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HallsofIvy · #2 November 3rd, 2009, 05:26 AM

Quote:

Originally Posted by Unenlightened

This question seems to be so basic that I can't find it anywhere... My grasp of linear algebra is embarrassing, and I'm working to remedy that, but I need this to produce something fairly sharpish...

Basically what I need is
the Kernel of, say, and how one gets it
$\left( \begin{array}{cccc} -1 & 0 & 0 & 1 \\ 1 & -1 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & -1 & 1 \\ 0 & -1 & 1 & 0 \end{array} \right)$

The definition of the "kernel" of the linear transformation, A, is "the space of all vectors, v, such that Av= 0.

Here , that means you are looking for <a, b, c, d> such that
$\left( \begin{array}{cccc} -1 & 0 & 0 & 1 \\ 1 & -1 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & -1 & 1 \\ 0 & -1 & 1 & 0 \end{array} \right)\begin{bmatrix}a \\ b\\ c\\ d\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0\\ 0\end{bmatrix}$
That gives you the equations -a+ d= 0, a- b= 0, -a+ c= 0, b- d= 0, and -b+ c= 0. Those do not have a single solution (if they did it would have to be a= b= c= d= 0) but can be solved for b, c, and d in terms of a. That gives a single vector basis for the kernel.

Quote:

and the Image of
$\left( \begin{array}{cccccc} -1 & -1 & 0 & -1 & 0 & 0 \\ 1 & 0 & -1 & 0 & -1 & 0 \\ 0 & 1 & 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array} \right)$ [/math]

Again, use the definition of "Image". The image of linear transformation, A, is the set of all vectors, v, such that Au= v for [b]some[\b] vector u. In other words, that would be the set of all v such that Au= v has a solution. If we let $v= \begin{bmatrix}a \\ b \\ c \\ d\end{bmatrix}$ then we are looking at the equation
$\left( \begin{array}{cccccc} -1 & -1 & 0 & -1 & 0 & 0 \\ 1 & 0 & -1 & 0 & -1 & 0 \\ 0 & 1 & 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array} \right)\begin{bmatrix}x \\ y \\ z \\ u\\ v\\ w\end{bmatrix}= \begin{bmatrix}a \\ b \\ c\\ d\end{bmatrix}$ .

You can "solve" that by row reduction of the augmented matrix
$\left( \begin{array}{ccccccc} -1 & -1 & 0 & -1 & 0 & 0 & a\\ 1 & 0 & -1 & 0 & -1 & 0 & b\\ 0 & 1 & 1 & 0 & 0 & -1 & c\\ 0 & 0 & 0 & 1 & 1 & 1 & d\end{array} \right)$
If there are any rows in which the first 6 terms are all 0, then the last term, which must be in terms of a, b, c, and d, must also be 0. That gives the equations you need to find the image. If there are no rows in which the first 6 numbers are 0, the kernel is all of $R^4$ .

Quote:

I need it to calculate the homology, ie. using Ker i / Im j

Many thanks...

The following users thank HallsofIvy for this useful post:
Unenlightened
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