A question in different methods of a solution

[transgalactic]

i added the question and both solutions in the link

the solution that i was presented is totally logical
but when i solved it in a standart way it gave me a different answer

http://img116.imageshack.us/my.php?image=img8272ib1.jpg

 

[HallsofIvy]

I hope it is no surprise that your book is correct. Basically the problem is this: you have a linear transformation from 2 by 2 matrices to 2 by two matrices defined by :
$$\left[\begin{array}{cc} a & b \\ c & d \end{array}\right]$$ mapped to $$\left[\begin{array}{cc} 0 & c \\ 0 & d \end{array}\right]$$

Yes, since a and b "disappear" they play no part in determining the kernel. The only non-zero terms in the resultant matrix are c and d. It they are 0, then the result is the zero matrix. What that means, then, is that a and b can be anything at all as long as c= 0 and d= 0. Taking a= 1, b= 0 gives the matrix
$$\left[\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right]$$
and taking a= 0, b= 1 gives the matrix
$$\left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right]$$
Those two matrices form a basis for the kernel.

I don't think it really makes sense to talk about "a and b don't matter in determining the Image". a and b are only in the matrix the transformation is applied to- not in the resultant matrix. Obviously, every matrix in the image have first column 0. The second column can be anything. A basis for the image is
$$\left[\begin{array}{cc} 0 & 1 \\ 0 & 1 \end{array}\right]$$
and
$$\left[\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right]$$

In your "totally logical" solution, you are applying a matrix to the 0 vector. The "kernel" of a transformation is the set of vectors that are mapped to the 0 vector, not that the 0 vector is mapped to. You are going the wrong way.

(Any linear transformation maps the 0 vector to the 0 vector.)

 

[transgalactic]

i can't understand
"not that the 0 vector is mapped to"

can you describe it a different words my mistake??

 

[HallsofIvy]

If Ax= y, then "A maps x to y".

x is in the kernel of A if and only if Ax= 0. That is, if "A maps x to 0".

For any linear transformation, A, A0= 0. That is "A always maps 0 to 0".

In your attempted solution you construct an 8 by 8 matrix and apply it to the 0 vector. Of course, you get the 0 vector. That has nothing to do with a "kernel" because that will always be true. It going the wrong way: you are seeing what 0 is changed into (mapped to) rather that seeing what gets changed to 0.