1
$\begingroup$

I have an equation of the following form:

$$BK=A$$

where $A\in\mathbb{R}^{n\times n}$, $B\in\mathbb{R}^{n\times l}$ and $K\in\mathbb{R}^{l\times n}$, with $l\leq n$. $B$ and $A$ are known, and I want to find some $K$ that satisfies the equation.

Due to the matrices $B$ and $K$ not being square (in general) the obvious problem is that $B$ cannot be inverted to solve the problem directly.

So, there are $n\times n$ equations and $n\times l$ unknowns. I'm trying to express this system in the usual, vector form:

$$\hat{B}k=a$$

with $\hat{B}$ being a square matrix, and $a$ and $k$ unidimensional vectors, to solve it as a common underdetermined linear system.

Is there a way to re-write the original equation in this way?

$\endgroup$
3
  • $\begingroup$ The equation has a solution if and only if each column of $A$ belongs to the column space of $B$. If you denote column $i$ of $A$ by $a_i$, this is equivalent to the solvability of each of the $\ell$ equations $B x = a_i$. $\endgroup$
    – Umberto P.
    May 30, 2018 at 16:48
  • $\begingroup$ @UmbertoP. What would $x$ mean? $\endgroup$
    – Tendero
    May 31, 2018 at 13:26
  • $\begingroup$ If $Bx = a_i$ has a solution for each $i$, then $x$ will be column $i$ of the matrix $K$. $\endgroup$
    – Umberto P.
    May 31, 2018 at 14:44

1 Answer 1

0
$\begingroup$

After some thought I managed to get what I wanted. I'll use an example to make it easier to understand. Suppose that:

$$ A= \left[ {\begin{array}{} a_1 & a_2 & a_3\\ a_4 & a_5 & a_6\\ a_7 & a_8 & a_9\\ \end{array} } \right] $$

$$ K= \left[ {\begin{array}{} k_1 & k_2 & k_3\\ k_4 & k_5 & k_6 \\ \end{array} } \right] $$

$$ B= \left[ {\begin{array}{} b_1 & b_2 \\ b_3 & b_4 \\ b_5 & b_6 \\ \end{array} } \right] $$

Then:

$$BK=A\implies \left[ {\begin{array}{} b_1 & b_2 \\ b_3 & b_4 \\ b_5 & b_6 \\ \end{array} } \right] \left[ {\begin{array}{} k_1 & k_2 & k_3\\ k_4 & k_5 & k_6 \\ \end{array} } \right] = \left[ {\begin{array}{} a_1 & a_2 & a_3\\ a_4 & a_5 & a_6\\ a_7 & a_8 & a_9\\ \end{array} } \right] $$

It can be seen that this set of equations can be also expressed as:

$$ \left[ {\begin{array}{} b_1 & b_2 & 0& 0& 0& 0\\ 0& 0&b_1 & b_2 & 0& 0\\ 0& 0 & 0& 0&b_1 & b_2\\ b_3 & b_4 & 0& 0& 0& 0\\ 0& 0&b_3 & b_4 & 0& 0\\ 0& 0 & 0& 0&b_3 & b_4\\ b_5 & b_6 & 0& 0& 0& 0\\ 0& 0&b_5 & b_6 & 0& 0\\ 0& 0 & 0& 0&b_5 & b_6\\ \end{array} } \right] \left[ {\begin{array}{} k_1 \\ k_4 \\ k_2\\ k_5 \\ k_3 \\ k_6 \\ \end{array} } \right] = \left[ {\begin{array}{} a_1 \\a_2\\ a_3\\ a_4 \\ a_5 \\ a_6\\ a_7 \\ a_8 \\ a_9\\ \end{array} } \right] $$

It's easy to take this to the general case where sizes are not $3$ and $2$ but $n$ and $l$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .