matrix for this system of non-linear equation?

Question

Let's consider the product of a square matrix and a vector: $ M{\times}v = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} {\times} \begin{bmatrix} a \\ b \\ c \end{bmatrix}= \begin{bmatrix} 1 {\times} a + 2 {\times} b + 3 {\times} c \\ 4 {\times} a + 5 {\times} b + 6 {\times} c \\ 7 {\times} a + 8 {\times} b + 9 {\times} c \end{bmatrix}$

How can I write $M$ such that $M{\times}v= \begin{bmatrix} 1 {\times} a + 2 {\times} b + a {\times} b \\ 4 {\times} a + 5 {\times} b + a {\times} c \\ 7 {\times} a + 8 {\times} b + b {\times} c \end{bmatrix}$

The difference is in the third column where it's not just some constant times $c$ but the product of two other inputs.

Update

Ben suggested using the input variables $a$ and $b$ directly in the matrix $M$, but then $M$ wouldn't be made of constants only (like in the first example)

Answer 1

You can't. Define $$v_1 = \left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right]$$ and $$v_2 = \left[ \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right]$$ By your definition: $$M v_1 = \left[ \begin{array}{c} 1 \\ 4 \\ 7 \end{array} \right]$$ and $$M v_2 = \left[ \begin{array}{c} 2 \\ 5 \\ 8 \end{array} \right]$$ However, $$M (v_1 + v_2) = \left[ \begin{array}{c} 4 \\ 9 \\ 15 \end{array} \right] \neq M v_1 + M v_2$$ Remember that matrix multiplication by a vector is a linear operator.

Answer 2

$\def\m#1{\left[\begin{array}{r}#1\end{array}\right]}$Let $$\eqalign{ P &= \m{1&2&3\\4&5&6\\7&8&9}\qquad a = \m{a\\b\\c} \\ }$$ Then what you have is a linear equation $(Mv = Pa)$ which you wish to solve for $M$. This can be accomplished using an arbitrary matrix $A,\,$ and $v^+$ the Moore-Penrose inverse of $v$. $$\eqalign{ M &= (Pa)v^+ + A(I-vv^+) \\ &= Pa\left(\frac{v^T}{v^Tv}\right) + A\left(I-\frac{vv^T}{v^Tv}\right) \\ &= \left(\frac{Pav^T+(v^Tv)A-Avv^T}{v^Tv}\right) \\ }$$ Note that this isn't a unique solution, it's a whole family of solutions; one for each $A$ matrix.