Matrix and Vector Magnitude

• January 16th 2011, 12:56 AM
tintin2006
Matrix and Vector Magnitude
Given matrix M and vector X where:
$

$M = \left( {\begin{array}{cc} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array} } \right)$

\textit{and}
X =
\left( {\begin{array}{cc}
x \\
y
\end{array} } \right)
$

Compute: $
\frac { \left|{MX}\right|}{\left|{X}\right|}
$

So I found that the result is 1.
The question is to explain the result and describe the action of M geometrically.
I think the result means that
$
$M = \left( {\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} } \right)$
$

which, geometrically, doesn't change the length of vector X. Would that be correct?
• January 16th 2011, 01:35 AM
FernandoRevilla
$M=\begin{bmatrix}{\;\;\;\cos \theta}&{\sin \theta}\\{-\sin \theta}&{\cos \theta}\end{bmatrix}=\begin{bmatrix}{\cos (-\theta)}&{-\sin (-\theta)}\\{\sin (-\theta)}&{\;\;\;\cos (-\theta)}\end{bmatrix}$

represents a rotation around $(0,0)$ and angle $-\theta$ so ,

$\left\|{MX}\right\|= \left\|{X}\right\|\; \forall{X}\in{\mathbb{R}^2}$

Fernando Revilla
• January 16th 2011, 06:54 AM
HallsofIvy
Another way of seeing that is to note that the determinant is 1:
$\left|\begin{array}{cc}cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta)\end{array}\right|= (cos(\theta))(cos(\theta))(- (sin(\theta))(-sin(\theta))= cos^2(\theta)+ sin^2(\theta)= 1$
and so the length of a vector is not changed.
• January 16th 2011, 11:43 AM
FernandoRevilla
But the length does not change by the fact of being $M$ orthogonal (in this case a rotation) and not by the fact of being $\det M=1$ . For example:

$M=\begin{bmatrix}{1}&{1}\\{0}&{1}\end{bmatrix}$

has determinant $1$ , however

$\left\|{(0,1)^t}\right\|=1\neq \sqrt{2}=\left\|{M(0,1)^t}\right\|$

Fernando Revilla
• January 16th 2011, 11:57 AM
HallsofIvy
Ouch! Good point!