$T: V \to V$ is a linear transformation such that $T^3 = T^2$, $T^2 \ne T$ and $T$ is not nilpotent. Is $T$ diagonalizable?
I haven't tried much since I don't really know where to start. Thanks in advance.
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Do you know about Jordan Normal Form? – Git Gud Sep 19 '13 at 22:37
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@GitGud No, I don't. – Nikki Sep 19 '13 at 22:42
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What about Cayley–Hamilton Theorem? – Git Gud Sep 19 '13 at 22:42
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Yes, I know that theorem. – Nikki Sep 19 '13 at 22:45
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How about $\large \left{0, 1\right}$ eigenvalues ? – Felix Marin Sep 19 '13 at 22:47
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@Nikki Use theo theorem in this answer. – Git Gud Sep 19 '13 at 22:50
2 Answers
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Hint: Use the fact that $T$ is diagonalizable if and only if its minimal polynomial has single roots. Now, translate each of the three pieces of information you have received into a statement about the minimal polynomial of $T$.
Christopher A. Wong
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Since $T^3 = T^2$, all eigenvalues $\lambda$ of $T$ must satisfy $\lambda^3 = \lambda^2$, so all eigenvalues are either $0$ or $1$. Suppose $T$ is diagonalizable. Then $T^2$ and $T^3$ are diagonalizable with the same eigenvectors and eigenvalues, and you have for any vector $x$ that $x = \sum_i a_i v_i$ where $v_i$ are the eigenvectors of $T$, and then because $T^3 = T^2$ you have $T^3(x) = \sum_i a_i \lambda_i^3 v_i = T^2(x) = \sum_i a_i \lambda_i^2 v_i$ where $\lambda_i$ are the eigenvalues of $T$. But this also equals $\sum_i a_i \lambda_i v_i = T(x)$, so $T = T^2$.
user2566092
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