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$$K=\ \begin{pmatrix} 0 & 0 & 1\\ -1 & 0 & 0\\ 0 & -1 & 0 \end{pmatrix}$$

Find the axis of rotation for the rotation matrix $K$ by INSPECTION.

This is from my other thread click here to view it

Everything you see below is me finding the axis of rotation by solving $Kv=v$. Just to show you how much working it requires:

Noting that the axis of rotation consists of vectors that remain unmoved. That is a vector $v$ satisfying $Kv = v$. Or, $Kv - Iv=0$ where $I$ is the $3\times3$ identity matrix. For matrix $K$ after solving the homogeneous equations given by $(K-I)v=0$ and showing the working:

$(K-I)v=0$

So $$K-I=\ \begin{pmatrix} 0 & 0 & 1\\ -1 & 0 & 0\\ 0 & -1 & 0 \end{pmatrix}-\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}=\ \begin{pmatrix} -1 & 0 & 1\\ -1 & -1 & 0\\ 0 & -1 & -1 \end{pmatrix}$$

therefore

$$\begin{pmatrix} -1 & 0 & 1\\ -1 & -1 & 0\\ 0 & -1 & -1 \end{pmatrix}v=0$$

writing out the components for $v$ gives

$$\begin{pmatrix} -1 & 0 & 1\\ -1 & -1 & 0\\ 0 & -1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}=0$$

Multiplying out gives three equations

$-x+z=0$

$-x-y=0$

$-y-z=0$

Since $$ v=\begin{bmatrix}x\\y\\z\end{bmatrix} $$ Here's the solution parametrically in terms of $x$ \begin{align*} z&= x\\ y&=-x\\ x&=x \end{align*} Hence the axis of rotation is given by the line $$ \begin{bmatrix} x\\-x\\x \end{bmatrix}=x\begin{bmatrix}1\\-1\\1\end{bmatrix}\quad x\in\Bbb R $$ That is, the axis of rotation is $$ \operatorname{Span}\left\{\begin{bmatrix}1\\-1\\1\end{bmatrix}\right\} $$

As you can see this was a lot of work so i would be so grateful if someone could please explain in simple english how to get the answer: $$ \operatorname{Span}\left\{\begin{bmatrix}1\\-1\\1\end{bmatrix}\right\} $$ by using Inspection?

Many thanks to all that helped so far particularly Brian Fitzpatrick in the last thread

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    $\begingroup$ Why didn't you row-reduce the matrix $K - I$ using Gaussian elimination? You can find your axis much faster that way. $\endgroup$ – André 3000 Apr 23 '14 at 22:28
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$\newcommand{\e}{\mathbf{e}}$Your matrix $K$ cyclically permutes the vectors $$ \e_{1} = (1, 0, 0),\quad -\e_{2} = (0, -1, 0),\quad \e_{3} = (0, 0, 1). $$ It should be visually apparent where the axis lies. ;)

Vectors cyclically permuted

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    $\begingroup$ This is how I immediately tried doing it in my head. Gorgeous picture, too! $\endgroup$ – Steven Stadnicki Apr 23 '14 at 22:20
  • $\begingroup$ I would love to understand this approach but i have no idea what "cyclically permutes the vectors" means, is there any way you could explain this in simple english?, and thank you for your reply. $\endgroup$ – BLAZE Apr 23 '14 at 22:37
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    $\begingroup$ Here, "cyclically permutes..." just means $K\mathbf{e}_{1} = -\mathbf{e}_{2}$, $K(-\mathbf{e}_{2}) = \mathbf{e}_{3}$, and $K\mathbf{e}_{3} = \mathbf{e}_{1}$ (as indicated by the arrows). These equations can be verified by computing the products of $K$ with the appropriate column vectors, or by noting that the $i$th column of $K$ is $K\mathbf{e}_{i}$ (a general property of matrix multiplication). $\endgroup$ – Andrew D. Hwang Apr 23 '14 at 22:46
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Neglect the sign for the moment and think of $$ K=\ \begin{pmatrix} 0 & 0 & 1\\ -1 & 0 & 0\\ 0 & -1 & 0 \end{pmatrix} $$ as a permutation matrix: $(1\to 3), (2\to 1), (3\to 2)$ or shorter $(312)$. So you permute your coordinate axes. This points to a rotation axis along one of the vectors $(\pm 1,\pm 1,\pm 1)^T$. Apply this set of vectors on $K$ to get $$ \begin{pmatrix} 0 & 0 & 1\\ -1 & 0 & 0\\ 0 & -1 & 0 \end{pmatrix} \begin{pmatrix} \color{red}{\pm 1} \\ \color{blue}{\pm 1} \\ \pm 1 \end{pmatrix}= \begin{pmatrix} \pm 1 \\ \color{red}{\mp 1} \\ \color{blue}{\mp 1} \end{pmatrix} $$ So you choose $\color{red}{+ 1},\color{blue}{- 1}$ and $+1$.

