Is the set of all singular $2\times2$ matrices a subspace of $R^3$?

Question

Is the set of all singular $2\times2$ matrices a subspace of $R^3$?

I guess that what must be proven is whether the addition of two singular $2\times2$ remains singular or could otherwise be nonsingular. In the case of scalar multiplication it must hold: any singular matrix A, has $det(A)=0$, then $det(\alpha A)=\alpha(ad-cb)=0$ since $\alpha$ would just be a common factor and the equality will remain true.

Answer 1

If $S_{2\times 2}(\mathbb{F})$, the set of $2\times 2$ singular matrices over the field $\mathbb{F}$, is not a subspace of $\mathbb{F}^4$, then it is not a subspace of $\mathbb{F}^3$. the vector space $M_{2\times 2}(\mathbb{F})$ of all $2\times 2$ matrices over $\mathbb{F}$ is isomorphic to $\mathbb{F}^4$. However, $S_{2\times 2}(\mathbb{F})$ is not a subspace of $\mathbb{F}^4$. For example, take $\mathbb{R}=\mathbb{F}$, and look at $\pmatrix{1&0\cr0&0\cr}\quad\hbox{and}\quad \pmatrix{0&0\cr0&1\cr}\ $. They're both singular, but their sum is not. So $S_{2\times 2}(\mathbb{F})$ isn't a subspace of $\mathbb{F}^4\implies S_{2\times 2}(\mathbb{F})$ isn't a subspace of $\mathbb{F}^3$.

Answer 2

$S$ is the set of all singular matrices that $\in \mathbb{R}^{2\times2}$

$C1:A+B \in S$

$\begin{bmatrix} \ a_1 \ a_2 \\ a_3 \ a_4 \end{bmatrix} $ $+$ $\begin{bmatrix} \ b_1 \ b_2 \\ b_3 \ b_4 \end{bmatrix} = \begin{bmatrix} \ a_1 + b_1 \ a_2 + b_2 \\ a_3 + b_3 \ a_4 + b_4 \end{bmatrix}$

$det(A)=0$ implies that $a_1a_4=a_3a_2$

$det(B)=0$ implies that $b_1b_4=b_3b_2$

then:

$det(A+B) = (a_1+b_1)(a_4+b_4)-(a_3+b_3)(a_2+b_2) \\det(A+B) = (a_1a_4 + a_1b_4 + b_1a_4 + b_1b_4) - (a_3a_2+a_3b_2+b_3a_2+b_3b_2)$

now, since $a_1a_4=a_3a_2$ and $b_1b_4=b_3b_2$

$det(A+B) = a_1b_4 + b_1a_4 - a_3b_2 - b_3a_2$ which could not be equal to zero and therefore $A+B$ may not b singular and $A+B \notin S$