I'm confused on how he wrote this image space answer, can someone explain thanks

[mr_coffee]

Hello everyone I got the answer, but I'm confused on how he wrote out the vectors, like I'm confused on what variables ur allowed to let be any number, and what variables ur actually representing in the vectors. Here is the problem and solution:
http://img233.imageshack.us/img233/5900/lastscan7cs.jpg

i'm confused on how he got
1 0 0
0 1 0
0 0 0
0 0 0

like what numbers did he let be 0, and what numbers did he assign the value too? to form the columns? thanks! maybe if u could write it out more, i would understand better!

 

[matt grime]

That isn't the problem... I don't see at any point you writing what it is you're asked to show, what it is you're doing... all i see is a load of scribble. so how about writing out the question in full?

either type in tex, or similar, eg a matrix could be expressed as

row 1= [1,0,0,0]
row 2= [0,0,1,0]
etc

 

[mr_coffee]

Sorry, all the problem says is:
Find null space and image space of the following Matrix:
1 1 1 1
1 -1 1 -1
0 0 0 1
2 0 2 1

I row reduced this matrix and got:
0 1 0 0
1 0 1 0
0 0 0 1
0 0 0 0

So I'm going to find the image space of the row reduced matrix.
so I'm going to assign x1 through x4 to the matrix and set it equal to a, b, c, d.
[0 1 0 0 ][x1]...[a]
[1 0 1 0][x2]...=
[0 0 0 1][x3]...[c]
[0 0 0 0][x4]...[d]
so now u have:
x2 = a;
x1 + x3 = b;
x4 = c;
then u have a row of 0's, so ur not sure what d is going to be, also x3 seems to be a free variable is that right?
the answer the professor wrote is:
Image Space of the Matrix is:
[1].[0].[0]
[0].[1].[0]
[0].[0].[1]
[0].[0].[0]

meaning, [1 0 0 0] is 1 set of vectors
[0 1 0 0] is another, and [0 0 1 0] is another. What I'm confused on is what did he let equal 1, and what did he let equal 0 to form those vectors out of these equations:
x2 = a;
x1 + x3 = b;
x4 = c;

for example, if he let x1,x2,x3 = 0, and x4 = 1, u would get
[0 0 1 0]
so is this all u do, and u just now let x3 = 1, and all the other x's equal 0?

 

[HallsofIvy]

I agree with Matt- it isn't clear what you are trying to do!

If the problem is to find the image space of that matrix (the image of any linear transformation is a subspace), then what your teacher did was apply that matrix to a "general" vector in R⁴, (a, b, c, d) and write out the result: (b, a+ c, d, 0) and set it equal to the resulting vector (x, y, z, w): x= b, y= a+ c, z= d, w= 0. Since (a,b,c,d) could be anything, a, b, c, d, separately could be anything. Now you can see what the matrix does to a basis for R⁴ by taking each to be 1 in turn while the others are 0. For example (calling the matrix "M") M(1, 0, 0, 0)= (0, 1, 0, 0), M(0, 1, 0, 0)= (1, 0, 0, 0), M(0, 0, 1, 0)= (0, 0, 0, 1, 0), and M(0, 0, 0, 1)= (0, 0, 0, 0). Of course, the zero vector, (0, 0, 0, 0) can't be in a basis but the others are independent: the image of the matrix is the subspace of R⁴ spanned by (1, 0, 0, 0), (0, 1, 0, 0), and (0, 0, 1, 0).
That is, of course, the space subspanned by the "column" vectors- taking the columns of the matrix separately to be vectors.

 

[mr_coffee]

okay i understand it now! thank u Ivy~ One last question, if the row of 0's were say in row 0(first row) insteed of row 3 (last), does that mean the top of each vector set will always be 0?
like i had:
[1].[0].[0]
[0].[1].[0]
[0].[0].[1]
[0].[0].[0]

becuase:
[0 1 0 0 ][x1]...[a]
[1 0 1 0][x2]...=
[0 0 0 1][x3]...[c]
[0 0 0 0][x4]...[d]

there was a row of 0's in the 3rd row, if u say first row is, row 0.

So insteed of it was:

[0 0 0 0][x1]...[a]
[1 0 1 0][x2]...=
[0 0 0 1][x3]...[c]
[0 1 0 0 ][x4]...[d]

would i get:
x1 = 0, all the time.
x1 + x3 = b;
x4 = c;
x2 = d;

would i get:
[0].[0].[0]
[0].[1].[0]
[0].[0].[1]
[1].[0].[0]

and would it still be correct? thanks!~

 

[HallsofIvy]

So insteed of it was:

[0 0 0 0][x1]...[a]
[1 0 1 0][x2]...=
[0 0 0 1][x3]...[c]
[0 1 0 0 ][x4]...[d]

would i get:
x1 = 0, all the time.
x1 + x3 = b;
x4 = c;
x2 = d;

No, you wouldn't. Doing that multiplication gives 0= a (not x1= 0- that may have been a typo), x1+ x3= b, x4= c, x2= d. Taking, again (1,0,0,0), (0,1,0,0),(0,0,1,0), and (0,0,0,1), you would get (0, 1, 0, 0), (0, 0, 0, 1), (0,1,0,0), and (0,0,1,0).

Since (0,1,0,0) appears twice in that, the image is the 3 dimensional subspace spanned by (0,1,0,0), (0,0,1,0), and (0,0,0,1). Again, those are the columns of the matrix.

 

[mr_coffee]

Thanks Ivy! I get it now!