# Show that S is a subspace of ${R}^{2\times2}$

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.

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 $$S$$.

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,

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

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 $$S=\ker f$$.