How do I determine whether or not this is a subspace?

I understand the basic axioms that must be proven to find a subspace:

1. Must contain 0 vector
2. Must be closed under addition
3. Must be closed under scalar multiplication

However, for this particular question I'm not sure how to prove it. Any working out you could show would be greatly appreciated.

$S = \{A \in \mathbb R^{2\times 2} : |A| = 0\}$

Supposed to say 2x2 matrix, still haven't gotten the hang of MathJax yet sorry :(

Solution 1:

To determine whether or not $S := \{A \in \mathbb{R}^{2\times 2} : |A|=0\}$ is a subspace of $\mathbb{R}^{2\times 2}$, let's just work through the criterions 1 through 3 you mentioned.

1. $S$ contains the vector $0 \in \mathbb{R}^{2\times 2}$. This is clearly true, since $\left|\begin{matrix}0 & 0\\0 & 0\end{matrix}\right| = 0$.

2. $S$ is closed under addition, that is, $\forall A_1,A_2 \in S:(A_1 +A_2 \in S)$.

This is not as straightforward as criterion 1, so let's think about whether or not this can be true. One way to reason about this is the following:

Let $A_1 = \begin{pmatrix}a & b\\c & d\end{pmatrix} \in S$ and $A_2 = \begin{pmatrix}a' & b'\\c' & d'\end{pmatrix} \in S$. $A_1$ and $A_2$ being in $S$ means that their determinants are $0$, that is, $ad-bc=a'd'-b'c'=0$.
Now let's take a look at the sum $A_1+A_2 = \begin{pmatrix}a + a' & b + b'\\c+c' & d+d'\end{pmatrix}$.

We get $|A_1 + A_2| = (a+a')(d+d')-(b+b')(c+c') = ad + ad' + a'd + a'd' - bc - bc' - b'c - b'c' = (ad - bc) + ad' + a'd + (a'd'- b'c') - bc' - b'c = ad' + a'd - bc' - b'c.$

Now, for $A_1 + A_2$ to be in $S$, its determinant must be $0$. However, we don't really have a reason to think that it's 0, so at this point it's well worth looking for a counterexample. Finding one is mostly just trial and error, trying out a few different matrices and seeing what happens.

Eventually, we might find that $A_1 = \begin{pmatrix}1 & 1\\0 & 0\end{pmatrix}$ and $A_2 = \begin{pmatrix}1 & 0\\1 & 0\end{pmatrix}$ are both in $S$, because $1 \cdot 0 - 1\cdot 0 = 1\cdot 0 - 0\cdot 1 = 0$. However, the determinant of $A_1 + A_2 = \begin{pmatrix}2 & 1\\1 & 0\end{pmatrix}$ is $2 \cdot 0 - 1 \cdot 1 = 0-1 = -1 \neq 0$ and thus $A_1 + A_2 \notin S$. Hence, criterion 2 does not hold and $S$ is not a subspace.

At this point, we are done. For the sake of completeness let's take a look anyway at

3. $S$ is closed under scalar multiplication, that is, $\forall A \in S \; \forall x \in \mathbb{R}: (Ax \in S)$.

This is in fact true, because for $A = \begin{pmatrix}a & b\\c & d\end{pmatrix} \in S$ we have $|A| = ad-bc = 0$. We then get $Ax = \begin{pmatrix}ax & bx\\cx & dx\end{pmatrix}$ and $|Ax| = axdx-bxcx = x^2(ad-bc) = x^2|A| = 0$. Therefore $Ax$ is in $S$ and $S$ is closed under scalar multiplication.

As already mentioned, after finding that criterion 2 does not hold, checking the other two is just unneeded extra work.

Solution 2:

both the matrix with a 1 in the upper left corner and the matrix with a 1 in the lower right corner (0's elsewhere) are singular, but their sum is the identity matrix, which is invertible. so the set is not closed under addition, so it is not a vector space.