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This is a problem from Hoffman and Kunze (Sec 3.4, problem 10, page 96).

If T is a linear operator on $R^2$ defined by $T(x_1,x_2)=(x_1,0)$ we see that it is represented in the standard basis by the matrix A such that $A(1,1)=1$ and rest of the entires are $0$. This operator satisfies $T^2=T$. Prove that if $S$ is any linear operator satisfying $S^2=S$, then $S=0$ or $S=I$ or there is an ordered basis $\beta$ for $R^2$ such that $[S]_{\beta}= A$ (above)

Any hints to solve this problem would be greatly appreciated.

Thank you.

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up vote 2 down vote accepted

Here is an outline of the proof:
1) Prove that if $S^2=S$ then for any $v\in {\rm{Im}}(S)$ you have $Sv=v$.
2) Assume that $S\neq 0$. Then you can find $v\neq 0$ such that $Sv=v$.
3) Assume that $S\neq Id$. Then you can find $w\neq 0$ such that $Sw=0$.
4) If $S\neq 0$ and $S\neq Id$, denote $B=(v,w)$ from the previous steps. $B$ is a basis for $\mathbb{R}^2$ (why?). Now compute $[S]_B$ (the matrix of $S$ represented w.r.t the basis $B$). You should get $A$ :)

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Thanks a lot for your detailed hint. –  Shibi Vasudevan Jun 11 '11 at 18:09
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Hint: You can distinguish three cases according as the dimension of the image of $S$ is $0$, $1$ or $2$.

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Thanks for your hint. –  Shibi Vasudevan Jun 11 '11 at 18:10
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