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  • $\begingroup$ @user144533 a matrix that tells you how to reshuffle your basis vectors... $\endgroup$ – draks ... Apr 24 '14 at 5:44
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Edit. For any $3\times3$ rotation matrix $K$, if it is not symmetric, you can read off the rotation axis directly from its skew-symmetric part $W=K-K^T$. More specifically, the rotation axis is parallel to $(w_{32}, w_{13}, w_{21})^T=-(w_{23}, w_{31}, w_{12})^T$.

Reason: any anticlockwise rotation for an angle $\theta$ about a unit vector $\mathbf u=(x,y,z)^T$ can be put into axis-angle form: $$ K=\begin{bmatrix} \cos\theta + x^2(1-\cos\theta) &xy(1-\cos\theta) - z\sin\theta &xz(1-\cos\theta) + y\sin\theta\\ yx(1-\cos\theta) + z\sin\theta &\cos\theta + y^2(1-\cos\theta) & yz(1-\cos\theta) - x\sin\theta\\ zx(1-\cos\theta) - y\sin\theta &zy(1-\cos\theta) + x\sin\theta & \cos\theta + z^2(1-\cos\theta) \end{bmatrix}. $$ When it is not symmetric, $\sin\theta\ne0$. Hence the skew-symmetric part of $K$ is given by $$ W=K-K^T=2\sin\theta\,\begin{bmatrix} 0&-z&y\\ z&0&-x\\ -y&x&0 \end{bmatrix}. $$ Thus $(w_{32}, w_{13}, w_{21})=2(\sin\theta)(x,y,z)=-(w_{23}, w_{31}, w_{12})$.

In your case, $$ W = K-K^T = \begin{pmatrix} \ast & \ast & 1\\ -1 & \ast & \ast\\ \ast & -1 & \ast \end{pmatrix}. $$ Therefore the rotation axis is parallel to $(-1,1,-1)^T$.

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  • $\begingroup$ Very impressive method you have there, to me inspection implies solve the problem in your head in one go. Do you think you could do that for this case? $\endgroup$ – BLAZE Apr 24 '14 at 20:11
  • $\begingroup$ I also want to read the proof of it, may I know what is the keyword to search that? $\endgroup$ – Jason May 4 at 22:33
  • $\begingroup$ @Jason See my new edit. I have also corrected an error in my old answer. $\endgroup$ – user1551 May 5 at 8:37
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I can offer a remark on how you could simplify in some sense the calculation, but I'm not sure whether you would call the final result "inspection" and not "calculation".

Sometimes it easier to do mental calculations with matrices by thinking of the result of matrix multiplication by a vector as a linear combination of the matrix rows. That is, $$ \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = x \begin{pmatrix} a_{11} \\ a_{21} \\ a_{31} \end{pmatrix} + y \begin{pmatrix} a_{12} \\ a_{22} \\ a_{32} \end{pmatrix} + z \begin{pmatrix} a_{13} \\ a_{23} \\ a_{33} \end{pmatrix}. $$

The result of the multiplication is what you get if you multiply $x$ by the first row, add to it $y$ multiplied by the second row and then $z$ by the third row.

In your case, we have $$ \begin{pmatrix} 0 & 0 & 1 \\ -1 & 0 & 0 \\ 0 & -1 & 0 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = x \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} + y \begin{pmatrix} 0 \\ 0 \\ -1 \end{pmatrix} + z \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} z \\ -x \\ -y \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}. $$

After some practice, you can do this in your head and then note that the first coordinate of the result is going to be $z$ and you want $z = x$. So you can take $z = x = 1$. Then, the second coordinate should be $-x = y$, so you can take $y = -1$.

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