quietrain said:
oh.. em..
ok let's say we have
(1 2) x (4 5)
(3 4) (6 7) = so this is just rows multiply by column to get a 2x2 matrix right? so what is the difference if i replace the x sign with the dot sign now. do i still get the same?
You can replace it by whatever symbol you like. As long as your multiplication is "matrix multiplication" you will get the same result.
i presume one is cross (x) product , one is dot (.) product?
No, just changing the symbol doesn't make it one or the other.
or is it for matrix there is no such things as cross or dot product? that's weird. my tutor tells us to know the difference between cross and dot matrix product
I suspect your tutor was talking about
vectors not matrices.
so for the case of the parallelpiped, what's the significance of the triple product (u x v) .w? why do we use x for u&v but . for w?
Because you are talking about
vectors not matrices!
is it just to tell us that we have to use sin and cos respectively? but if u v and w were square matrix, then there won't be any sin and cos to use? so we just multiply as usual rows by columns?
They are NOT matrices, they are vectors!
You can think of vectors as "row matrices" (n by 1) or "column matrices" (1 by n) but they still have properties that matrices in general do not have.
oh by definition . so that means |k.k| = (k)(k)cos(0) = (1)(1)cos(0) = 1
so |i.k| =(1)(1)cos(90) = 0 ?
Yes, that is correct.
so if i x k gives us -j by the right hand rule, then does it mean the magnitude, which is |i.k| = 0 is 0? in the direction of the -j?? or are they 2 totally different aspects?
No, the
length of i x k is NOT |i.k|, it is |i||j|= 1.
In general, the length of \vec{u}\times\vec{v} is |u||v| sin(\theta) where \theta is the angle between \vec{u} and \vec{v}.
btw, sry for another question,
why is e(w)(A),
where A =
(0 -1)
(1 0)
can be expressed as
( cosw -sinw)
( sinw cosw)
which is the rotational matrix anti-clockwise about the x-axis right?
thanks
For objects other than numbers, where we have a notion of addition and multiplication, we define higher functions by using their "Taylor series", power series that are equal to the functions. In particular, e^x= 1+ x+ (1/2)x^2+ \cdot\cdot\cdot+ (1/n!)x^n+ \cdot\cdot\cdot.
It should be easy to calculate that
A^2= \begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}
A^3= \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}
and, since that is the identity matrix, it all repeats:
A^4= \begin{pmatrix}0 & -1 \\ 1 & 0}\end{pmatrix}= A
etc.
That gives
e^{Aw}= \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}+ \begin{pmatrix}0 & -w \\ w & 0\end{pmatrix}+ \frac{1}{2}\begin{pmatrix}-w^2 & 0 \\ 0 & -w^2\end{pmatrix}+ \frac{1}{3!}\begin{pmatrix}0 & w^3 \\ -w^3 & 0\end{pmatrix}+ \frac{1}{4!}\begin{pmatrix}w^4 & 0 \\ 0 & w^4\end{pmatrix}+ \cdot\cdot\cdot
= \begin{pmatrix}1- \frac{1}{2}w^2+ \frac{1}{4}w^4+ \cdot\cdot\cdot & -w+ \frac{1}{3!} w^3+ \cdot\cdot\cdot \\ w- \frac{1}{3!}w^3+ \cdot\cdot\cdot & -1+ \frac{1}{2}w^2- \frac{1}{4}w^4+ \cdot\cdot\cdot\end{pmatrix}
and you should be able to recognise those as the Taylor's series about 0 for cos(w) and sin(w).