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Let v= (1,2)$^T$ be a given vector, and let $S$ = {$A$${ \mathbb{R} }^{2\times2}$ | a$_1$$\bot$v}. (I.e., $S$ is the set of all 2x2 real matrices with column 1 orthogonal to the given vector v.)

How do I show that $S$ is a subspace of $\mathbb{R}^{2\times2}$?

I know that in order for $S$ to be a subspace of $\mathbb{R}^{2\times2}$, $S$ has to satisfy the following conditions:

  1. $S$ can not be an empty set.
  2. If $s$$S$ and $a$ is a scalar, then $as$$S$.
  3. If $s_1, s_2$$S$, then $s_1 + s_2$$S$.

I understand this much, but I have no idea how to go about showing that the following hold true in this case.

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Yes, that's exactly what you must do.

  1. The subspace is obviously not empty, e.g. $$ \begin{pmatrix} 2 & x \\ -1 & y \\ \end{pmatrix} $$ for any $x,y$.
  2. and 3. I've already given you a hint for these two. Your matrices are characterised through the orthogonality of the first column with $$ v= \begin{pmatrix} 1 \\ 2 \\ \end{pmatrix} $$ You can derive from that the form of the column and then check if the sum and scalar product also lie in $S$.
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Hint: If $A\in S$, it is easy to check that $A$ has the following format:

$$A=\begin{bmatrix}2a&b\\-a&c\end{bmatrix},\ a,b,c\in\mathbb{R}.$$

Indeed,

$$a_1^T\cdot v=\begin{bmatrix}2a&-a\end{bmatrix}\cdot\begin{bmatrix}1\\2\end{bmatrix}=2a-2a=0\Rightarrow a_1\perp v.$$

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Hint: prove that, setting $e_1=(1,0)^T$, $$ S=\{A\in\mathbb{R}^{2\times2}:v^T\!Ae_1=0\}$$

Now the zero matrix obviously belongs to $S$. If $A,B\in S$, then $$ v^T(A+B)e_1=\dots $$ Do similarly for $A\in S$ and a scalar $a$.

This is essentially the same as proving that the function $$ f\colon \mathbb{R}^{2\times2}\to\mathbb{R},\qquad f(A)=v^T\!Ae_1 $$ is linear and $S=\ker f$.

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