Find 4x3 matrix B such that AB = I

Question

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 $\bf AB = I_3$

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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!

Answer 1

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 $$ \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} $$

Answer 2

1. One way to do this is by Kronecker products, where you write matrix multiplication as a matrix and then solve an equation system.
2. Another is by using some Pseudo-inverse, for example Moore-Penrose pseudoinverse. For this example you can calculate it like:

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

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

Answer 3

Denote the $j$-th columnn of $A$ by $a_j$.

1. 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 $\Bbb R^3$.

2. 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}

Answer 4

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.