# Thread: Can this matrix be ...

1. Fmatrix.gif

Can this 10x10 matrix be represented as the the outer product of two vectors?

2.

3. Originally Posted by ReMakeIt
Fmatrix.gif

Can this 10x10 matrix be represented as the the outer product of two vectors?
You can pull the mass factors m out into a vector, but the x will be a major problem because of the cube factor. I'm sure it can be done somehow ( binomial expansion ? ) but the result will be very ugly from an algebraic point of view. Do you have access to a computer algebra system like Mathematica or Maple ? You could try that way.

4. Is there some way to factor the vector X out of this?

x1 = \vect{x}
x2
x3
.
.
.
.
x10

5. ick the attachment looks like that?

I dont have a computer to do the dirty work. The vectors should make some sort of physical sense don't you think? Perhaps I need to look at the vector

X(r, Phi, theta)

r1 phi1 theta1
r2 phi2 theta2
...
...
...
r10 phi10 theta10

Somehow that division of magnitude squared and pointer should fall out...

Magnitude can be written X1TX2 instead of (X1-X2)2

6. Originally Posted by ReMakeIt
ick the attachment looks like that?

I dont have a computer to do the dirty work. The vectors should make some sort of physical sense don't you think? Perhaps I need to look at the vector

X(r, Phi, theta)

r1 phi1 theta1
r2 phi2 theta2
...
...
...
r10 phi10 theta10

Somehow that division of magnitude squared and pointer should fall out...

Magnitude can be written X1TX2 instead of (X1-X2)2
Like I said, I'm not sure that this is possible since all the x are different vectors.

7. $\large \dpi{150} \fn_cm \begin{bmatrix} X_{11} & X_{12} & X_{13} \\ X_{21} & X_{22} & X_{33} \\ . & . & . \\ X_{n1} & X_{n2} & X_{n3} \end{bmatrix} \bullet \begin{bmatrix} X_{11} & X_{21}& .. & X_{n1}\\ X_{21}& X_{22}& .. & X_{n2}\\ X_{31} & X_{32} & .. & X_{n3} \end{bmatrix} = \begin{bmatrix} \vec{x}_1 ^T\vec{x}_1 & \vec{x}_1 ^T\vec{x}_2 & ...& \vec{x}_1 ^T\vec{x}_n\\ : & : & :& :\\ \vec{x}_n ^T\vec{x}_1 &\vec{x}_n ^T\vec{x}_2 & ..& \vec{x}_n ^T\vec{x}_n \end{bmatrix}$

8. $\large \dpi{150} \fn_cm \vec{A}_\left ( \vec{x} \right ) = G \bullet \sum \frac{m_i} {\vec{X} ^T \vec{X_i}}$

in matrix form

$\large \dpi{150} \fn_cm \vec{A} = \vec{M} \bullet \left( \vec_X^T\vec_X \right )^{-1}$

Mass matrix M and acceleration matrix A:

$\large \dpi{150} \fn_cm \vec{M} = \begin{bmatrix} M_1 & 0 & 0 & .. & 0\\ 0 & M_2 & 0 & & 0\\ :& : & :& & :\\ 0 & 0 & 0 & .. & M_n \end{bmatrix}$ ; $\large \dpi{150} \fn_cm \vec{A} = \begin{bmatrix} A_{11} & A_{12} & .. & A_{1n}\\ A_{21} & A_{22} & & A_{2n}\\ :& : & & :\\ A_{n1} & A_{n2} & .. & A_{nn} \end{bmatrix}$

And

$\large \dpi{150} \fn_cm \begin{bmatrix} X_{11} & X_{12} & X_{13} \\ X_{21} & X_{22} & X_{33} \\ . & . & . \\ X_{n1} & X_{n2} & X_{n3} \end{bmatrix} \bullet \begin{bmatrix} X_{11} & X_{21}& .. & X_{n1}\\ X_{21}& X_{22}& .. & X_{n2}\\ X_{31} & X_{32} & .. & X_{n3} \end{bmatrix} = \begin{bmatrix} \vec{x}_1 ^T\vec{x}_1 & \vec{x}_1 ^T\vec{x}_2 & ...& \vec{x}_1 ^T\vec{x}_n\\ : & : & :& :\\ \vec{x}_n ^T\vec{x}_1 &\vec{x}_n ^T\vec{x}_2 & ..& \vec{x}_n ^T\vec{x}_n \end{bmatrix}$

$\large \dpi{150} \fn_cm \vec{X}^T\vec{X}\bullet \vec{A} = \vec{M}$

9. Originally Posted by ReMakeIt
$\large \dpi{150} \fn_cm \vec{A}_\left ( \vec{x} \right ) = G \bullet \sum \frac{m_i} {\vec{X} ^T \vec{X_i}}$

in matrix form

$\large \dpi{150} \fn_cm \vec{A} = \vec{M} \bullet \left( \vec_X^T\vec_X \right )^{-1}$

Mass matrix M and acceleration matrix A:

$\large \dpi{150} \fn_cm \vec{M} = \begin{bmatrix} M_1 & 0 & 0 & .. & 0\\ 0 & M_2 & 0 & & 0\\ :& : & :& & :\\ 0 & 0 & 0 & .. & M_n \end{bmatrix}$ ; $\large \dpi{150} \fn_cm \vec{A} = \begin{bmatrix} A_{11} & A_{12} & .. & A_{1n}\\ A_{21} & A_{22} & & A_{2n}\\ :& : & & :\\ A_{n1} & A_{n2} & .. & A_{nn} \end{bmatrix}$

And
$\large \dpi{150} \fn_cm \begin{bmatrix} X_{11} & X_{12} & X_{13} \\ X_{21} & X_{22} & X_{33} \\ . & . & . \\ X_{n1} & X_{n2} & X_{n3} \end{bmatrix} \bullet \begin{bmatrix} X_{11} & X_{21}& .. & X_{n1}\\ X_{21}& X_{22}& .. & X_{n2}\\ X_{31} & X_{32} & .. & X_{n3} \end{bmatrix} = \begin{bmatrix} \vec{x}_1 ^T\vec{x}_1 & \vec{x}_1 ^T\vec{x}_2 & ...& \vec{x}_1 ^T\vec{x}_n\\ : & : & :& :\\ \vec{x}_n ^T\vec{x}_1 &\vec{x}_n ^T\vec{x}_2 & ..& \vec{x}_n ^T\vec{x}_n \end{bmatrix}$

$\large \dpi{150} \fn_cm \vec{X}^T\vec{X}\bullet \vec{A} = \vec{M}$
Is this not different from the original matrix? I don't see the cube factors coming into this.
Also, where is the differential operator ? These aren't differential equations any more.

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