Vtyjkyuk

Q: Determine if the following sets are subspaces of $mathbbR^4$ and if they are affine spaces. (Suppose regular operations on the elements of the sets such as addition and skalar multiplication in $mathbbR^4$.)
beginequation
beginsplit
M_1 & = xin mathbbR^4: x_1x_3=0 \
M_2 & = xin mathbbR^4: x_1^2+x_2^2=1
endsplit
endequation

I struggle to determine if $M_1$ and $M_2$ are affine spaces.

a) Let $x,yin M_1$ $alpha,betainmathbbR$. If $M_1$ is a subspace of $mathbbR^4$ then $alpha x+beta y$ should satisfy the equation that define $M_1$.
beginequation
(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)
endequation
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 $mathbbR^4$.

$M_1$ is affine if $M_1=u+V$ for a fix vector $uinmathbbR^4$ and a subspace $VsubsetmathbbR^4$

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 $mathbbR^4$

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 $mathbbR^n$, $M$ 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.

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

• $u_1=u_2$


• $5u_1^2-4u_1^2+3=0 implies u_1=frac4pm sqrt-4410$

which show that such vector $vec u$ doesn't exist.