Modified from a Linear Algebra exercise
$S$ is a subspace of $\mathbb{R}^4$ where a basis of $S$ is $\{ {\begin{pmatrix} 1 \\ 0 \\ 1\\1 \end{pmatrix}}, {\begin{pmatrix} 5 \\ 0 \\ 2\\2 \end{pmatrix}} \}$. Give a basis for a subspace $U$ of $\mathbb{R}^4$ such that $S\cap U = {\bf\underset{\sim}{0}}$ and $S+U=\mathbb{R}^4$
My attempt:
Firstly the basis of $S$ can be simplified to the form $\{ {\begin{pmatrix} 1 \\ 0 \\ 0\\0 \end{pmatrix}}, {\begin{pmatrix} 0 \\ 0 \\ 1\\1 \end{pmatrix}} \}$ since ${\begin{pmatrix} 5 \\ 0 \\ 2\\2 \end{pmatrix}}$ can be expressed as a linear combination of the above 2 basis vectors.
For $S+U=\mathbb{R}^4$, Basis of $S \cup$ Basis of $U$ is a spanning set. Thus there must be 2 basis vectors $U$ in the form $\begin{pmatrix} 0 \\ a \\ 0\\0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 0 \\ b\\0 \end{pmatrix}$ for constants $a,b$
Since $S \cap U = {\bf\underset{\sim}{0}}$, the basis vectors of $U$ and negative vectors of $S$ must be linearly independent, as for intersection:
$x_1\begin{pmatrix} 0 \\ a \\ 0\\0 \end{pmatrix} + x_2\begin{pmatrix} 0 \\ 0 \\ b\\0 \end{pmatrix} = y_1\begin{pmatrix} 1 \\ 0 \\ 0 \\0 \end{pmatrix} + y_2\begin{pmatrix} 0 \\ 0 \\ 1 \\1 \end{pmatrix}$
Checking $a=b=1$, vectors are linearly independent as determinant $= 1>0$.
Thus basis vector : $\{ \begin{pmatrix} 0 \\ 1 \\ 0\\0 \end{pmatrix} ,\begin{pmatrix} 0 \\ 0 \\ 1 \\0 \end{pmatrix} \}$
Question: it is the right answer, but I am not convinced that my method is right.
Firstly, $dim(S+U)=dim(S)+dim(U)-dim(S\cap U)$ so $dim(S)=3$, don't we require 3 basis vectors. How are we certain that $dim(U)=2$ as shown
Secondly, is there a rigorous way to show that the basis vectors of $U$ is of that form? I guessed the form of the basis vectors without proof. My idea as that the second column of the basis of $S$ is 0 so the basis vector must be non-zero for the second entry for it to span the vector space. Also, to span it, the third and fourth column must not be always the same value which is the case for $S$, thus for $U$, one is different than the other.
Thank you!
