So the given subspaces are
$ S = \mathbf{span} \{ (1,0,1,-1)^{T}, (0,1,0,-1)^{T} \} $
and
$ T = \mathbf{span}\{(-1,1,3,1)^{T}\} $
Using identities
$ (S + T)^{0} = S^{0} \cap T^{0}$ and $ (S \cap T)^{0} = S^{0} + T^{0}$
I can determine basis for $S^{0}$ and $T^{0}$ first.
So for S^{0} I find
$B_{1} = \{x_{1}-x_{3}, x_{2}+x_{3}+x_{4}\}$ to be a basis.
Now for $T^{0}$ is where I start having problems.
$T^{0}$ has dimension 3.
Since $ T = \mathbf{span}\{(-1,1,3,1)^{T}\} $
I rewrite its anhilator as
$T^{0} = \mathbf{span}\{\phi \in T^{*}: \phi(-1,1,3,1)^{T}=0\}$ (is this right?)
So I can just find three functionals in $T^{*}$ that vanish in $(-1,1,3,1)$
So a basis of $T^{0}$ is $ \{x_{1}+x_{3}, x_{2}-x_{4}, 3x_{2}-x_{3}\}$
Now to find a basis of $S^{0} + T^{0}$, I take the span of the union of $B_{1}$ and $B_{2}$. Which I guess is correct, but this isnt an explicit answer, how can I fix this? I know I must eliminate one vector from the union, since the dimension of this sum of subspaces is at most 4.
Now for the basis of $ (S + T)^{0} = S^{0} \cap T^{0}$ Where I find the basis to be $\mathbf{span} \{ x_{1} \}$
Can I get some feedback on this? I really appreciate any help. Thanks.