Kernel and Image?

Unenlightened · #1 $Old$ November 3rd, 2009, 01:08 AM

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
$\left( \begin{array}{cccc} -1 & 0 & 0 & 1 \\ 1 & -1 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & -1 & 1 \\ 0 & -1 & 1 & 0 \end{array} \right)$

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

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

Many thanks...

HallsofIvy · #2 $Old$ November 3rd, 2009, 06:26 AM

Quote:

Originally Posted by Unenlightened $View Post$

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...

Unenlightened · #3 $Old$ November 3rd, 2009, 01:04 PM

So for the kernel a=b=c=d, so the basis is (1, 1, 1, 1) ?

The augmented matirx reduces to
1 0 -1 0 -1 0 b
0 1 1 0 0 -1 c
0 0 0 1 1 1 d
0 0 0 0 0 0 a+b+c+d

...how does this give me the equations I need to find the image?

math2009 · #4 $Old$ November 3rd, 2009, 06:46 PM

generally, steps of calculating kernel & image are getting rref(matrix) at first

Image:

$rref\left[ \begin{array}{cccc} -1 & 0 & 0 & 1 \\1 & -1 & 0 & 0 \\-1 & 0 & 1 & 0 \\0 & 1 & 0 & -1 \\0 & 0 & -1 & 1 \\0 & -1 & 1 & 0 \end{array} \right]=\left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 0 & 1 \\0 & 0 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 0 \end{array} \right]$
find columns including leading 1's in rref(matrix), these columns are linearly independent.

$im(A)=span(column1, \cdots , column4)$

Kernel:

$rref\left[ \begin{array}{cccccc} -1 & -1 & 0 & -1 & 0 & 0 \\1 & 0 & -1 & 0 & -1 & 0 \\0 & 1 & 1 & 0 & 0 & -1 \\0 & 0 & 0 & 1 & 1 & 1 \end{array} \right]=\left[ \begin{array}{cccccc} 1 & 0 & -1 & 0 & -1 & 0 \\0 & 1 & 1 & 0 & 0 & -1 \\0 & 0 & 0 & 1 & 1 & 1 \\0 & 0 & 0 & 0 & 0 & 0 \\\uparrow &\uparrow & &\uparrow & & \end{array} \right]$

Obviously, there are only 3 columns which are linearly independent.
The remainder 3 columns will spanned by linearly independent columns.
$\vec{v}_3=-\vec{v}_1+\vec{v}_2 \longrightarrow \vec{v}_1-\vec{v}_2+\vec{v}_3=0 \longrightarrow A\left[ \begin{array}{cccccc} 1\\-1\\1\\0\\0\\0 \end{array} \right]=0$

find another 2 vectors by the same method,

$ker(A)=span(\begin{bmatrix} 1\\-1\\1\\0\\0\\0 \end{bmatrix},\begin{bmatrix} 1\\0\\0\\-1\\1\\0 \end{bmatrix},\begin{bmatrix} 0\\1\\0\\-1\\0\\1 \end{bmatrix})$

Unenlightened · #5 $Old$ November 3rd, 2009, 08:02 PM

Quote:

Originally Posted by math2009 $View Post$

The remainder 3 columns will spanned by linearly independent columns.
$\vec{v}_3=-\vec{v}_1+\vec{v}_2 \longrightarrow \vec{v}_1-\vec{v}_2+\vec{v}_3=0 \longrightarrow A\left[ \begin{array}{cccccc} 1\\-1\\1\\0\\0\\0 \end{array} \right]=0$

I'm sorry, I can't see where you got that v_3 = -v_1 + v_2 from...

math2009 · #6 $Old$ November 3rd, 2009, 08:07 PM

$\vec{v}_i$ represent rref(matrix) column

Unenlightened · #7 $Old$ November 4th, 2009, 01:26 AM

So
v_1 = (1 0 0 0) (transposed - it's easier to write like this)
v_2 = (0 1 0 0)
v_3 = (-1 1 0 0) or (0 0 1 0) ? Either way, I don't see where you get the v_3 = -v_1 + v_2 from

HallsofIvy · #8 $Old$ November 4th, 2009, 03:42 AM

Quote:

Originally Posted by Unenlightened $View Post$

So for the kernel a=b=c=d, so the basis is (1, 1, 1, 1) ?

Yes, that is correct.

Quote:

The augmented matirx reduces to
1 0 -1 0 -1 0 b
0 1 1 0 0 -1 c
0 0 0 1 1 1 d
0 0 0 0 0 0 a+b+c+d

...how does this give me the equations I need to find the image?

The last row corresponds to an equation like 0x+ 0y+ 0z+ 0u+ 0v+ 0w= a+b+c+d. That is only possible, as I said before, if the entire row is 0: that is, if a+ b+ c+ d= 0. That reduces to, say, a= - b- c- d so any such vector is of the form <-b-c-d, b, c, d>= b<-1, 1, 0, 0>+ c<-1, 0, 1, 0>+ d<-1, 0, 0, 1>.

Unenlightened · #9 $Old$ November 4th, 2009, 04:03 AM

So the homology = Ker/Im gives me
$\frac{<1,1,1,1>}{<-1,1,0,0>,<-1,0,1,0>,<-1,0,0,1>}$ ...?

Given that it's homology I'm looking for, this answer should be an integer or group... have I gone wrong somewhere, or is there some way to solve this fraction that I'm not seeing? (Neither of which would surprise me very much

)

hjortur · #10 $Old$ November 4th, 2009, 08:58 AM

Sorry about bringing up another question in this thread.

Quote:

generally, steps of calculating kernel & image are getting rref(matrix) at first

Image:
$rref\left[ \begin{array}{cccc} -1 & 0 & 0 & 1 \\1 & -1 & 0 & 0 \\-1 & 0 & 1 & 0 \\0 & 1 & 0 & -1 \\0 & 0 & -1 & 1 \\0 & -1 & 1 & 0 \end{array} \right]=\left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 0 & 1 \\0 & 0 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 0 \end{array} \right]$
find columns including leading 1's in rref(matrix), these columns are linearly independent.

im(A)=span(column1, \cdots , column4)

I wanted to ask about this. When you find rref(matrix) you generally change the column space but not the row space.
The image is the span of the original columns. So would you not have to use gauss method on (matrix)' <--- (supposed to mean transpose)
and then the image is the row space of rref((matrix)') ?