Simple fundamental subspaces problem stumping me

[kostoglotov]

So the matrix is just a row vector

$$\begin{bmatrix}3 & 4 & 0\end{bmatrix}$$

My problem, is that I get the nullspace as having to 2 dimensions, and the row space as having 2 dimenions, but that adds up to 4 dimensions, when it should add up to three. What simple thing am I missing?

Null space base vectors

$$\begin{bmatrix}-4\\3\\0\end{bmatrix},\begin{bmatrix}0\\0\\1\end{bmatrix}$$

Row space base vectors

$$\begin{bmatrix}1\\0\\0\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix}$$

 

[HallsofIvy]

Taking the matrix to be the single row $\begin{bmatrix}3 & 4 & 0\end{bmatrix}$ then the null space is all $\begin{bmatrix}x \\ y \\ z \end{bmatrix}$ such that $\begin{bmatrix}3 & 4 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= 3x+ 4y= 0$. Yes, that is of dimension 2 having $\{\begin{bmatrix}-4 \\ 3 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix}\}$ as basis. However, the row space is one dimensional, having the vector $\begin{bmatrix}3 & 4 & 0 \end{bmatrix}$, the single row making up the matrix, as basis.

 

[Geofleur]

Those two last vectors cannot be in the row space, because a linear combination of them will give the first null space vector, $[-4 \ 3 \ 0]^T$. In fact, the row space is the space spanned by the rows of the matrix, considered as vectors. So the row space consists of all multiples of $[3 \ 4 \ 0]^T$.

EDIT: oops, HallsofIvy beat me to it :-)

 

[kostoglotov]

I realize what I was missing now.

I kept thinking of the row space as all those vector that can be operated on by the matrix without giving a zero vector result. Except that's not what the rowsapce is! It's the vector/s that characterize/s the rows of the matrix. Thanks @HallsofIvy and @Geofleur!