How to describe the space of all $2 \times 2$ matrices of rank $1$?

Question

This problem came with a statement which I already proved.

$M(n;\mathbb{R})$ has a linear subspace $V$ of dimension $n(n-1)$ which consists entirely of singular matrices.

So, I tried to describe the set $B$ which consists entirely of $2 \times 2$ matrices of rank $1$ in terms of dimension and isomorphisms with the statement above.

Let the set $A = \{ M \in M(2,\mathbb{R}) | \det(M)=0 \}$.

Then, by the statement, $\dim(A)=2$.

But, note that every $2 \times 2$ matrices that have rank of $1$ are in the form $$\begin{bmatrix} a & ka \\ b & kb \\ \end{bmatrix}$$ where $k \in \mathbb{R}$, and "$a \neq 0$ or $b \neq 0$."

Then, we can see that the determinants of such matrices are always $0$.

Now, let the set $B$ be the set of all such matrices.

Clearly, $B \subset A$.

As we want to show that $B$ is a space in $A$, we can apply the subspace test to $B$.

Let $m_1 = \begin{bmatrix} a & k_1a \\ b & k_1b \\ \end{bmatrix}$, and $m_2 = \begin{bmatrix} c & k_2a \\ d & k_2b \\ \end{bmatrix}$. Then, $m_1 + m_2 = \begin{bmatrix} a+c & k_1a+k_2c \\ b+d & k_1b + k_2d \\ \end{bmatrix}$.

I had to stop here because the second column of this matrix may not depend on the first column.

Questions:

• How do we show a set is a space? (It is a fundamental question, but I want to clarify the distinction between the space and the subspace.) I also went over the axioms of vector space, but this also does not work for the same reason.
• How should be the statement applied to this problem? I.e: Assuming we have a dimension of the set(or space) $B$, what can we say about $B$ in terms of dimension?

Edit: Thanks to all of the help, I concluded the set (call it $M$) as

$M=\bigcup_{k \in \mathbb{R}}\text{span} \left\{ \begin{bmatrix} 1 & k \\ 0 & 0 \\ \end{bmatrix}, \begin{bmatrix} 0 &0 \\ 1 &k \\ \end{bmatrix} \right\} \setminus \{0\}$, which is an infinite union of subsets of $A$ with each $k \in \mathbb{R}$.

I hope this is the right description of $M$.

Another question:

• However, I'm still wondering if this set still could be categorized as a space.

Answer 1

Hint

Your idea to consider matrices of the form $$\begin{pmatrix} a & ka\\ b & kb\end{pmatrix}$$ goes in the right direction. If you fix $k$ and let $a,b$ be variable in $\mathbb R$, you then get a dimension $2$ linear subspace. Which can be proven using the subspace axioms.

Even simpler,

$$V=\left\{\begin{pmatrix} a & a\\ b & b\end{pmatrix} \mid (a,b) \in \mathbb R^2\right\}$$ is a linear subspace of $M(2;\mathbb R)$ of dimension equal to $2$ (can you find a basis?).