There is a 7-by-6 matrix $H$ given. Its rank is 6. I'd like to design a 6-by-5 matrix $D$ such that the following holds:

$ \left[ \begin{array}{l} l_1(a_1, a_2, a_3, a_4) \\ l_2(a_1, a_2, a_3, a_4) \\ l_3(a_1, a_2, a_3, a_4) \\ l_4(a_1, a_2, a_3, a_4) \\ l_5(l_1, a_5) \\ l_6(l_1, a_5) \\ l_7(l_1, a_5) \end{array} \right] $ = $H$$D$ $\left[ \begin{array}{l} a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \end{array} \right]$,

where $l_k(a,b,c)$ denotes a linear combination of $a, b, c$, and subscript $k$ is to denote different linear combinations.

The top four linear combinations of $a_1, a_2, a_3, a_4$ are independent. Two out of the bottom three linear combinations of $l_1, a_5$ are independent.

This probem arises from a communication problem where a transmitter has multiple antennas, and the total number of antennas of multiple receivers is greater that the number of antennas the transmitter has. The transmitter wishes to transmit different symbols to different receivers.

If my description is not clear enough, please ask. I need help from who are familiar with matrix decomposition, linear transformation, etc.


  • $\begingroup$ Do you know anything else about the given matrix $H$? If not, then its possible for the $i$-th row of $H$ to be all $0$'s, in which case, the $i$-th row of $HD\vec{a}$ will also be $0$ for any matrix $D$. $\endgroup$ – JimmyK4542 Dec 20 '14 at 4:34
  • $\begingroup$ @JimmyK4542 Hi, thanks for your comment. All entries of $H$ will be non-zero. $\endgroup$ – PurplePenguin Dec 20 '14 at 4:35

Let $L$ be the $7 \times 5$ matrix such that $L\begin{bmatrix}a_1\\a_2\\a_3\\a_4\\a_5\end{bmatrix} = \begin{bmatrix}l_1(a_1, a_2, a_3, a_4) \\l_2(a_1, a_2, a_3, a_4) \\l_3(a_1, a_2, a_3, a_4) \\l_4(a_1, a_2, a_3, a_4) \\l_5(l_1, a_5) \\l_6(l_1, a_5) \\l_7(l_1, a_5)\end{bmatrix}$.

We wish to find a $6 \times 5$ matrix $D$ such that $HD\vec{a} = L\vec{a}$ for every $\vec{a}$, i.e. $HD = L$.

A solution (if one exists) is $D = H^{+}L$ where the $H^{+}$ is the pseudoinverse of $H$.

Since $H$ is a $7 \times 6$ matrix with rank $6$, $H$ has linearly independent columns.

Hence, $H^{+} = (H^TH)^{-1}H^T$, and thus, $D = (H^TH)^{-1}H^TL$.

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  • $\begingroup$ Thanks. So, we can't guarantee the existence of a solution? $\endgroup$ – PurplePenguin Dec 20 '14 at 4:59
  • $\begingroup$ It depends on what $H$ is. For instance, suppose the top $4$ rows of $H$ are linearly dependent (which is possible since $H$ has rank $6$). Then, the top $4$ rows of $HD$ will be linearly independent while the top $4$ rows of $L$ are linearly independent. So, there won't be a solution in this case. Of course, if $H$ is "nice" we won't have this problem. $\endgroup$ – JimmyK4542 Dec 20 '14 at 5:05
  • $\begingroup$ What if we assume all possible submatrices have either full column rank or full row rank depending on the size of them? In other words, if a submatrix is of size $m$-by-$n$, then its rank is $\min (m, n)$. $\endgroup$ – PurplePenguin Dec 20 '14 at 5:08
  • $\begingroup$ That might work, I'd have to think about it for a bit. Regardless, a necessary and sufficient condition for the existence of a solution $D$ to $HD = L$ is that the columns of $L$ lie in the span of the columns of $H$. Also, if there isn't a solution, then $D = H^{+}L$ gives you the matrix $D$ which minimizes $\|HD-L\|_F^2$ (Frobenius norm), i.e. that will be the next best thing to a solution. $\endgroup$ – JimmyK4542 Dec 20 '14 at 5:29
  • $\begingroup$ Thanks. I should think about it too based on your comments. Linear combinations of $L$ could be anything depending on $H$, as long as the dependencies between the linear combinations of $L$ hold the same as mentioned. If you have come up with an idea to achieve the goal, please let me know. You've been a lot of help. Thanks again! $\endgroup$ – PurplePenguin Dec 20 '14 at 5:39

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