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the question is as above, I don't know how the rotation matrix works in this question. After giving a formula of proving linear independence, I have no more ideas.

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    The rotation matrix is for $\Bbb R^2$, for $\Bbb R^3$ or something else? – 2017-02-15
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    I don't know!! no more information in the question – 2017-02-15

2 Answers 2

1

$R_\theta = \left( \begin{array}{cc} \cos (\theta) & -\sin (\theta) \\ \sin (\theta) & \cos (\theta) \\ \end{array} \right)$

To be linearly dependent, we must be able to express one row as a (real) multiple $\alpha$ of the other row. That would mean:

  • $\alpha \cos \theta = \sin \theta$
  • $-\alpha \sin \theta = \cos \theta$

This implies that $\alpha = \sqrt{-1}$, and hence is not real. Hence these rows are not dependent, and hence are independent.

2

The rotation matrix is invertible, with $(R (\theta))^{-1} = R(-\theta)$.

Therefore, suppose that the columns of $R(\theta)$ are $C_1,\ldots, C_n$.

Then, if $a_1C_1 + \ldots + a_nC_n = 0$ for some constants $a_i$, then applying $R(-\theta)$ to this equation on both sides, we get $a_1e_1 + a_2e_2 + \ldots + a_ne_n = 0$ (where $e_i$ are the " basis vectors in which the matrix form is expressed". This happens because the inverse returns the columns to the identity matrix, which has columns $e_i$).

This implies that $a_i = 0$ for all $i$ (as the $e_i$ are linearly independent), which then implies that $C_i$ are linearly independent.

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    @DavidG.Stork I don't see the problem. Can you elaborate? – 2017-02-15