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So I'm doing some review for a class I have, going over some linear algebra and I don't remember a lot about rotation. Here's what's up:

I constructed a rotation matrix to transform vector components from respect with $\lt \hat x,\hat y,\hat z \gt$ to respect with $\lt \hat u,\hat v, \hat w \gt$ (where $\hat u = \lt 6/7, -3/7, 2/7 \gt, \hat v = \lt 2/7, 6/7$, $3/7 \gt$, and $\hat w = \lt -3/7,6/7,3/7 \gt$).

This is the rotation matrix I obtained: $$ \begin{bmatrix} \hat u \cdot \hat x & \hat u \cdot \hat y & \hat u \cdot \hat z \\ \hat v \cdot \hat x & \hat v \cdot \hat y & \hat v \cdot \hat z \\ \hat w \cdot \hat x & \hat w \cdot \hat y & \hat w \cdot \hat z \\ \end{bmatrix} = \begin{bmatrix} \left(\frac{6}{7}\right) & \left(\frac{-3}{7}\right) & \left(\frac{2}{7}\right) \\ \left(\frac{2}{7}\right) & \left(\frac{6}{7}\right) & \left(\frac{3}{7}\right)\\ \left(\frac{-3}{7}\right) & \left(\frac{-2}{7}\right) & \left(\frac{6}{7}\right) \end{bmatrix} $$ Then I find the new components of several vectors by vector-matrix multiplication. The one vector that caused me problems was vector $v = \lt 2,-2,-2 \gt$. This is what I did: $$A * v=v'$$ $$ \begin{bmatrix} \left(\frac{6}{7}\right) & \left(\frac{-3}{7}\right) & \left(\frac{2}{7}\right) \\ \left(\frac{2}{7}\right) & \left(\frac{6}{7}\right) & \left(\frac{3}{7}\right)\\ \left(\frac{-3}{7}\right) & \left(\frac{-2}{7}\right) & \left(\frac{6}{7}\right) \end{bmatrix} \begin{bmatrix} 2 \\ -2 \\ -2 \\ \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \\ -2 \\ \end{bmatrix} $$ Thus I found that $v=v'$.

My question is this: I don't understand what this means geometrically. I'm having trouble thinking of how I would conceptualise this rotation.

Is $v'$ in the same location as $v$? which was my initial reaction, but that doesn't seem right, since I changed the old components to new components with the rotation matrix. Does any one have any tips to lead me in the right direction to conceptualise this, because I am well and truly stuck.

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What this means geometrically is that $v$ lies on the axis about which the rotation matrix $A$ rotates other points in $\Bbb R^3$. If we try to rotate any point lying on that axis, then we get back the same point after applying the transformation.

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Imagine this, the earth is rotating, but the poles stay stationary.

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Geometrically, the answer by AOrtiz is pretty much spot on. Your rotation matrix doesn't rotate just arbitrarily, it rotates about some line, and things on that line get fixed.

You can notice quite readily that by linearity, this is true fore evry vector on the line spanned by $v$, and so this is the simplest example of an eigenvector of a linear map.

https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors

http://setosa.io/ev/eigenvectors-and-eigenvalues/

You should definitely try exploring this latter page to feel your way around this new concept. It is quite important in not just linear algebra, but throughout mathematics.

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