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A linear operator $T$ on a complex vector space $V$ has characteristic polynomial $x^3(x-5)^2$ and minimal polynomial $x^2(x-5)$.

Show that the operator induced by $T$ on the quotient space $V/\operatorname{ker} (T-5I)$ is nilpotent

My try:

The Jordan Canonical form of the matrix of $T$ will consist of a Jordan block of order 2 corresponding to $\lambda =0$ and another Jordan block of order $1$.

Similarly The Jordan Canonical form of the matrix of $T$ will consist of a Jordan block of order $1$ corresponding to $\lambda =5$ and another Jordan block of order $1$.

How can I find from here the the operator induced by $T$ on the quotient space $V/\operatorname{ker} (T-5I)$ ?

Please give some hints.

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Your assertions about the Jordan form of $T$ are correct. That is, the Jordan form is $$ J= \left[\begin{array}{rrrrr} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0 & 5 \end{array}\right] $$ Note that $\dim\ker(T^k-\lambda I)=\dim\ker(J^k-\lambda I)$ for any $k$ and $\lambda$ since $T$ is similar to its Jordan form. In particular, \begin{align*} \dim\ker(T^2) &= 3 & \dim\ker(T-5\,I)=2 \end{align*} Of course, $\ker(T^2)\cap\ker(T-5\,I)=\{\mathbf 0\}$ so $V$ has a basis $\{v_1,v_2,v_3,w_1,w_2\}$ where $\{v_1,v_2,v_3\}$ is a basis for $\ker T^2$ and $\{w_1,w_2\}$ is a basis for $\ker(T-5\,I)$.

The quotient space $V/\ker(T-5\,I)$ then has a basis $\{[v_1],[v_2],[v_3]\}$ and the map $V/\ker(T-5\,I)\to V/\ker(T-5\,I)$ induced by $T$ is $T([x])=[T(x)]$.

Is this induced map nilpotent? To answer this question, note that every $[x]$ in the quotient space decomposes as $$ [x]=\alpha\cdot[v_1]+\beta\cdot[v_2]+\gamma\cdot[v_3] $$ for constants $\alpha$, $\beta$, and $\gamma$. What can we say about $T^k([x])$?

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  • $\begingroup$ I don't get why $dim ker T^2=3$ even if it is true why is the intersection null $\endgroup$ – Learnmore Apr 30 '16 at 2:42
  • $\begingroup$ Can you please explain the facts $\endgroup$ – Learnmore Apr 30 '16 at 2:42
  • $\begingroup$ @learnmore To see why the intersection is trivial see math.stackexchange.com/questions/1557676/… $\endgroup$ – Brian Fitzpatrick Apr 30 '16 at 8:42
  • $\begingroup$ you did not say why $\dim \ker T^2=3$ $\endgroup$ – Learnmore May 1 '16 at 4:34

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