Find 4x3 matrix B such that AB = I I got stuck with this problem.
\begin{matrix}
        1 & 1 & 1 & 1\\
        0 & 1 & 1 & 0\\
        0 & 0 & 1 & 1\\
        \end{matrix}
Consider the $3\times 4$ matrix $\bf A$ (above). Do the columns of $\bf A$ span $\mathbb R^3$? 
Prove your answer. Also, Find a $4\times 3$ matrix $\bf B$, such that $\bf AB = I_3$
--
I know that the columns of $\bf A$ span $\mathbb R^3$ as there more columns than rows. But I cannot understand how to find matrix $\bf B$ because I cannot implement "super-augmented" matrix and do Gauss-Jordan elimination. Looks like I need to do something with 4th column of $\bf A$ and 4th row of $\bf B$. What do you think?
Thanks!
 A: You can use the augmented matrix method!
\begin{align}
\left[\begin{array}{cccc|ccc}
1 & 1 & 1 & 1 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 1 & 0 & 0 & 1
\end{array}\right]
&\to
\left[\begin{array}{cccc|ccc}
1 & 1 & 0 & 0 & 1 & 0 & -1 \\
0 & 1 & 0 & -1 & 0 & 1 & -1 \\
0 & 0 & 1 & 1 & 0 & 0 & 1
\end{array}\right]
\\&\to
\left[\begin{array}{cccc|ccc}
1 & 0 & 0 & 1 & 1 & -1 & 0 \\
0 & 1 & 0 & -1 & 0 & 1 & -1 \\
0 & 0 & 1 & 1 & 0 & 0 & 1
\end{array}\right]
\end{align}
Now the left part represents the equations and each column in the right part represents the constant terms in a linear system.
For instance, the first system is
$$
\begin{cases}
x_1+x_4=1\\
x_2-x_4=0\\
x_3+x_4=0
\end{cases}
$$
so the first column of a right inverse is (choosing $x_4=0$)
$$
\begin{bmatrix}1\\0\\0\\0\end{bmatrix}
$$
Thus a right inverse is
$$
\begin{bmatrix}
1 & -1 & 0 \\
0 & 1 & -1 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{bmatrix}
$$
A: Denote the $j$-th columnn of $A$ by $a_j$.


*

*The columns of $A$ span $\Bbb R^3$ because the first three columns $a_1,a_2,a_3$ are clearly linearly independent, and $\dim(\Bbb R^3) = \bf 3$, so $\{a_1,a_2,a_3\}$ are a linearly independent spanning subset of $\Bbb R^3$.

*Observe that $e_2 = a_2 - a_1$ and $e_3 = a_3 - a_2$, where $e_i$ is the $i$-th standard unit vector.  Then, by letting $B = \begin{bmatrix}
1 & -1 & 0 \\
0 & 1 & -1 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{bmatrix}$, we have
\begin{align}
AB =& 
\begin{bmatrix}
a_1 & a_2 & a_3 & a_4
\end{bmatrix}
\begin{bmatrix}
1 & -1 & 0 \\
0 & 1 & -1 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{bmatrix} \\
=& \begin{bmatrix}
a_1 & -a_1 + a_2 & - a_2 + a_3
\end{bmatrix} \\
=& \begin{bmatrix}
e_1 & e_2 & e_3
\end{bmatrix} \\
=& I_3.
\end{align}
A: *

*One way to do this is by Kronecker products, where you write matrix multiplication as a matrix and then solve an equation system.

*Another is by using some Pseudo-inverse, for example Moore-Penrose pseudoinverse. For this example you can calculate it like:


$${\bf A}^{+}=({\bf AA}^T)^{-1}{\bf A}$$


*Use Singular Value Decomposition (SVD) : $$\bf A = U\Sigma V^*$$
If $\bf AB=I_3$, what must hold for $\bf B$ in terms of $\bf U,\Sigma,V$?

A: B is a 4*3 matrix. we will find the columns of B using the fact that the columns of AB are linear combinations of columns of A with column elements of B.
$\\1\begin{bmatrix}
1\\ 
0\\ 
0
\end{bmatrix}+0\begin{bmatrix}
1\\ 
1\\ 
0
\end{bmatrix}+0\begin{bmatrix}
1\\ 
1\\ 
1
\end{bmatrix}+0\begin{bmatrix}
1\\ 
0\\ 
1
\end{bmatrix}=\begin{bmatrix}
1\\ 
0\\ 
0
\end{bmatrix}$ So the first column of B is given by $ \begin{bmatrix}
1\\ 
0\\ 
0\\ 
0
\end{bmatrix}$
 similarly $ \\-1\begin{bmatrix}
1\\ 
0\\ 
0
\end{bmatrix}+1\begin{bmatrix}
1\\ 
1\\ 
0
\end{bmatrix}+0\begin{bmatrix}
1\\ 
1\\ 
1
\end{bmatrix}+0\begin{bmatrix}
1\\ 
0\\ 
1
\end{bmatrix}=\begin{bmatrix}
0\\ 
1\\ 
0
\end{bmatrix}$ so the second column of B is given by $\begin{bmatrix}
-1\\ 
1\\ 
0\\ 
0
\end{bmatrix}$
$\\0\begin{bmatrix}
1\\ 
0\\ 
0
\end{bmatrix}-1\begin{bmatrix}
1\\ 
1\\ 
0
\end{bmatrix}+1\begin{bmatrix}
1\\ 
1\\ 
1
\end{bmatrix}+0\begin{bmatrix}
1\\ 
0\\ 
1
\end{bmatrix}=\begin{bmatrix}
0\\ 
0\\ 
1
\end{bmatrix}$ So the third column of B is given by $\begin{bmatrix}
0\\ 
-1\\ 
1\\ 
0
\end{bmatrix}$
$B=\begin{bmatrix}
1 &-1  &0 \\ 
 0& 1 &-1 \\ 
 0& 0 &1 \\ 
 0& 0 & 0
\end{bmatrix}$.
Note that the matrix B is not unique. 
