<p>To align vectors, you are applying a rotation.</p>

<p>If you consider a simple rotation matrix, e.g. rotation of angle $\theta$ around Z axis, then you will trivially see that its inverse matrix, a rotation of $-\theta$, is the transpose,  since $sin(-\theta)=-sin(\theta)$ and $cos(-\theta)=cos(\theta)$.</p>

<p>An arbitrary rotation matrix, R, can be constructed by multiplication of other rotations, e.g $R = A \cdot B$.  If the inverse of A and B exist, then $R^{-1} = B^{-1} \cdot A^{-1}$.<br>
Similarly we know  $(A \cdot B)^T = (B^T \cdot A^T)$.</p>

<p>Put these together and you have your answer.</p>

CC BY-SA 3.0
2018-01-05T17:20:28.070Z
2018-01-05T17:26:08.587Z
2018-01-05T17:26:08.587Z
Simon F
<p>$A_{v}^{-1}$ = $A_{v}^{T}$<br/><br/>
$A_{v}$ = Allignment matrix that alligns vector v with z axis<br/>
$=$\begin{bmatrix}
    \frac{\lambda}{|v|}       &amp; \frac{-ab}{\lambda|v|} &amp; \frac{-ac}{\lambda|v|} &amp;   0 \\
    0 &amp; \frac{c}{|v|} &amp; \frac{-b}{|v|} &amp;   0 \\
    \frac{a}{|v|}       &amp; \frac{b}{|v|} &amp; \frac{c}{|v|} &amp;   0 \\
0 &amp; 0 &amp; 0 &amp; 1
\end{bmatrix}<br/>
here, $\lambda = \sqrt{b^2+c^2} $<br/>
$|v| = \sqrt{a^2+b^2+c^2} $<br/>
And vector $v = (a, b, c)$</p>

CC BY-SA 3.0
2018-01-05T17:08:05.990Z
Are you asking "Why is the inverse of a rotation matrix the same as its transpose"?
Simon F
CC BY-SA 3.0
2018-01-05T21:26:36.693Z
That is a rotation or reflection matrix. You can check that the four columns are orthogonal to each other (the dot product of the columns is zero). Also they are of unit length, so the determinant is $\pm$1. A well known property of orthonormal square matrices is that transpose is equal to the inverse.
Mauricio Cele Lopez Belon
CC BY-SA 3.0
2018-01-06T06:51:13.260Z
@MauricioCeleLopezBelon your explanation was truly helpful. Thanks.
Bhattacharjee
<matrix><vector-graphics>
Why inverse of an allignment matrix is the same as its transpose?
Bhattacharjee
# Why inverse of an allignment matrix is the same as its transpose?

(Asked by: username_0 on 2018-01-05)

$A\_{v}^{-1}$ = $A\_{v}^{T}$  
  
$A\_{v}$ = Allignment matrix that alligns vector v with z axis  
$=$\begin{bmatrix}
\frac{\lambda}{|v|} & \frac{-ab}{\lambda|v|} & \frac{-ac}{\lambda|v|} & 0 \\
0 & \frac{c}{|v|} & \frac{-b}{|v|} & 0 \\
\frac{a}{|v|} & \frac{b}{|v|} & \frac{c}{|v|} & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}  
here, $\lambda = \sqrt{b^2+c^2} $  
$|v| = \sqrt{a^2+b^2+c^2} $  
And vector $v = (a, b, c)$

**Tags:** matrix, vector-graphics

> **Comments:**
> **username_1**: Are you asking "Why is the inverse of a rotation matrix the same as its transpose"?  
> **username_2**: That is a rotation or reflection matrix. You can check that the four columns are orthogonal to each other (the dot product of the columns is zero). Also they are of unit length, so the determinant is $\pm$1. A well known property of orthonormal square matrices is that transpose is equal to the inverse.  
> **username_0**: @username_2 your explanation was truly helpful. Thanks.  

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## 1 answers

### Answer by username_1 on 2018-01-05. Score: 2 | Accepted answer

To align vectors, you are applying a rotation.

If you consider a simple rotation matrix, e.g. rotation of angle $\theta$ around Z axis, then you will trivially see that its inverse matrix, a rotation of $-\theta$, is the transpose, since $sin(-\theta)=-sin(\theta)$ and $cos(-\theta)=cos(\theta)$.

An arbitrary rotation matrix, R, can be constructed by multiplication of other rotations, e.g $R = A \cdot B$. If the inverse of A and B exist, then $R^{-1} = B^{-1} \cdot A^{-1}$.  
Similarly we know $(A \cdot B)^T = (B^T \cdot A^T)$.

Put these together and you have your answer.


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