Given are now two matrices and I have to discuss what the given functions are doing (geometrically). Maybe you can revise/add the following:

given are the matrices

$ A = \begin{pmatrix} \cos(a) & -\sin(a) & 0 \\ \sin(a) & \cos(b) & 0 \\ 0 & 0 & 1 \end{pmatrix}$

$ B = \begin{pmatrix} \cos(b) & 0 & -\sin(b) \\ 0 & 1 & 0 \\ \sin(b) & 0 & \cos(b) \end{pmatrix}$.

(1) What does $f: \mathbb{R^3} \to \mathbb{R^3}, x \to Ax$ mean?

It means a rotation around the z-axis.

I do $A*(1,0,0) = (\cos(a), \sin(a), 0)$. This is a rotation around z-axis counterclockwise.

I do $A*(0,1,1) = (-\sin(a), \cos(a), 0)$. This is a rotation around z-axis counterclockwise.

I do $A*(0,0,1) = (0,0,1)$. It remains.

(2) What does $g: \mathbb{R^3} \to \mathbb{R^3}, x \to Bx$ mean?

It means a rotation around y-axis.

I do $B*(1,0,0) = (\cos(b), 0, \sin(b))$. This is a rotation around y-axis clockwise.

I do $B*(0,1,0) = (0, 1, 0)$. It remains.

I do $B*(0,0,1) = (-\sin(b), 0, \cos(b))$. This is a rotation around y-axis clockwise.

(3) What does $h: \mathbb{R^3} \to \mathbb{R^3}, x \to ABx$ mean?

I do A*B and then

I do $AB*(1,0,0) = \begin{pmatrix} \cos(a)cos(b) \\ \sin(a)\cos(b) \\ \sin(b) \end{pmatrix}$. This is.. ?!

I do $AB*(0,1,0) = \begin{pmatrix} -\sin(a) \\ \cos(a) \\ 0 \end{pmatrix}$. So this should be again a rotation around z-axis for (a). Regardless of what degree b is!

I do $AB*(0,0,1) = \begin{pmatrix} -\cos(a)\sin(b) \\ -\sin(a)\sin(b) \\ \cos(b) \end{pmatrix}$. ?!

And so on.

  • $\begingroup$ $AB(1,0,0)$ means that we have a rotation of the vector $(1,0,0)$ first around the $y$ axis by an angle $b$ and then a second rotation around the $z$ axis by an angle $a$. $\endgroup$ – user63181 Mar 18 '14 at 9:24
  • $\begingroup$ @SamiBenRomdhane: Thank you, but: why is this true? I mean, how can I see/understand that AB(1,0,0) is rotation around y-axis with angle b and z-axis with angle a? $\endgroup$ – Vazrael Mar 18 '14 at 9:33
  • $\begingroup$ Well, simply this is the meaning of the composition! If we have $f\circ g(x)$ so this means that $g$ acts on $x$ to give $g(x)$ and then $f$ acts on $g(x)$ to give $f(g(x))$. $\endgroup$ – user63181 Mar 18 '14 at 9:37

I think the answer to your question is Euler's rotation theorem which states that the composition of two rotations in $3$-d space is also a rotation. Thus, the rotations have the structure of a group.

For example, in your case, composing rotation $A$ with $B$ is another rotation along a different axis by this theorem.


A rotation in 3-dimensional space is specified by two pieces of data: its axis and amount (angle) of rotation. To find out which is the axis for composition of two rotations A and B, find the eigenspace of eigenvalue 1:. That is Solve the system $(AB-I)v=0$.

Then the angle is $\cos^{-1}\frac12( tr(AB)-1) )$

  • $\begingroup$ This sounds nice. I tried calculating (AB-I)v=0. I'm stuck :D I need to solve $\begin{pmatrix} (cos(a)cos(b)-1)x -sin(a)y -cos(a)sin(b)z \\ sin(a)cos(b)x + (cos(a)-1)y -sin(a)sin(b)z \\ sin(b)x + (cos(b)-1)z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\0 \end{pmatrix}$. Is this correct? Now, how is the solution of this related to the axis of rotation? $\endgroup$ – Vazrael Mar 18 '14 at 10:39
  • $\begingroup$ $AB-I$ is a $3\times 3$ matrix $\endgroup$ – P Vanchinathan Mar 18 '14 at 10:40
  • $\begingroup$ @P Vanchinathan: What does that mean? :( $\endgroup$ – Vazrael Mar 18 '14 at 10:48
  • $\begingroup$ You have already given A and B as $3\times3$ matrix. Choose id matrix of the same size, then for the product matrix $AB$ which is a rotation whose axis is the line(infinitely many vectors, multiple of a single vector) got by solving the homogeneous system corr. to the $3\times 3$ matrix: $(AB-I)$ $\endgroup$ – P Vanchinathan Mar 18 '14 at 10:54
  • $\begingroup$ @ P Vanchinathan; Sorry, I really don't get it. Could you please help me, for example, finding the eigenvalues/eigenvectors for matrix A which then should help me finding the rotation axis etc.? $\endgroup$ – Vazrael Mar 18 '14 at 11:22

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