Find matrix $B$ from kernel of $A$ Here's the question:
For the matrix 
$$
A=\begin{bmatrix}
1 & 1 & 0 & 1 & 1\\
1 & 0 & 1 & 1 & 0\\
0 & 1 & 1 & 1 & 1\\
\end{bmatrix}
$$
find a matrix B such that $\text{Im}(\phi_B)=\text{Ker}(\phi_A)$ and  $\text{Ker}(\phi_B)=0$
I have found the null space of $A$ which must be the image of B but I have no idea how to proceed from here. Any tips?
 A: I like using the language of column spaces and null spaces when discussing matrices. 
Presumably, the notation $\phi_M$ refers to the linear map $\phi_M:\Bbb R^n\to \Bbb R^m$ given by $\phi_M(\vec{x})=M\vec{x}$ where $M$ is an $m\times n$ matrix. Then, in our notation, for any matrix $M$ we have
\begin{align*}
\operatorname{Ker}(\phi_M) &= \operatorname{Null}(M) & \operatorname{Im}(\phi_M) &= \operatorname{Col}(M)
\end{align*}
I think this language is useful because it hints at how we might construct our desired matrix $B$. 
We are given the matrix 
$$
A=\left[\begin{array}{rrrrr}
1 & 1 & 0 & 1 & 1 \\
1 & 0 & 1 & 1 & 0 \\
0 & 1 & 1 & 1 & 1
\end{array}\right]
$$
and wish to find $B$ so that $\operatorname{Null}(A)=\operatorname{Col}(B)$. Since $\operatorname{Col}(B)$ is the span of the columns of $B$, we can start by finding a basis of $\operatorname{Null}(A)$ and then inserting these basis vectors into the columns of $B$.
To find a basis of $\operatorname{Null}(A)$, note that
$$
\operatorname{rref}(A)
=
\left[\begin{array}{rrrrr}
1 & 0 & 0 & \frac{1}{2} & 0 \\
0 & 1 & 0 & \frac{1}{2} & 1 \\
0 & 0 & 1 & \frac{1}{2} & 0
\end{array}\right]
$$
This means that every solution $\vec{x}$ to $A\vec{x}=\vec{O}$ is of the form
$$
\vec{x}
=\left[\begin{array}{r}
x_{1} \\
x_{2} \\
x_{3} \\
x_{4} \\
x_{5}
\end{array}\right]
= \left[\begin{array}{r}
-\frac{1}{2} \, x_{4} \\
-\frac{1}{2} \, x_{4} - x_{5} \\
-\frac{1}{2} \, x_{4} \\
x_{4} \\
x_{5}
\end{array}\right]
= x_4\left[\begin{array}{r}
-\frac{1}{2} \\
-\frac{1}{2} \\
-\frac{1}{2} \\
1 \\
0
\end{array}\right]+x_5\left[\begin{array}{r}
0 \\
-1 \\
0 \\
0 \\
1
\end{array}\right]
$$
This gives the basis
$$
\operatorname{Null}(A)=\operatorname{Span}\left\{\left[\begin{array}{r}
-\frac{1}{2} \\
-\frac{1}{2} \\
-\frac{1}{2} \\
1 \\
0
\end{array}\right], \left[\begin{array}{r}
0 \\
-1 \\
0 \\
0 \\
1
\end{array}\right]\right\}
$$
Inserting these two vectors into the columns of a matrix $B$ gives
$$
B=\left[\begin{array}{rr}
-\frac{1}{2} & 0 \\
-\frac{1}{2} & -1 \\
-\frac{1}{2} & 0 \\
1 & 0 \\
0 & 1
\end{array}\right]
$$
How could we verify that this $B$ has our desired properties?
A: Hint:
The column vectors of $B$ will be a basis of $\ker(\phi_A)$.  With this condition satisfied, the matrix $B$ will have full rank, hence will be the matrix of a (linear)  injection.
