Is it possible to give an example of an operator such that $2$ and $3$ are the only eigenvalues and $T^2-5T+6I\neq0?$
My try:We know that if $2$ and $3$ are the only eigen values and $T^2-5T+6I=0$,then $T$ is self adjoint.Thank you.
Is it possible to give an example of an operator such that $2$ and $3$ are the only eigenvalues and $T^2-5T+6I\neq0?$
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linear-algebra
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1Observe that $\;x^2-5x+6=(x-2)(x-3)\implies\;$ you need an operator $\;T\;$ such that the above quadratic **is not** its minimal polynomial (and thus the operator is non-diagonalizable), so **any** non-diagonalizable operator will do. – 2017-02-23
1 Answers
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$$ \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} $$
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0It is also written that $T\in L(C^3)$ – 2017-02-23
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2@MatheMagic ...and that operator can be thought of as in $ L(\mathbb C^3) $... – 2017-02-23
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0can you plz explain what does that $L(C^3)$ mean ? – 2017-02-23
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1@MatheMagic It's the complex vector space of linear maps $ \mathbb C^3 \to \mathbb C^3 $. – 2017-02-23