# Matrices Equation 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.

• look at the equations row by row. It turns out you are trying to obtain 4 linearly independent vectors as combinations of 3 vectors. Commented Aug 9 at 17:47
• What do you mean by 4 linearly independent vectors？ Commented Aug 9 at 17:52
• OK I understand what you‘re saying. thanks. Commented Aug 9 at 18:04
• My solution (exhibiting a contradiction) wouldn't work with a fourth matrix $D$ as you are doing because the rank of the span (refer to my answer) could be 4, therefore no such vector as "my" $W$ could be found... Commented Aug 9 at 21:28

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

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.

• [+1] Clever solution. But you should recall the theorem $\operatorname{rank}(AB) \le \min(\operatorname{rank} A, \ \operatorname{rank} B)$ Commented Aug 9 at 19:34
• I think I have another solution. Commented Aug 9 at 19:45
• @JeanMarie Also, we need the result that row rank is same as column rank and this common value is called the rank of the matrix. I'll update this answer so that it's accessible to more people. I would love to see your solution, professor. Commented Aug 9 at 19:56
• I don't think you need to invoke here "row rank=column rank". Commented Aug 9 at 21:25

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.