at 6:56
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Let $\mu \in F[t]$ be the minimal polynomial of $T$. It satisfies $\mu(T) = 0$, and divides any polynomial with the same property. and To say that $T$ is diagonalizable is the same thing as saying that all the roots of $\mu$ are in $F$, and that all these roots are distinct.

If $W$ is a subspace of $T$ which is stable under $T$, then consider the minimal polynomial $g \in F[t]$ of $T|_W$. Since $\mu(T|_W) = 0$, we can conclude that $g$ divides $\mu$. Then all the roots of $g$ are in $F$, and these roots are distinct, so $T|_W$ is diagonalizable.

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If $T$ is diagonalisable, then $E=E_1\oplus E_2\oplus\cdots\oplus E_k$ where the $E_j$ are the eigenspaces of $T$. Then prove that any $W$ which is preserved by $T$ is a sum $W_1+\cdots+W_k$ where $W_j\subseteq E_j$. It may help to note that the projection maps from $E$ to $E_j$ are polynomials in $T$.

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