Nullspaces relation between components and overall matrix

[worryingchem]

Homework Statement

If matrix ## C = \left[ {\begin{array}{c} A \\ B \ \end{array} } \right]## then how is N(C), the nullspace of C, related to N(A) and N(B)?

Homework Equations

Ax = 0; x = N(A)

The Attempt at a Solution

First, I thought that the relation between A and B with C is ## C = A + B ## so then I thought that ## N(C) = N(A) + N(B) ##.
But when I checked the solution it said N(C) = N(A) ∩ N(B)
and the only explanation is that ## Cx = \left[ {\begin{array}{c} Ax \\ Bx \ \end{array} } \right] = 0. ##
Can someone explain the solution to me?

 

[Dick]

Homework Statement

If matrix ## C = \left[ {\begin{array}{c} A \\ B \ \end{array} } \right]## then how is N(C), the nullspace of C, related to N(A) and N(B)?

Homework Equations

Ax = 0; x = N(A)

The Attempt at a Solution

First, I thought that the relation between A and B with C is ## C = A + B ## so then I thought that ## N(C) = N(A) + N(B) ##.
But when I checked the solution it said N(C) = N(A) ∩ N(B)
and the only explanation is that ## Cx = \left[ {\begin{array}{c} Ax \\ Bx \ \end{array} } \right] = 0. ##
Can someone explain the solution to me?

I'm not sure why you aren't getting the explanation. If ##\left[ \begin{array}{c} Ax \\ Bx \ \end{array} \right]## is the zero vector doesn't that mean that BOTH ##Ax## and ##Bx## must be zero vectors? Not just one or the other? ##C## isn't equal to ##A+B##, it's equal to ##\left[ \begin{array}{c} A\\ 0 \ \end{array} \right]+\left[ \begin{array}{c} 0 \\ B \ \end{array} \right]##. Even if it were the null space of ##A+B## is not equal to the null space of ##A## plus the null space of ##B##.

 

[worryingchem]

I didn't understand how they went from ## Cx = \left[ {\begin{array}{c} Ax \\ Bx \ \end{array} } \right] = 0 ## to how N(C) is the intersection of N(A) and N(B). I just saw that the nullspaces were zero, but doesn't all nullspaces contain zero.

After some more thought, I did manage to visualize it if I use row vectors for ## A = \left[ {\begin{array}{cc} 1 & 1 \ \end{array} } \right] ## and ## B = \left[ {\begin{array}{cc} 1 & 2 \ \end{array} } \right] ##.
Then ## N(A) = \left[ {\begin{array}{c} 1 \\ -1 \ \end{array} } \right] ## and ## N(B) = \left[ {\begin{array}{c} 2 \\ -1 \ \end{array} } \right] ##. If I think of the nullspace as column vectors, then it's easier to visualize the nullspaces, and that they are perpendicular to the original matrix and intersect at 0.
When I make C, it would be ## \left[ {\begin{array}{cc} 1 & 1 \\ 1 & 2 \ \end{array} } \right] ## and ## N(C) = \left[ {\begin{array}{c} 0 & 0 \\ 0 & 0 \ \end{array} } \right] ##, the intersection of N(A) and N(B). Is this example right?

When I try to picture a plane though, I don't know how to define a plane in matrix notation and would the nullspace be a second plane with a normal vector orthogonal to the first plane's normal vector?

 

[Dick]

I didn't understand how they went from ## Cx = \left[ {\begin{array}{c} Ax \\ Bx \ \end{array} } \right] = 0 ## to how N(C) is the intersection of N(A) and N(B). I just saw that the nullspaces were zero, but doesn't all nullspaces contain zero.

After some more thought, I did manage to visualize it if I use row vectors for ## A = \left[ {\begin{array}{cc} 1 & 1 \ \end{array} } \right] ## and ## B = \left[ {\begin{array}{cc} 1 & 2 \ \end{array} } \right] ##.
Then ## N(A) = \left[ {\begin{array}{c} 1 \\ -1 \ \end{array} } \right] ## and ## N(B) = \left[ {\begin{array}{c} 2 \\ -1 \ \end{array} } \right] ##. If I think of the nullspace as column vectors, then it's easier to visualize the nullspaces, and that they are perpendicular to the original matrix and intersect at 0.
When I make C, it would be ## \left[ {\begin{array}{cc} 1 & 1 \\ 1 & 2 \ \end{array} } \right] ## and ## N(C) = \left[ {\begin{array}{c} 0 & 0 \\ 0 & 0 \ \end{array} } \right] ##, the intersection of N(A) and N(B). Is this example right?

When I try to picture a plane though, I don't know how to define a plane in matrix notation and would the nullspace be a second plane with a normal vector orthogonal to the first plane's normal vector?

You don't have to visualize it that precisely. If ##x## is the nullspace of ##C##, then ##Cx=\left[ {\begin{array}{cc} Ax \\ Bx \ \end{array} } \right]=\left[ {\begin{array}{cc} 0 \\ 0 \ \end{array} } \right]##, right? Doesn't that mean ##Ax=0## and ##Bx=0##? And doesn't that mean the ##x## is in the nullspace of both ##A## and ##B##?

 

[worryingchem]

Ah, I see.
Then, ## C*N(A) = \left[ {\begin{array}{c} A*N(A)\\ B*N(A) \ \end{array} } \right]=\left[ {\begin{array}{c} 0 \\ n \ \end{array} } \right] ##.
And ## C*N(B) = \left[ {\begin{array}{c} A*N(B)\\ B*N(B) \ \end{array} } \right]=\left[ {\begin{array}{c} n \\ 0 \ \end{array} } \right] ##.
n can be some non-zero numbers.
So in order to make ##\left[ {\begin{array}{c} 0 \\ 0 \ \end{array} } \right] ##, C has to be multiply by something that exists in both N(A) and N(B), the intersection point of the two, and that is x.