# Row and null complements of x; need clarity...

Tags:
1. Oct 15, 2015

### kostoglotov

I've managed to distill the rambling into just this question, posted here and at the end of my digressive thoughts as well:

"Will we always be able to split x up in such a way that we have a nullspace component and a non-row space component?"

Take a matrix
$$A = \begin{bmatrix}1 & 2\\ 3 & 6\end{bmatrix} \ \text{with} \ \vec{x}=\begin{bmatrix} 4 \\ 3 \end{bmatrix}$$

Split $\vec{x}$ into it's null-space and row-space components.

Now, the example in the text is fine

First splitting of x:

$$\begin{bmatrix}1 & 2\\ 3 & 6\end{bmatrix}\left(\begin{bmatrix} 2 \\ -1 \end{bmatrix}+\begin{bmatrix} 2 \\ 4 \end{bmatrix}\right) = \begin{bmatrix} 10 \\ 30 \end{bmatrix}$$

I can easily see that one x vec is in the nullspace and one x vec used is in the row space.

But this other set of two x's also works

Second splitting of x:

$$\begin{bmatrix}1 & 2\\ 3 & 6\end{bmatrix}\left(\begin{bmatrix} -2 \\ 1 \end{bmatrix}+\begin{bmatrix} 6 \\ 2 \end{bmatrix}\right) = \begin{bmatrix} 10 \\ 30 \end{bmatrix}$$

The first vec in this second splitting is still in the nullspace, but the second one is not in the row-space.

After all, the row space of A, $C(A^T)$ just contains combinations of the vector

$$\begin{bmatrix} 1 \\ 2 \end{bmatrix}$$

I can't see how any linear combination of that vector could produce the (6,2) vec I used in the second splitting of x.

Is this a problem? I don't see how it could be since both splittings produce the same answer, an answer that is in the column space of A.

However, this is also taking place in $R^2$, and by the fundamental theorem, shouldn't that whole space be partitioned by the subspaces? NO...that's dumb. Row space is just a line through zero in the xy and nullspace is a line through zero in xy perp to row space...that's not a partition, it's subsets.

But does that THEN mean, that any vector in xy, that is NOT in the null space, always dot products with row space vectors to produce vectors in the column space? Yes...of course...it has to.

Will we always be able to split x up in such a way that we have a nullspace component and a non-row space component?

2. Oct 15, 2015

### andrewkirk

Yes, because the null and row spaces are orthogonal complements, and hence have intersection {0}.

So if we have $\vec{v}=\vec{n}+\vec{r}$ where the two vectors on the RHS are in the null and row spaces respectively, then we can write $\vec{v}=2\vec{n}+(\vec{r}-\vec{n})$ and the first vector on the RHS is in the nullspace but the second is in neither the null nor the row space (the proof of that is left as an exercise).