Would a matrix remain un-rotated if it is multiplied by an orthonormal rotation matrix on right side and transpose of same rotation matrix on the left side?
[Math] matrix multiplied by rotation matrix on right side and transpose(rotation) on left side
linear algebramatricesrotations
Related Solutions
Sometime one has to left-multiply, sometimes one has to right-multiply. This really depends.
Prerequisites:
You are performing scaling, rotation, and translation. So let us assume we have linear point transformation of the general form:
$$\mathtt T = \left[ \begin{array}{cc} s\mathtt R & \mathbf t \\ \mathtt O& 1\end{array} \right]$$ which first rotates a point by $\mathtt R$ , then scales it by $s$ and then adds the translation $\mathbf t$:
$$\mathtt T \cdot \left( \begin{array}{c} \mathbf x \\ 1\end{array} \right) = \left[ \begin{array}{c} s(\mathtt R\cdot \mathbf x) + \mathbf t \\ 1\end{array} \right]$$
(Note that rotation and scaling commutes: $s(\mathtt R\cdot \mathbf x)=\mathtt R(s\cdot \mathbf x)$)
From now on we will assume that all points $\mathbf y$ are homogenous points ($\mathbf y= (\mathbf x, 1)^\top $).
Mind the reference frames: In order to make it clear whether you need a left or right multiplication, it is important to highlight in which reference frame your points are!
Let us assume, we have points $\mathbf y_a$ in reference frame $a$, and you want to transform them into reference frame $b$, you do
$$ \mathbf y_b = \mathtt T_{ba} \mathbf y_a$$ where $\mathtt T_{ba}$ is a transformation to $b$ from $a$. Note that the indices must match!
Now, let us look at a more complicated example. One might be interested in:
$$\mathbf y_a = \mathtt T_{ab}\mathtt T_{bc}\mathtt T_{cd}\mathbf y_d$$
Further, let's assume that we receive the poses in order (First $\mathtt T_{ab}$, then $\mathtt T_{bc}$...).
We would calculate in an algorithm:
$\mathtt T_{ai} := \mathtt T_{ab}$
(thus, $i=b$)
$\mathtt T_{ai} := \mathtt T_{ai}\cdot \mathtt T_{bc}$
(now, $i=c$)
$\mathtt T_{ai} := \mathtt T_{ai}\cdot \mathtt T_{cd}$
($i=d$)
Thus, we right-multiplied and $\mathtt T_{ai}$ means now $\mathtt T_{ad}$, the transformation from $d$ to $a$. Finally, we can transform our points:
$$\mathbf y_a := \mathtt T_{ad} \mathbf y_d $$
However, if one really wants to left-multiply, this is possible too! Note that $\mathtt T_{ia}=\mathtt T_{ai}^{-1}.$ Thus, we can do:
$\mathtt T_{ia} := \mathtt T_{ab}^{-1}$
($i=b$)
$\mathtt T_{ia} := \mathtt T_{bc}^{-1} \mathtt T_{ia}$
($i=c$)
$\mathtt T_{ia} := \mathtt T_{cd}^{-1}\mathtt T_{ia}$
($i=d$)
Thus, we have $\mathtt T_{ia} = \mathtt T_{da}$, and therefore we can transfrom the point from $d$ to $a$ using the inverse:
$$\mathbf y_a := \mathtt T_{da}^{-1} \mathbf y_d $$
Background: It is very convenient to use the following notation:
$\mathtt R_{UV}$ is a rotation matrix which transform points from reference frame $V$ into the reference frame $U$. Thus: $\mathbf x_U = \mathtt R_{UV} \mathbf x_V$. The inverse rotation is $\mathtt R_{VU} = \mathtt R_{UV}^{-1}=\mathtt R_{UV}^\top$.
Lets, call the initial frame $O$. I assume you mean with "rotated from the same initial frame" a change in the observer frame/passive transformation (http://en.wikipedia.org/wiki/Active_and_passive_transformation).
Thus, you have the rotation matrices $\mathtt R_{OA}$ and $\mathtt R_{OB}$ (which describe the motion of the observer from $O$ to $A$/$B$, or in other words maps points from $A$/$B$ to $O$). Now, I assume you are interested in $\mathtt R_{AB} = \mathtt R_{AO}\mathtt R_{OB} = \mathtt R_{OA}^\top\mathtt R_{OB}$.
Finally, convert $\mathtt R_{AB}$ into axis-angle...
(In case you have $\mathtt R_{AO}$ and $\mathtt R_{BO}$ and want to calculate $\mathtt R_{BA}$ , you get $\mathtt R_{BA}=\mathtt R_{BO}\mathtt R_{AO}^\top$.)
Best Answer
As @hardmath suggested, by doing what you say you'd get a similarity transformation: suppose that you have a matrix $A$ and an invertible matrix $P$. Then, the matrix $B = P^{-1}AP$ is said to be similar to A.
And here's a counterexample. Suppose that you have the Rotation matrix
$R = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\[0.3em] -\frac{1}{2} & \frac{\sqrt{3}}{2} \\[0.3em] \end{bmatrix}$
and its inverse (transpose)
$R^{-1} = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\[0.3em] \frac{1}{2} & \frac{\sqrt{3}}{2} \\[0.3em] \end{bmatrix}$
Then, if you consider a matrix
$A = \begin{bmatrix} 1 & 1 \\[0.3em] 1 & 0 \\[0.3em] \end{bmatrix}$
and you left-multiply it by $R^{-1}$ and right-multiply it by $R$, you get
$\begin{bmatrix} \frac{\sqrt{3}}{4}(-2+\sqrt{3}) & \frac{1}{4}(2+\sqrt{3}) \\[0.3em] \frac{1}{4}(-1+\sqrt{3}(1+\sqrt{3})) & \frac{1}{4}(1+ 2\sqrt{3}) \\[0.3em] \end{bmatrix}$
which is clearly a completely different matrix!