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Let $S = \{v_1,\ldots,v_n\} \subset \mathbb{R}^n$ and let $T = \{w_1,\ldots,w_n\} \subset \mathbb{R}^n$ be such that the angle between $v_i$ and $v_j$ is equal to the angles between $w_i$ and $w_j$. Suppose we know $S$ is linearly independent.

  1. Can we say $T$ is linearly independent?

  2. Can we say $T$ is the image of $S$ under an orthogonal transformation? (see edit)

  3. Is $T$ just $S$ rotated by some angle? (see edit)

This isn't homework, I'd just like to know this since it's related to a problem about reflection groups that I'm working on. For what it's worth, I'm thinking of $S$ as a simple system in a root system.

EDIT: OK, clearly questions 2 and 3 are not true, as the comments make clear. However, I think something may be able to be said about them ... maybe.

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  • $\begingroup$ What I can tell you is: $T$ isn't just $S$ rotated by an angle. Consider in $\mathbb{R}^3$, $S = \{ (1,0,0), (0,1,0), (0,0,1) \}$ and $T = \{ (2,0,0), (0,2,0), (0,0,2) \}$, for instance. My intuition says that $T$ will be linearly independent, but I'll think more about it. $\endgroup$ – Ivo Terek May 24 '14 at 0:46
  • $\begingroup$ Questions 2 and 3 are easy "no"s, unless you wish to add another hypothesis, such as unit length. $\endgroup$ – Lee Mosher May 24 '14 at 0:46
  • $\begingroup$ Up to a scalar multiplication of each of the $w_i$s, $T$ is an orthogonal transformation of $S$. I have edited my answer to explain this. $\endgroup$ – 6005 May 24 '14 at 3:54
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Response to edit.

  • If you additionally assume that $\| v_i \| = \| w_i \|$ for all $i$, then $T$ is linearly independent, and moreover, $T$ is $S$ multiplied by an orthogonal matrix. (Multiplication by an orthogonal matrix is an isometry.) To see this, note that $\| v_i \| = \| w_i \|$ for all $i$ along with the angles between vectors being the same implies that $v_i \cdot v_j = w_i \cdot w_j$ for all $i, j$, and apply the proposition below.

  • Without this additional assumption, then $T$ is still linearly independent, because we can scale each $w_i$ to have the same norm as $v_i$ and then apply the first bullet point. But $T$ will not necessarily be a orthogonal transformation of $S$.

In summary, the answer to question (1) is yes, and the answer to (2) is yes up to scalar multiplication of each vector.


Proposition. Let $S = \{v_1, \ldots, v_n\}$ and $T = \{w_1, \ldots, w_n\}$, such that $v_i \cdot v_j = w_i \cdot w_j$ for all $i, j$ not necessarily distinct. Suppose further that $S$ is linearly independent. Then $T$ is linearly independent, and there is an orthogonal matrix $Q$ such that $w_i = Qv_i$ for all $i$.

Proof. Let $$ A = \begin{bmatrix} \quad v_1 \quad \\ \quad v_2 \quad \\ \cdots \\ \quad v_n \quad \\ \end{bmatrix} \;, \quad B = \begin{bmatrix} \quad w_1 \quad \\ \quad w_2 \quad \\ \cdots \\ \quad w_n \quad \\ \end{bmatrix} $$

Since, $v_i \cdot v_j = w_i \cdot w_j$ for all $i, j$, we have $$ A A^T = B B^T $$ Since $S$ is linearly independent, $\det A \ne 0$, i.e. $A$ is invertible. The above implies $(\det A)^2 = (\det B)^2$, so $\det B \ne 0$ also. We conclude $B$ is invertible, and we also conclude the first statement that $T$ is linearly independent.

Now consider $$ Q = B^{-1} A $$ Note that \begin{align*} Q Q^T &= B^{-1} A A^T (B^{-1})^T = B^{-1} B B^T (B^T)^{-1} = (I)(I) = I \\ \end{align*} so $Q$ is orthogonal. Moreover, $$ Q(A^T) = B^{-1} A A^T = B^{-1} B B^T = B^T $$ i.e. $Qv_i = w_i$ for all $i$.

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If you assume, in addition, that $||v_i||=||w_i||$ for all $i$ then you can derive that $W$ is linearly independent in another way to @6005's answer: if $W$ is linearly dependent then there is some combination $\sum_{i=1}^n a_i w_i=0$ where not all $a_i$ are zero.

$$ 0= (\sum_{i=1}^n a_i w_i,\sum_{i=1}^n a_i w_i) = \sum_{i=1}^n\sum_{j=1}^n a_i a_j(w_i,w_j) = \sum_{i=1}^n\sum_{j=1}^n a_i a_j(v_i,v_j) = (\sum_{i=1}^n a_i v_i,\sum_{i=1}^n a_i v_i) $$

Thus $\sum_{i=1}^n a_i v_i=0$ where not all $a_i=0$ in contradiction to $S$ being linearly independent.

This approach has the advantage that $|S|$ and $|W|$ do not have to be the dimension of the space.

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