# Thread: Method for working out if subspace is closed under Vec+/Scal.x

1. ## Method for working out if subspace is closed under Vec+/Scal.x

Hi, I understand what these two ideas mean, but I'm a bit lost on how to actually sit down and work out algebraically whether a set contains both properties or not.

An example to illustrate my point:

Determine whether the set of all 2x2 matrices whose determinant is 0 is a subspace of M22 (all 2x2 matrices).
My thinking went something like this

Matrix 2. FG-HJ

hence if (A+F)(D+G)-(B+H)(C+J)=0 then the subset was close under vector addition. What is wrong with my thought train here?

For multiplication I figured something like

hence subset was closed under scalar multiplication, is this correct?

Nic

2. Its been a while but if I remember correctly you have to prove the following:

1. For any given matrix $A\in F^{2\times 2}$ (F is a scalar field, usually $\mathbb{R}$) whose determinant $det A=0$, and for every scalar $\lambda \in F$, the matrix $\lambda\cdot A$ is also element of the space $\mathbb{F}^{2\times 2}$ and has the same property as the matrix A: $det\left(\lambda \cdot A\right)=0.$.

This one is easy to prove. First, it is obvious that: $\forall A\in F^{2\times 2}, \,\, \forall \lambda \in F$ the matrix $\lambda A \in F^{2\times 2}$, no doubt about that. Just recall the manner in which matrices are multiplied by scalars. Now one has to see if the matrix $\lambda A$ has the same property as the matrix A. If one can prove that $det\left(\lambda A \right)=0$ then we proved that subset is closed under multiplication by scalars. And this is easy too:
$det\left( \lambda A \right)=\lambda^2 \cdot \det A=\lambda^2\cdot 0=0.$ ( $\lambda^2$ because you extract one $\lambda$ per row of the determinant).

2. One still has to prove that for any two matrices A and B, both having zero determinant values, matrix A+B will also have a zero determinant value. Now, I hope someone will correct me if I'm wrong but I think this is not the case. A counterexample comes to mind:
Matrices $A=\left[\begin{array}{cc} 3 & 2\\ 6 & 4\end{array}\right]$ and $B=\left[\begin{array}{cc} 8 & 6\\ 4 & 3\end{array}\right]$ both have zero determinants, but the matrix $A+B=\left[\begin{array}{cc} 11 & 8\\ 10 & 7\end{array}\right]$ does not. $det\left(A+B\right)=-2$.

So it turns out that it is not a subspace closed under matrix/vector addition.

I hope I didn't make a mistake somewhere. As I said, its been a while...

3. Originally Posted by Nicholas
Hi, I understand what these two ideas mean, but I'm a bit lost on how to actually sit down and work out algebraically whether a set contains both properties or not.

An example to illustrate my point:

Determine whether the set of all 2x2 matrices whose determinant is 0 is a subspace of M22 (all 2x2 matrices).
My thinking went something like this

Matrix 2. FG-HJ

hence if (A+F)(D+G)-(B+H)(C+J)=0 then the subset was close under vector addition. What is wrong with my thought train here?
Yes, IF that is equal to 0, then the subset would be closed under vector addition- but haven't shown that it is 0! Mathoman gives an example showing that it is not.

For multiplication I figured something like