Understanding "dimension of row space" of a matrix using an example

Question

Qn: If the null space of a 7x5 matrix A is 4-dimensional, what is the dimension of the row space A?

Ans: dimension of row A = 1

I feel like my confusion comes from understanding what the question want when they say dimension, got validated by this thread. Please can you correct/point out any of my knowledge gap?

I'd like to verify the following:

1. My intuitive definition of "dimension of row space" is n.o of rows in A. Am I correct here? (Wikipedia seem to confirm that row space is n.o of rows in A).

2. Confusingly, also according to wikipedia, "dimension of row space" = "rank" = "n.o of linearly independent columns of A". I'm assuming this is for square matrix only and does not apply here?...

3. Given the answer being 1 and considering the above, I try and work backwards, set the following with an example of how I can achieve a single row, is this correct?

4. Feels like a roundabout way to address this question. Is there a more efficient way to tackle this question?

$A\vec{v}=\vec{0}$

$\begin{pmatrix}1&a&b&c&d\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\end{pmatrix}\begin{pmatrix}v_1\\ v_2\\ v_3\\ v_4\\ v_5\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\\ 0\\ 0\end{pmatrix}$

$v_1+av_2+bv_3+cv_4+dv_5 = 0 \rightarrow v_1 = -av_2-bv_3-cv_4-dv_5$

$Nullspace (\vec{v}) = v_2\begin{pmatrix}-a\\ 1\\ 0\\ 0\\ 0\end{pmatrix}+v_3\begin{pmatrix}-b\\ 0\\ 1\\ 0\\ 0\end{pmatrix}+v4\begin{pmatrix}-c\\ 0\\ 0\\ 1\\ \:0\end{pmatrix}+v_5\begin{pmatrix}-d\\ 0\\ \:0\\ \:0\\ 1\end{pmatrix}$ , such that $v_2, v_3,v_4,v_5$ are $\in \mathbb{R}$.

Answer 1

I assume that you are familiar with the concept of elimination. With that in mind

1. My intuitive definition of "dimension of row space" is n.o of rows in A. Am I correct here?

No, the dimension of $A$'s row space (also that of its column space) is the number of pivots you have after elimination, i.e., the rank. This can certainly be smaller than $m$.

2. Confusingly, also according to wikipedia, "dimension of row space" = "rank" = "n.o of linearly independent columns of A". I'm assuming this is for square matrix only and does not apply here?

No, you can do elimination on rectangular matrices to get the pivots just as well.

3. Given the answer being 1 and considering the above, I try and work backwards, set the following with an example of how I can achieve a single row, is this correct?

• Dimension of nullspace = number of columns without pivots
• Dimension of row/column space = number of pivots

Their sum obviously equals the number of elements in a row. So yes the answer is $5 - 4 = 1$.

Your single non-zero row example is valid but trivial. You can have multiple non-zero rows in $A$ that all become zero rows (and thus produce no pivots) after elimination. This is the case when the rows are linearly dependent.

For example, the dimension of the row space of the following matrix is also 1.

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

4. Feels like a roundabout way to address this question. Is there a more efficient way to tackle this question?

Imo the pivot view is one of the most intuitive. Another way is to see from a vector space viewpoint:

• $A$'s columns are vectors of $R^m$, but don't necessarily spans all $R^m$. It's like many $R^3$ vectors that all lie on the same plane/line (linearly dependent) cannot produce all vectors of $R^3$.
• $A$'s rows are vectors of $R^n$, but don't necessarily spans all $R^n$
• The dimensionality of $A$, i.e., $dim(A)$, is the number of linearly independent $R^m$ vectors among its columns, or linearly independent $R^n$ vectors among its rows. Either ways this number is the same.

The two viewpoints agree nicely, as the number of pivots signal the number of linearly independent vectors among $A$'s rows/cols.

I suggest going through "Introduction to Linear Algebra" by Gilbert Strang to get a firm hold of the basics. Most of what I write here was taken from his book.