# Nullspaces relation between components and overall matrix

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1. May 18, 2015

### worryingchem

1. The problem statement, all variables and given/known data
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)?

2. Relevant equations
Ax = 0; x = N(A)

3. 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?

2. May 18, 2015

### Dick

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 $B$.

Last edited: May 18, 2015
3. May 19, 2015

### 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?

4. May 20, 2015

### Dick

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 $B$?

5. May 20, 2015

### 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.