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I am given 2 linear maps: $T,S$, both from $V$ to $V$, satisfying $T^2 = S^2$.

$T,S \ne id$, and $T,S \ne 0$.

The question given is: Does it necessarily mean that $T=S$ or $T=-S$? (or not both?). Prove!

I think that this is not necessarily true, but I can't find a counterexample to support my claim.

Any ideas? thanks.

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4 Answers 4

up vote 8 down vote accepted

It's not true. Let $V=\mathbb R^3$ and consider $T(x,y,z) = (x,-y,z)$, $S(x,y,z)=(x,y,-z)$. Now $S^2=T^2=I$, but $S\neq T$ and $S\neq -T$.

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No.

Hint: Think of operators defined by $Tv=Av,Sv=Bv$ for diagonal $A,B$ with $\pm1$ on the diagonal. Can you construct a counter example ?

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By definition the square of every symmetry is the identity hence the case where $T$ and $S$ are symmetries disproves that $T=\pm S$, say $T=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}$ and $S=\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}$.

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@BenjaLim Right. Thanks. Corrected now. –  Did Jan 14 '13 at 11:35
    
I fell into the same trap before but now I've edited my answer. –  user38268 Jan 14 '13 at 11:37

The following will show you how to construct a big family of examples. Let $V$ be a two dimensional real vector space. Define endomorphisms

$$S = \left[\begin{array}{cc} 0 & 3 \\ 0 & 0 \end{array}\right], T = \left[\begin{array}{cc} 0 & 4 \\ 0 & 0 \end{array}\right]$$

which both square to the zero matrix, but clearly $S \neq T$. Notice this way you can construct a vast amount of matrices that give an example to what you want. Furthermore you can see how this generalises in the case when the vector space is $n$ - dimensional and not just $2$ - dimensional.

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