- #1
kamui8899
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The question is:
For each of the following subspaces, find the dimension and a basis:
{(x,y,z) are elements of R^3: 7x - 3y + z = 0}
I had actually posted about this before, but I'm confused as to what the Null space is here.
So, z = -7x + 3y, so there is one dependent variable and two free variables. However, since the equation is equal to 0, the solutions are in the Null Space (or something like that)? So where:
-7 3
1 0
0 1
Is in the Null space, it also provides two independent solutions to the equation. Basically, what I'm trying to say is, how many dimensions does the Null Space have, and is the basis of the Null Space also a basis for the subspace defined by the above equation (if that makes sense)?
Furthermore, written as a matrix:
(7 -3 1)
It goes from R^3 -> R^1 (I can say that right?). So if the dimension of the Null Space is 2, then the dimension of the image is 1, but what is the basis of the image, is that needed in the solution to my question, and graphically what is the image?
I'm also having trouble with Row Reduction of a Matrix, or at least interpreting it. The original matrix is:
2 5 1 0 8 -4
0 3 4 3 2 2
0 6 2 1 3 4
I then reduced it to get:
1 0 0 (-5/36) (101/36) (-11/3)
0 1 0 (-1/9) (4/9) (2/3)
0 0 1 (5/6) (1/6) 0
Now, I need to find a basis for and the dimension of the column Space, the row space and the null space.
So the dimension of the row and column space are = 3 and the dimension of the Null Space = 3.
The basis for the row space is (The above reduced matrix)
The basis for the column Space is:
1 0 0
0 1 0
0 0 1 (?)
Now, the basis for the Null Space is:
-5/36 101/36 -11/3
-1/9 4/9 2/3
5/6 1/6 0
1 0 0
0 1 0
0 0 1
So when I plug in these values into the original equation, I should get 0 right?
So (refer back to the original matrix):
2*(-5/36) + (-1/9)*5 + (5/6)*1 + 1*0 = 0 (or at least it should)
However, these values do not give me 0 when plugged back into the equation, which leads me to believe I'm checking the basis of the Null Space incorrectly, can anyone show me how to check it correctly?
If something came across as unclear feel free to ask me to clarify.
Thanks for the help.
For each of the following subspaces, find the dimension and a basis:
{(x,y,z) are elements of R^3: 7x - 3y + z = 0}
I had actually posted about this before, but I'm confused as to what the Null space is here.
So, z = -7x + 3y, so there is one dependent variable and two free variables. However, since the equation is equal to 0, the solutions are in the Null Space (or something like that)? So where:
-7 3
1 0
0 1
Is in the Null space, it also provides two independent solutions to the equation. Basically, what I'm trying to say is, how many dimensions does the Null Space have, and is the basis of the Null Space also a basis for the subspace defined by the above equation (if that makes sense)?
Furthermore, written as a matrix:
(7 -3 1)
It goes from R^3 -> R^1 (I can say that right?). So if the dimension of the Null Space is 2, then the dimension of the image is 1, but what is the basis of the image, is that needed in the solution to my question, and graphically what is the image?
I'm also having trouble with Row Reduction of a Matrix, or at least interpreting it. The original matrix is:
2 5 1 0 8 -4
0 3 4 3 2 2
0 6 2 1 3 4
I then reduced it to get:
1 0 0 (-5/36) (101/36) (-11/3)
0 1 0 (-1/9) (4/9) (2/3)
0 0 1 (5/6) (1/6) 0
Now, I need to find a basis for and the dimension of the column Space, the row space and the null space.
So the dimension of the row and column space are = 3 and the dimension of the Null Space = 3.
The basis for the row space is (The above reduced matrix)
The basis for the column Space is:
1 0 0
0 1 0
0 0 1 (?)
Now, the basis for the Null Space is:
-5/36 101/36 -11/3
-1/9 4/9 2/3
5/6 1/6 0
1 0 0
0 1 0
0 0 1
So when I plug in these values into the original equation, I should get 0 right?
So (refer back to the original matrix):
2*(-5/36) + (-1/9)*5 + (5/6)*1 + 1*0 = 0 (or at least it should)
However, these values do not give me 0 when plugged back into the equation, which leads me to believe I'm checking the basis of the Null Space incorrectly, can anyone show me how to check it correctly?
If something came across as unclear feel free to ask me to clarify.
Thanks for the help.