I got problem in understanding the Im finding method

1. Feb 9, 2008

transgalactic

i am puzzled about the method of finding the Im

in one axample i see the method of thaking the column withound the term of the zero
row
and exract the vectors from the column

the other way is taking the "b" term that make the whole row zeros
and extract from it

the third way i making the coulmns zeros

i am realy confused about the method of finding Im
because there is 2 example the contaridct each other

http://img504.imageshack.us/my.php?image=img8271ip1.jpg

2. Feb 9, 2008

HallsofIvy

Staff Emeritus
You seem to be consistently having trouble distinguishing between "kernel" and "image". If A is a linear transformation from vector space U to vector space V, then the kernel of A is a subspace of U and the image of A is a subspace of V.

In the problem before, where you were seeking the kernel, you found A0, a vector in V, when you should have been solving Au= 0 to get a vector in U.

Now, you appear to be solving the equation
$$\left[\begin{array}{ccc} 1 & 1 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 1\end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]= \left[\begin{array}{c} b_1 \\ b_2 \\ b_3 \end{array}\right]$$
when you should be calculating
$$\left[\begin{array}{ccc} 1 & 1 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 1\end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]= \left[\begin{array}{c} x+ y \\ 0 \\ 2x+ z\end{array}\right]$$

It is obvious that the the second component is always 0 and is easy to see that, since x, y, z are arbitrary, that the first and third components can be any numbers: the image is the set of vectors <x, 0, z> or, equivalently, the subspace with basis <1, 0, 0> and <0, 0, 1>.