Explanation from Linear combination and basis transformation

Let me explain this from another point of view. This explanation doesn't need null space, but, it lacks a rigorous proof. However, we have a proof by null space and so I think this is correct. For me this is a more intuitive explanation.

From the independence of the columns of

*A*, the independent vectors will be organized as the following. You can also see the how the resulting columns are computed.

Since

*A^T*is a linear operator,

This becomes 0 when

*A^{T}*degenerates. However, this is also the same form of basis transformation. Each basis is each column. Because,

**a**_**A^{T}*'s row are independent (since

*A*'s column are independent, transposing

*A*exchanges columns with rows.) Each independent row vector is projected to the each independent columns. The column vector's independence is preserved unless

*A^{T}*degenerates.

**a**_*Unfortunately, I have no proof of degeneration. But, what happens is an independent basis is transformed into another independent basis. I don't see how this independence breaks under this condition.

By the way, this matrix is called metric tensor. My favorite explanation of this is in the book ``Mathematical sciences of graphics'' by Koukichi Sugihara, Chapter 1. (I have the first edition, the third printing. This has a typo in Equation 1.20, page 19. But, the introducing equations are correct and only the result equation has a typo.)

Next I would like to have an appendix, explanation about null space, column space, and existence of inverse. This might not be necessary, so the story about

*A^T A*ended here.

## No comments:

Post a Comment