Calculate matrix A from null space basis of $A-4I$ How to find a matrix $A$ when you are given some parameters and the basis for the null space?
The problem I've been scratching my head over is this. The basis  for the null space of $A-4I$ is
$$\left\{
\begin{pmatrix} 1\\0\\0\end{pmatrix}, 
\begin{pmatrix} 1\\1\\1\end{pmatrix}
\right\}
$$
We also know that the matrix $A$ is square.
I'm pretty confused about what to do; this seems to be implying that there are two free variables but I'm not sure how. 
 A: It sounds from the context like you're looking for a square matrix.
One way to get a matrix with the right nullspace is to work backwards.  That is, if you used row-reduction (Gaussian elimination) to get the matrix into a reduced form, what would it have to look like in order to have this null-space?
In fact, the only row-reduced $3 \times 3$ matrix with the correct null space is
$$
M = \pmatrix{0&1&-1\\0&0&0\\0&0&0}
$$
So, if we had $A - 4I = M$, then $A$ would be a matrix that works.
Perhaps you can take it form there.
A: That's not actually a good basis to work with. If your vectors are $\vec v_1, \ldots, \vec v_k$, then there should be distinct integers $n_1,\ldots,n_k$ such that:


*

*The $n_k$'th entry of $\vec v_k$ is $1$.

*The $i$'th entry of $\vec v_k$ is $0$, for $i>n_k$.

*The $n_k$'th entry of $\vec v_j$ is $0$, for $j\not=k$.


So I will use the basis 
$\left\{\left[\begin{array}{c} 1\\0\\0\end{array}\right], \left[\begin{array}{c} 0\\1\\1\end{array}\right] \right\}$ instead. (Subtract your first vector from the second vector.)
The key is to build the RREF $R$ of $A$. By inspection, the last $1$ occurs in row $1$ of the first vector, and the last $1$ occurs in row $3$ of the second vector. Thus, columns $1$ and $3$ of $R$ do not have pivots; column $2$ does. That means we know $R=\left[\begin{array}{ccc}0&1&?\\0&0&0\\ 0&0&0\end{array}\right]$.
Any question marks at this point are found by taking appropriate entries (above the pivots of $N$) and multiplying them by $-1$. Thus, $?=-1$. (This comes from the second entry of the second vector in my null space basis.) 
Additional example. If the null space basis is $\left\{\left[\begin{array}{c} 2\\1\\0\\0\end{array}\right], \left[\begin{array}{c} -3\\0\\4\\1\end{array}\right] \right\}$ instead, and the matrix is to have $3$ rows, then we proceed as follows:
There are pivots in columns $1$ and $3$ of $R$ only; thus
$$R = \left[\begin{array}{cccc}1&?&0&?\\ 0&0&1&?\\0&0&0&0\end{array} \right].$$
The first entries in the basis vectors are $2$ and $-3$; we multiply each of these by $-1$ and put them in the first row.
$$R = \left[\begin{array}{cccc}1&-2&0&3\\ 0&0&1&?\\0&0&0&0\end{array} \right].$$
The entries in the third row of each basis vector are multiplied by $-1$ to get $0$ and $-4$. The zero is dropped -- it would end up in row $2$, column $2$ -- and the $-4$ replaces the final $?$:
$$R = \left[\begin{array}{cccc}1&-2&0&3\\ 0&0&1&-4\\0&0&0&0\end{array} \right].$$
(This is the second time in the past couple of weeks where a "how do I find a matrix such that ..." has come up. I'm writing up some notes about how to generate random matrices that are "nice" and which have certain useful properties. That's why I had this algorithm in the back of my mind.)
