Determine if the following sets are subspaces of $\mathbb{R}^4$ and if they are affine spaces. Q: Determine if the following sets are subspaces of $\mathbb{R}^4$ and if they are affine spaces. (Suppose regular operations on the elements of the sets such as addition and skalar multiplication in $\mathbb{R}^4$.)
\begin{equation}
\begin{split}
M_1 & = \{ x\in \mathbb{R}^4: x_1x_3=0 \} \\
M_2 & = \{ x\in \mathbb{R}^4: x_1^2+x_2^2=1 \}
\end{split}
\end{equation}
I struggle to determine if $M_1$ and $M_2$ are affine spaces.
a) Let $x,y\in M_1$ $\alpha,\beta\in\mathbb{R}$. If $M_1$ is a subspace of $\mathbb{R}^4$ then $\alpha x+\beta y$ should satisfy the equation that define $M_1$.
\begin{equation}
  (\alpha x_1+\beta y_1)(\alpha x_3+\beta y_3)=\alpha^2x_1x_3+\alpha \beta(x_1y_3+x_3y_1) + \beta^2y_1y_3= \alpha \beta(x_1y_3+x_3y_1)
\end{equation}
But $\alpha x+\beta y \notin M_1$ if for example $\alpha,\beta \neq 0$ and $x_1y_3 \neq 0$ or $x_3y_1 \neq 0$, therefore $M_1$ is not a subspace of $\mathbb{R}^4$.
$M_1$ is affine if $M_1=u+V$ for a fix vector $u\in\mathbb{R}^4$ and a subspace $V\subset\mathbb{R}^4$
b) For instance $(-1,0,0,0)+(1,0,0,0)=(0,0,0,0) \notin M_2$ proves that $M_2$ is not closed under addition and is not a subspace of $\mathbb{R}^4$
 A: For the first one since $\vec 0 \in M_1$ and it is not a subspace we can conclude also that it is not an affine space.
For the second one consider


*

*$(a_1,a_2,a_3,a_4)\in M_2 \implies a_1^2+a_2^2=1$


and


*

*$(b_1,b_2,b_3,b_4)=k[(a_1,a_2,a_3,a_4)-\vec u]+\vec u=(ka_1,ka_2,ka_3,ka_4)+(1-k)\vec u$


then we need to check that the following holds
$$(ka_1+(1-k)u_1)^2+(ka_2+(1-k)u_2)^2=1$$
and for $a_1=1$ and $a_2=0$ we have
$$2k(1-k)u_1+(1-k)^2u_1^2+(1-k)^2u_2^2 =1-k^2$$
and for $a_1=0$ and $a_2=1$ we have
$$(1-k)^2u_1^2+2k(1-k)u_2+(1-k)^2u_2^2 =1-k^2$$
and for $k=-1$


*

*$-4u_1+4u_1^2+4u_2^2 =0$

*$4u_1^2-4u_2+4u_2^2 =0$


that is


*

*$u_1=u_2$

*$-u_1+2u_1^2=0 \implies u_1=0 \quad u_1=\frac12$


and for $k=2$


*

*$-4u_1+u_1^2+4u_2^2 =-3$

*$u_1^2-4u_2+u_2^2 =-3$


that is


*

*$u_1=u_2$

*$5u_1^2-4u_1^2+3=0 \implies u_1=\frac{4\pm \sqrt{-44}}{10}$


which show that such vector $\vec u$ doesn't exist.
A: I'm not sure what your definition of affine space is, but I think this condition is correct: a subset $M$ of a vector space $V$ is an affine subspace if for all $x$ and $y$ in $M$, and scalars $\alpha$ and $\beta$ with $\alpha + \beta = 1$, we have $\alpha x + \beta y \in M$.  
Geometrically, if you draw $V$ as $\mathbb{R}^n$, $M$ is an affine subspace if the line between any two points of $M$ is contained in $M$.  
Negating this definition, $M$ is not an affine subspace if there exist $x$ and $y$ in $M$, and scalars $\alpha$ and $\beta$ with $\alpha + \beta = 1$, such that $\alpha x + \beta y \notin M$.
It shouldn't be too hard to come up with such counterexamples in these two cases.
Since regular subspaces are also affine subspaces, showing that $M_1$ and $M_2$ are not affine subspaces is sufficient.  
