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Let V be a finite-dimensional vector space and let $f \in End(V)$ be such that $f^3 = 9f$ Show that $f$ is diagonalizable.

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Please say something about where you're getting these questions, as explained in comments on your earlier questions. –  Noah Snyder Nov 3 '12 at 11:54
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Spaces aren't diagonalizable, linear transformations are or aren't. Think of changing your question's title. –  DonAntonio Nov 3 '12 at 12:02
    
How do we show that $X^2 - 9X$ is actually the minimal polynomial though? –  user48557 Nov 7 '12 at 20:22
    
We don't. We do know that the minimal polynomial divides the above polynomial which is sufficient to show that it splits into linear factors. That's all we need. –  EuYu Nov 7 '12 at 20:31
    
Thank you for replying nonetheless! So we know that the minimal polynomial factorizes into distinct monic linear factors because it divides X^3 - 9X which factorizes neatly... –  user48557 Nov 7 '12 at 21:18
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The correct wording of Flute's answer (now deleted) is: $f$ satisfies a polynomial equation where the polynomial (here $X^3-9X$) factors into distinct monic linear factors (here $X$, $X-3$, $X+3$); therefore $f$ must be diagonalizable.

If you know the theorem that says $f$ is diagonalizable if and only if its minimal polynomial splits into distinct monic linear factors, then this follows because the minimal polynomial must divide the polynomial of the equation satisfied by $f$, and therefore also split similarly.

One can actually prove the initial statement (with "$f$ satisfies a polynomial equation...") by explicitly showing that any vector $v$ is a sum of eigenvectors, where thoses eigenvectors can be found inside the subspace spanned by $v,f(v),f^2(v),\ldots,f^{d-1}(v)$ where $d$ is the degree of the polynomial. The argument specialized to the polynomial $X^3-9X$ is as follows.

The vectors $v,f(v),f^2(v),f^3(v)$ are linearly dependent because $f^3(v)-9f(v)=0$, and the first linear dependence involves a polynomial that divides $X^3-9X$; suppose for concreteness it is the polynomial $X^3-9X$ itself, in other words that $v$, $f(v)$ and $f^2(v)$ are linearly independent (the cases for proper divisors of $X^3-9X$ are similar but simpler). Now $$ X^3-9X=X(X^2-9)=(X-3)(X^2+3X)=(X+3)(X^2-3X). $$ The second factors in these decompositions (i.e., $X^2-9$, $X^2+3X$ and $X^2-3X$), evaluated at $X=f$ and then applied to $v$ (so the first one is $f^2(v)-9v$), give nonzero (because $v$, $f(v)$ and $f^2(v)$ are linearly independent) vectors that are annihilated respectively by $f$, by $f-3I$ and by $f+3I$. For instance $(f-3I)(f^2(v)+3f(v))=f^3(v)-9f(v)=0$ for the second one. Therefore they are eigenvectors for eigenvalues $0,3,-3$ respectively, and these eigenvectors span the same space as $v,f(v),f^2(v)$ do; in particular $v$ is a sum of eigenvectors.

Of course this is in general proved more abstractly, which is less messy as well, but it is good to know this is in principle very explicit. And it does not depend on the space $V$ being finite dimensional.

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Let P(X) = X^3 - 9. By hypothesis we know that P(f) = 0. and we know that mf divides P. But P(X) = X(X^2-9) = X(X-3)(X+3). So mf is a product of some of the polynomials X, X-3, X+3. Written precisely, mf(X) = X^r(X -3)^s(X +3)^t with r, s, t ∈{0,1}. It follows that mf is a product of distinct polynomials of degree 1, so f is diagonalizable (and its eigenvalues, being roots of mf are among 0, 3,-3, but we may not have all of them)

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