Matrices Equation Question

Question

Does there exist three $2\times4$ real matrices $A,B,C$ and reals $a_1,a_2,b_1,b_2,c_1,c_2$ such that: \begin{aligned}a_1A+b_1B+c_1C=\left(\begin{matrix}1&0&0&0\\0&0&1&0\end{matrix}\right)\\a_2A+b_2B+c_2C=\left(\begin{matrix}0&1&0&0\\0&0&0&1\end{matrix}\right)\end{aligned} with $\displaystyle \text{rank}(A)=\text{rank}(B)=\text{rank}(C)=1$?

My attempt The only thing I know is that there exist four $2\times4$ real matrices $A,B,C,D$ and reals $a_1,a_2,b_1,b_2,c_1,c_2,d_1,d_2$ such that: \begin{aligned}a_1A+b_1B+c_1C+d_1D=\left(\begin{matrix}1&0&0&0\\0&0&1&0\end{matrix}\right)\\a_2A+b_2B+c_2C+d_2D=\left(\begin{matrix}0&1&0&0\\0&0&0&1\end{matrix}\right)\end{aligned} with $\displaystyle \text{rank}(A)=\text{rank}(B)=\text{rank}(C)=\text{rank}(D)=1$. This is obvious since we can just let: $a_1=b_1=c_2=d_2=1$ and $a_2=b_2=c_1=d_1=0$ with \begin{aligned}A=\left(\begin{matrix}1&0&0&0\\0&0&0&0\end{matrix}\right)\\B=\left(\begin{matrix}0&0&0&0\\0&0&1&0\end{matrix}\right)\\C=\left(\begin{matrix}0&1&0&0\\0&0&0&0\end{matrix}\right)\\D=\left(\begin{matrix}0&0&0&0\\0&0&0&1\end{matrix}\right)\end{aligned} However, I'm curious about whether we can satisfy the above two equations using only three or fewer matrices. Any help will be appreciated.

Answer 1

Your system is equivalent to:

$\begin{pmatrix}a_1I & b_1I & c_1 I\\ a_2I & b_2I & c_2 I\end{pmatrix}_{4\times 6}\begin{pmatrix}A\\ B\\ C\end{pmatrix}_{6\times 4} = \pmatrix{1&0&0&0\\ 0&0&1&0\\0&1&0&0\\ 0&0&0&1}_{4\times 4}\tag*{}$

where $I$ is the $2\times 2$ identity matrix.

You restrict $A$, $B$ and $C$ to be rank-one. This implies that the maximum possible row-rank of the $6\times 4$ matrix in the above equation is $3$.

To reconcile yourself with the above implication, think in terms of row span. Rows of $A$ span a one-dimensional space. So do rows of $B$ and $C$. So when we stack $A$, $B$ and $C$ one over the other, the over all row span is at most a three-dimensional space.

If $X$ is an $m\times n$ matrix and $Y$ is an $n\times p$ matrix then $\text{rank}(XY)\leq \min(\text{rank } X, \text{rank } Y).\tag*{}$ See this for reference.

Using this result, $3$ is the maximum possible rank of the product in the LHS. However, the RHS is a rank $4$ matrix. (Contradiction!)

So the answer to the Does there exist... question is No.

Answer 2

If the ranks of $A, B, C$ are all equal to 1, it means they can be written resp. in the following product form :

$$A=P_{2 \times 1}Q^T_{1 \times 4}, \ \ \ B=R_{2 \times 1}S^T_{1 \times 4}, \ \ \ C=U_{2 \times 1}V^T_{1 \times 4}$$

for certain matrices with the indicated sizes.

As the rank of $Span(Q,S,V)\subset \mathbb{R^4}$ is at most $3$, there exist a non-zero vector $W \in \mathbb{R^4}$ orthogonal to $Q,S,V$, i.e., such that :

$$Q^TW=0, \ \ S^TW=0, \ \ V^TW=0 \tag{1}$$

As a consequence, right-multiplying LHS and RHS of your first equation by $W$ gives, using (1) :

$$a_1P\underbrace{Q^TW}_0+a_2R\underbrace{S^TW}_0+a_3U\underbrace{V^TW}_0= \underbrace{\pmatrix{1&0&0&0\\ 0&0&1&0}}_{M_1}W$$

In short :

$$M_1W=0$$

For the same reason, using your second equation, $M_2W=0$ with $M_2:=\pmatrix{1&0&0&0\\0&0&1&0}$.

But, grouping the two constraints, it would mean that $MW=0$ where $M=\pmatrix{M_1\\M_2}_{4 \times 4}$ which is impossible because the rank of $M$ is $4$ (i.e., its kernel is reduced to $\{0\}$).

Remark : $M$ is the same matrix as used in the solution by @Nothing special.