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Math Help - Matrices

  1. #1
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    Matrices

    M=\left(\begin{array}{cc}a&b\\c&d\end{array}\right  )
    r_1=x_1i+y_1j and r_2=x_2i+y_2j are any two vectors, and when multiplied by M they are transformed into s_1 and s_2 respectively. Given that a,b,c,d satisfy ab+cd=0 and a^2+c^2=b^2+d^2, show that the angle between s_1 and s_2 is the same as between r_1 and r_2.
    If the two sets of vectors have the same angle between them the difference between the gradients would be the same, am I right?
    Thanks
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  2. #2
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    Quote Originally Posted by arze View Post
    M=\left(\begin{array}{cc}a&b\\c&d\end{array}\right  )
    r_1=x_1i+y_1j and r_2=x_2i+y_2j are any two vectors, and when multiplied by M they are transformed into s_1 and s_2 respectively. Given that a,b,c,d satisfy ab+cd=0 and a^2+c^2=b^2+d^2, show that the angle between s_1 and s_2 is the same as between r_1 and r_2.
    If the two sets of vectors have the same angle between them the difference between the gradients would be the same, am I right?
    Thanks
    This is just algebraic manipulation. Slow and plodding, but quite simple:

    We have \left(\begin{array}{cc}a&b\\c&d\end{array}\right)\  left(\begin{array}{c}x_i\\y_i\end{array}\right)=\l  eft(\begin{array}{c}ax_i+by_i\\cx_i+dy_i\end{array  }\right).

    So \textbf{s}_1\cdot\textbf{s}_2=(ax_1+by_1)(ax_2+by_  2)+(cx_1+dy_1)(cx_2+dy_2)

    =(a^2+c^2)x_1x_2+(ab+cd)(x_1y_2+x_2y_1)+(b^2+d^2)y  _1y_2, which, because of our hypotheses,

    =(a^2+c^2)(x_1x_2+y_1y_2)

    =(a^2+c^2)(\textbf{r}_1\cdot\textbf{r}_2).

    So \cos\theta_2=\frac{\textbf{s}_1\cdot\textbf{s}_2}{  |\textbf{s}_1||\textbf{s}_2|}

    =\frac{(a^2+b^2)(\textbf{r}_1\cdot\textbf{r}_2)}{\  sqrt{(ax_1+by_1)^2+(cx_1+dy_1)^2}|\textbf{s}_2|}

    =\frac{(a^2+b^2)(\textbf{r}_1\cdot\textbf{r}_2)}{\  sqrt{(a^2+c^2)x_1^2+2(ab+cd)x_1y_1+(b^2+d^2)y_1^2}  |\textbf{s}_2|}

    =\frac{(a^2+b^2)(\textbf{r}_1\cdot\textbf{r}_2)}{\  sqrt{(a^2+c^2)(x_1^2+y_1^2)}|\textbf{s}_2|}

    =\frac{(a^2+b^2)(\textbf{r}_1\cdot\textbf{r}_2)}{\  sqrt{(a^2+c^2)^2(x_1^2+y_1^2)(x_2^2+y_2^2)}}

    =\frac{\textbf{r}_1\cdot\textbf{r}_2}{\sqrt{(x_1^2  +y_1^2)(x_2^2+y_2^2)}}

    =\frac{\textbf{r}_1\cdot\textbf{r}_2}{|\textbf{r}_  1||\textbf{r}_2|}

    =\cos \theta_1,

    that is, \theta_1=\theta_2.
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  3. #3
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    Quote Originally Posted by hatsoff View Post

    So \textbf{s}_1\cdot\textbf{s}_2=(ax_1+by_1)(ax_2+by_  2)+(cx_1+dy_1)(cx_2+dy_2)
    Thanks! but I don't understand how you got this part.
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  4. #4
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    Quote Originally Posted by arze View Post
    Thanks! but I don't understand how you got this part.
    It's the expansion of a dot product. Remember that the binary operation \cdot is defined by

    \textbf{u}\cdot\textbf{v}=\sum_{i=1}^nu_iv_i, where \textbf{u}=(u_1,u_2,\cdots,u_n) and \textbf{v}=(v_1,v_2,\cdots,v_n).
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  5. #5
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    Quote Originally Posted by arze View Post
    M=\left(\begin{array}{cc}a&b\\c&d\end{array}\right  )
    r_1=x_1i+y_1j and r_2=x_2i+y_2j are any two vectors, and when multiplied by M they are transformed into s_1 and s_2 respectively. Given that a,b,c,d satisfy ab+cd=0 and a^2+c^2=b^2+d^2, show that the angle between s_1 and s_2 is the same as between r_1 and r_2.
    If the two sets of vectors have the same angle between them the difference between the gradients would be the same, am I right?
    Thanks

    u_i:=Mr_i=\begin{pmatrix}a&b\\c&d\end{pmatrix}\beg  in{pmatrix}x_i\\y_i\end{pmatrix}=\begin{pmatrix}ax  _i+by_i\\cx_i+dy_i\end{pmatrix}\,,\,\,i=1,2 . Now, let us go for the calculations to get \cos\theta\,,\,\,\theta= the angle between u_1:=Mr_1,\,u_2:=Mr_2 ( and you

    check these are correct taking into account the given info about a,b,c,d ! I'm assuming the usual, euclidean inner product in \mathbb{R}^2 ):

    u_1\cdot u_2=\left(a^2+c^2\right)x_1x_2+(ab+cd)(x_1y_2+x_2y  _1)+\left(b^2+d^2 \right)y_1y_2=\left(a^2+c^2 \right)(x_1x_2+y_1y_2)

    \|u_i\|=\left(a^2+c^2\right)\left(x_i^2+y_i^2 \right)

    Well, now just check that indeed \frac{r_1\cdot r_2}{\|r_1\|\,\|r_2\|}=\frac{u_1 \cdot u_2}{\|u_1\|\,\|u_2\|}

    Tonio
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