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Math Help - Matrix and Vector Magnitude

  1. #1
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    Matrix and Vector Magnitude

    Given matrix M and vector X where:
    <br /> <br />
\[M = <br />
\left( {\begin{array}{cc}<br />
cos\theta & sin\theta \\<br />
-sin\theta & cos\theta<br />
\end{array} } \right) \]<br /> <br />
\textit{and}<br />
X = <br />
\left( {\begin{array}{cc}<br />
x \\<br />
 y<br />
\end{array} } \right)<br />
    Compute: <br />
\frac { \left|{MX}\right|}{\left|{X}\right|}<br />
    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
    <br />
\[M = <br />
\left( {\begin{array}{cc}<br />
  1 & 0 \\<br />
  0 & 1<br />
\end{array} } \right) \]<br />
    which, geometrically, doesn't change the length of vector X. Would that be correct?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    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
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  3. #3
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    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.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    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
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  5. #5
    MHF Contributor

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    Ouch! Good point!
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