find two vectors $ b_1 \in S$ and $ \ b_2 \in S^T$ such that $ b_1+b_2=b=(1,1,1,1).$ Consider the subspace $S$ of $ \ \mathbb{R}^4$ and its orthogonal complement $ \ S^T$ having bases $ \ \{(1,0,0,1), \ (0,1,0,1), \ (0,0,1,1) \}$ and $ \ \{(1,1,1,-1) \}$ respectively. 
Then find two vectors $ b_1 \in S$ and $ \ b_2 \in S^T$ such that
$$ b_1+b_2=b=(1,1,1,1).$$
Answer:
Since $ b_1 \in S$ and $ b_2 \in S^T$, we have 
$b_1 \cdot b_2=0$.
But I can't find out it.
Help me
 A: In this particular question, since $S^T$ is a one-dimensional subspace, we know that $b_2 \in S^T$ must be equal to $\lambda (1, 1, 1, -1)$ for some $\lambda$.  For brevity, let us denote $v := (1, 1, 1, -1)$.
Since we are given that $S^T$ is the orthogonal complement of $S$, then
$$ \langle b, v \rangle = \langle b_1 + b_2, v \rangle = \langle b_1, v \rangle + \langle b_2, v \rangle = 0 + \lambda \langle v, v \rangle. $$
Now substituting in the known values of $b$ and $v$, it is easy to solve for $\lambda$.
Then, once you know $\lambda$, it should be easy from there to calculate what $b_2$ and then $b_1$ must be.
A: Hint: A vector in $S$ is $a(1,0,0,1)+b(0,1,0,1)+c(0,0,1,1)$, and a vector in $S^T$ is $d(1,1,1,-1)$. If the sum of these is to equal $(1,1,1,1)$, can you construct and solve a linear system that gives you the answer?
A: Consider a matrix whose columns are the bases of $S$ and $S^T$ that is$$M=\begin{bmatrix}1&0&0&1\\0&1&0&1\\0&0&1&1\\1&1&1&-1\end{bmatrix}$$then for some vector $v$ we must have $$Mv=\begin{bmatrix}1\\1\\1\\1\end{bmatrix}$$using elementary row operations we obtain from$$\begin{bmatrix}1&0&0&1&1\\0&1&0&1&1\\0&0&1&1&1\\1&1&1&-1&1\end{bmatrix}$$to$$\begin{bmatrix}1&0&0&1&1\\0&1&0&1&1\\0&0&1&1&1\\0&0&0&-4&-2\end{bmatrix}$$which leads to $$v_4={1\over 2}\\v_1=v_2=v_3={1\over 2}$$therefore$$b_1={1\over 2}(1,0,0,1)+{1\over 2}(0,1,0,1)+{1\over 2}(0,0,1,1)\\b_2={1\over 2}(1,1,1,-1)$$
