# subspaces of the given real vector spaces

Decide which of the following are subspaces of the given real vector spaces. Justify your answers by using the subspace theorem or by giving a specific counterexample to show it is not a subspace.

(a) $$A = \left\{\begin{bmatrix}2 & a_{12} \\ a_{21} & 0\end{bmatrix} :a_{12}, a_{21} \in \mathbb{R}\right\} \subseteq M_{2,2}(\mathbb{R})$$

Solution:

A is not empty

$$\begin{bmatrix}2 & 0 \\ 0 & 0\end{bmatrix} \epsilon \ \mathbb{R}$$

let B = $$\begin{bmatrix}2 & a'_{12} \\ a'_{21} & 0\end{bmatrix}$$

then A + B = $$\begin{bmatrix}2+2 & a_{12}+a'_{12} \\ a_{12}+a'_{21} & 0+0\end{bmatrix}$$

since $$a_{12} + a'_{12} \ \epsilon \ \mathbb{R}$$ and $$a_{21} + a'_{21} \ \epsilon \ \mathbb{R}, \ \ \ A+B \ \ \epsilon \ \ \ M_{2,2}$$

Closure under scalar multiplication

let $$\alpha \ \ \epsilon \ \ \mathbb{R}$$

then $$\alpha A = \begin{bmatrix}\alpha2 & \alpha a_{12} \\ \alpha a_{21} & \alpha0\end{bmatrix}$$

since $$\alpha a_{12}, \alpha a_{21} \ \ \epsilon \ \ \mathbb{R},$$ we have scalar multiplication closure as well.

A satisfies the subspace theorem and is a subspace of $$M_{2,2}\mathbb{R}$$

Updated solution on the advise of Dietrich:

let $$B = \begin{bmatrix}2 & a'_{12} \\ a'_{21} & 0\end{bmatrix}$$

then $$A+B = \begin{bmatrix}2+2 & a_{12}+a'_{12} \\ a_{21}+a'_{21} & 0\end{bmatrix}$$

Since 2+2$${\neq}2$$, A is not a subspace of $$M_{2,2}(\mathbb{R})$$

• But $\alpha\cdot 2\neq 2$ in general, so how can it be closed under scalars? Anyway, $2+2\neq 2$, so how can it be closed under addition? Obviously $A+B$ is not in the space. The left upper entry is not equal to $2$, as it should be. – Dietrich Burde Sep 8 '19 at 16:10
• thank you for having a look. I will make the required changes – John Smith Sep 8 '19 at 16:14

It is not a subspace, e.g. because multiplying by any $$\lambda\neq1$$ gives a matrix whose upper-left element is not a $$2$$.