<s>[INST] How to find a basis and dimension of two subspaces together with their intersection space? 
Given a vector $[a,b,c,d]^T$ ,find a basis for the subspace $S$ of $\Bbb{R}^4$ such that $a+c+d = 0$
and for the subspace $T$ such that $a+b = 0$ and $c = 2d$.
What is the dimension of the intersection $S∩T$?


My approach, it is known that to form a basis the vectors must be linearly independent, and this should be a spanning set. 
Basis for subspace $S$,
$\begin{align}&\{(a,b,c,d)| a+c+d=0\}\\&=\{(a,b,-a-d,d) | a+c+d=0\}\\&=\{a(1,0,-1,0)\} +\{b(0,1,0,0)\}+\{d(0,0,-1,1)\}\end{align}$
Hence $\dim = 3$ and basis is $\{(1,0,-1,0),(0,1,0,0),(0,0,-1,1)\}$
Basis for subspace $T$,
$\begin{align}&\{(a,b,c,d): a+b=0 \text{ and } c= 2d\}\\&=\{(a,-a,2d,d)\}
\\&=\{a(1,-1,0,0) + d(0,0,2,1)\}\end{align}$
$\dim T= 2$  and  basis is $\{(1,-1,0,0),(0,0,2,1)\}$.
Hence
$\dim(S∩T)= 1 $
I'd appreciate corrections or if there is a more technical way to approach this. I am little confused on the transpose part as well, I want to understand the significance of this representation. Might be it wanted to reflect that those are column vectors. I need validation on the answer. thanks!
Note: Here $[a, b, c, d]^T$ means the column vector $\{\begin{pmatrix}a\\b\\c\\d\end{pmatrix}\}$
 [/INST] $S=\textrm{null}\begin{pmatrix}1&0&1&1\end{pmatrix}$
$\begin{align}\mathcal{B}_S&=\{\begin{pmatrix}0\\1\\0\\0\end{pmatrix},\begin{pmatrix}-1\\0\\1\\0\end{pmatrix},  
\begin{pmatrix}-1\\0\\0\\1\end{pmatrix}\}\end{align}$
$\dim S=3$

$T=\textrm{null}\begin{pmatrix}1&1&0&0\\0&0&1&-2\end{pmatrix}$
$\mathcal{B}_T=\{\begin{pmatrix}-1\\1\\0\\0\end{pmatrix},\begin{pmatrix}0\\0\\2\\1\end{pmatrix}\}$
$\dim T=2$

$S\cap T=\textrm{null}\begin{pmatrix}1&0&1&1\\1&1&0&0\\0&0&1&-2\end{pmatrix}$
$\begin{align}\textrm{rref}\begin{pmatrix}1&0&1&1\\1&1&0&0\\0&0&1&-2\end{pmatrix}&=\begin{pmatrix}1&0&0&3\\0&1&0&-3\\0&0&1&-2\end{pmatrix}\end{align}$
$\mathcal{B}_{S\cap T}=\{\begin{pmatrix}-3\\3\\2\\1\end{pmatrix}\}$
$\dim(S\cap T) =1$
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