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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} $

Closure under vector addition

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 $M_{2,2}\mathbb{R}$

Can someone please help me verify if my approach is the correct one?

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 $M_{2,2}(\mathbb{R})$

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    $\begingroup$ 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. $\endgroup$ – Dietrich Burde Sep 8 '19 at 16:10
  • $\begingroup$ thank you for having a look. I will make the required changes $\endgroup$ – John Smith Sep 8 '19 at 16:14
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It is not a subspace, e.g. because multiplying by any $\lambda\neq1$ gives a matrix whose upper-left element is not a $2$.

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