Ok i got to the end of this whole Im theory but

[transgalactic]

ok i got to the end of this whole I am theory but...

i have two cases that i need guidense

in the link i showed a basic case and how we solve it using this method

http://img145.imageshack.us/my.php?image=img8275vc2.jpg

than i presented the first case that i can't solve
i which i tried to copy a case that we have delt with before
(with the b2=0)

than i presented the second case that i can't solve like our (b2=0) example

basically may method say to make A*x=b

then to find the "b"s value for which we get a solution for the matrix

in the end we copy the "b" column with the conclution the gives us an answer for the matrix.

in both cases i get to part of the conclution but i can't go further
to wright down the "b" column with this conclutions and to
exract the vectors of the basis

?

(by the way)
i am using this method because i was tought that it works for every basis
unlike the tangugation mathod that works only for standart basis

 

[HallsofIvy]

How many different problems do you have here?

The first appears to be "Find the kernel of linear transformation A where A is the matrix
\left[\begin{array}{cccc} 1 & 0 & 0 & -3 \\ 2 & 0 & 4 & -5 \end{array}\right].
from R⁴ to R². It is easy to see that, multiplied by &lt;x_1, x_2, x_3, x_4&gt; gives <x_1- 3x_4, 2x_1+ 4x_3- 5x_4>[/itex] it is equally easy to see that, for arbitrary x_1, x_2, x_3, x_4 those can give any numbers. The image of A is all of R² and so < 1, 0> and <0, 1> are basis vectors. The rest of what you have there, I simply don't understand. And I have no idea what "tangugation" means! Translate please.

 

[transgalactic]

all of my questions are about finding the Im

i have there two cases that i don't know how to find their Im

i also tried to solve them using your method
but i have troubles with it

http://img167.imageshack.us/my.php?image=img8276dz3.jpg

and tangugation i think its the operation of transforming column into rows

i don't have a problem with the first metrix
i just showed an example of how this method is working
the problem is with the second metrix and the last metrix
in the middle i gave another example of how to find the I am
using this method

 

[HallsofIvy]

Your first operator changes the general (x, y, z, t) into (0, -y+ z, 0, 0). Isn't it obvious, that we can pick y and z to make -y+ z anything? (More technically, we can always solve -y+ z= a for any a- and it has an infinite number of solutions.) The image is simply the one dimensional subspace (0, a, 0, 0). You were almost there in writing y(0, -1, 0, 0)+ z(0, 1, 0, 0) since that is equal to (-y+ z)(0, 1, 0, 0). The image is the subspace having basis (0, 1, 0, 0) and, of course, is one-dimensional.

But you say the answer in your book is (0, 1, -1, 0). I don't know how they got that. The "augmented" matrix you show
\left(\begin{array} {ccccc}0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; -1 &amp; 1 &amp; 0 &amp; b_2 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; b_3+ b_2 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \end{array}\right)
Looks like it is the result of row reducing another augmented matrix: it looks like it started as
\left(\begin{array}{ccccc} 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; -1 &amp; 1 &amp; 0 &amp; b_2 \\ 0 &amp; 1 &amp; -1 &amp; 0 &amp; b_3 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \end{array}\right)
In other words, that the original transformation "mapped" (x, y, z, t) to (0, -y+ z, y- z, 0). If that is the case,since y- z= -(-y+ z) and, again, for y and z arbitrary, y- z can be anything, that is of the form (0, a, -a, 0)= a(0, 1, -1, 0) and, in this case, the image is the subspace having basis {(0, 1, -1, 0)}.

So which is, in fact, the correct transformation? (x, y, z, t)-> (0, -y+ z, 0, 0) which is what you give initially, or (x, y, z, t)-> (0, -y+ z, y- z, 0) which is what your book seems to be working with.

For the second problem where T(x, y, z)= (2x- y- z, -y+ z, 0), since 2x-y-z and y- z are arbitrary (there exist at least one solution to 2x- y- z= a and y-z= b for all a and b), any vector in the image if of the form (a, b, 0)= a(1, 0, 0)+ b(0, 1, 0). The image has basis {(1, 0, 0), (0, 1, 0)}.


Swapping rows and columns, in English, at least, is "transposition". What by, the way, is your native language? Your English is excellent.

 

[transgalactic]

ok i understood your method
the only problem that remains for me is
"when to start this process?"

of making A*(x,y,z)

in any general problem i get a metrix like

(a b c)
(d e f)
(s t y)

from substituting A(e1) A(e2) A(e3) into the general formula

than i try to transform it so it will contain as much zeros as possible

(a b c)
(0 e f)
(0 0 y)

it can take several operations of row reduction for me to get to this resolt
i can go further and make this metrix to look like this

(a b 0)
(0 e 0)
(0 0 y)

i got the impression that we can get many I am resolts
because our matrix can change many time.
for example when you worked with
0 0 0 0
0-1 1 0
0 0 0 0

you got an answer that differs the answer in my book
so you took the previos form of this matrix (one step before)
and then you made A*(x,y,z) and then you got the right answer

what is the guide line for me to know
for stopping row reduction procces and starting with A*(x,y,z) process
thanks?