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Let $V$ be an $n$ dimensional real vector space. Let $S = \{ v_1, ... , v_m \}$ be a spanning set for $V$ not containing the zero vector. Suppose that $\phi_i: V \rightarrow V, i = 1, 2$ are linear transformations which send $\alpha := v_1$ to $-\alpha = -v_1$, which fix pointwise a subspace $W_i$ of dimension $n-1$, and which map $\{v_1, ... , v_m\}$ into itself. Why is it the case that $\phi_1 = \phi_2$?

Attempt: Let $T = \{v_t, ... , v_m\}$ be a maximal subset of $S$ whose span $W'$ does not contain $\alpha$. Then $W'$ is $n-1$ dimensional, and we have a linear transformation $\phi': V \rightarrow V$ which fixes $W'$ pointwise and sends $\alpha$ to $-\alpha$.

So we just have to show that if $\phi$ is a linear operator sending $\alpha$ to $-\alpha$, fixing pointwise a subspace $W$ of dimension $n-1$, and permuting $S$, then $\phi = \phi'$.

The only thing I can think of is to suppose that $W' \neq W$. Then by a dimension argument, $V = W + W'$. We can write $$\alpha = w + \sum\limits_{i=t}^m c_i \alpha_i$$ so $-\alpha = w + \sum\limits_i c_i\alpha_{\sigma i}$, where $\sigma$ is the permutation of $S$ given by $\phi$. Adding these equations, we get that $w$ is in the span of $\alpha_i, \alpha_{\sigma i}, i = t, ... , m$.

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  • $\begingroup$ It might be helpful to notice that both $\phi_i$ are diagonalizable with $V_{-1}(\phi_i) = \langle e_1 \rangle$ and $V_1(\phi_i) = W_i$. $\endgroup$ Jun 11, 2016 at 3:39
  • $\begingroup$ Sorry, edited. $\alpha$ is $v_1$ ${}$ $\endgroup$
    – D_S
    Jun 11, 2016 at 12:28
  • $\begingroup$ So the $\phi_1, \phi_2$ can be represented by the same diagonal matrix with one $-1$ and $(n-1)$ $1$s on the diagonal. If I can show they commute, then I could simultaneously diagonalize them, and that just about would show they are equal $\endgroup$
    – D_S
    Jun 11, 2016 at 12:30

1 Answer 1

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Let $G$ be the subgroup of $GL(V)$ generated by $\phi_1$, $\phi_2$. $G$ is finite, as its action on $S$ defines an injection $G \to Aut(S)$ into a the finite group of permutations of $S$. Hence there is a $G$-invariant scalar product $\langle-,-\rangle \colon V \times V \to \mathbb{R}$. Thus $\phi_1$ and $\phi_2$ must fix the orthogonal complement of $v_1$ with respect to $\langle-,-\rangle$, thus they are equal.

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  • $\begingroup$ Thanks for answering. So just to be clear, if $W_i$ is a subspace of dimension $n-1$ fixed pointwise by $\phi_i$, and $w \in W_i$, then $$\langle v_1, w \rangle = \langle \phi_i(v_1), \phi_i(w) \rangle = \langle - v_1, w \rangle = - \langle v_1, w \rangle$$ whence $\langle v_1, w \rangle = 0$. Thus $W_1, W_2$ are both contained in the orthogonal complement $W$ of $v_1$ with respect to $\langle - , -\rangle$. Since they all have the same dimension, $W_1 = W_2 = W$, so $\phi_1 = \phi_2$. Is this correct? $\endgroup$
    – D_S
    Jun 22, 2016 at 15:19
  • $\begingroup$ Yes, I think that is correct. $\endgroup$ Jun 23, 2016 at 11:58

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