1. ## 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

2. Originally Posted by arze
$M=\left(\begin{array}{cc}a&b\\c&d\end{array}\right )$
$r_1=x_1i+y_1j$ 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, $\theta_1=\theta_2$.

3. Originally Posted by hatsoff

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.

4. Originally Posted by arze
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 $\textbf{v}=(v_1,v_2,\cdots,v_n)$.

5. Originally Posted by arze
$M=\left(\begin{array}{cc}a&b\\c&d\end{array}\right )$
$r_1=x_1i+y_1j$ 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