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I have a system of $n$ equations which follows a particular pattern as follows (showing the case $n=3$):

$$\phi = a_1 + \psi_2 a_2 + \psi_3 a_3 \\ \phi = \psi_1 a_1 + a_2 + \psi_3 a_3\\ \phi = \psi_1 a_1 + \psi_2 a_2 + a_3$$ The scalars $\phi$ and the $a_i$ are known, and I need to solve for the $\psi_i$. All scalars are $\in \mathbb{R}$.

So I am trying to rewrite it into a matrix equation, and getting mixed up with constructing the matrix, and then solving, somewhat similar to this and this previous questions.

So I've got this far: $$\phi \begin{Bmatrix} 1\\1\\1 \end{Bmatrix} = \begin{bmatrix} 1 & \psi_2 & \psi_3 \\ \psi_1 & 1 & \psi_3 \\ \psi_1 & \psi_2 & 1 \end{bmatrix} \begin{Bmatrix} a_1\\a_2\\a_3\ \end{Bmatrix} $$ which more generally can become: $$\phi\boldsymbol{\unicode{x1D7D9}}_{n\times 1} = \boldsymbol{\Psi}_{n\times n} \boldsymbol{a}_{n\times 1}$$ so to solve for the vector $\boldsymbol{\psi} = \left[\psi_1, ..., \psi_n\right]^T$ and inspired by this I try to decompose $\boldsymbol{\Psi}$ as follows:

$$ \boldsymbol{\Psi} = \left[ \textrm{diag}(\boldsymbol{\psi} )(\boldsymbol{\unicode{x1D7D9}}_{n\times n}) - \boldsymbol{I}_{n\times n}\right]^T + \boldsymbol{I}_{n\times n} $$ but this both feels messy, and also doesn't seem to make finding $\boldsymbol{\psi}$ any easier. The system might also be overdetermined.

Any ideas appreciated!

Colin
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2 Answers2

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$ \def\o{{\tt1}} \def\M{\Psi} \def\LR#1{\left(#1\right)} \def\BR#1{\Big(#1\Big)} \def\op#1{\operatorname{#1}} \def\Diag#1{\op{Diag}\LR{#1}} \def\a{{\o a^T}} $Let $x$ denote the unknown vector and $\o$ the all-ones vector, then $$\eqalign{ \M &= I + \o x^T - \Diag{x} \\ }$$ This decomposition allows your linear system to be algebraically manipulated and solved $$\eqalign{ \phi\o &= \M a \\ &= a + \o\LR{x^Ta} - \Diag{x}\,a \\ &= a + \LR{\a}x - \Diag{a}\,x \\ \LR{\phi\o-a} &= \BR{\a - \Diag{a}}\,x \\ x &= \BR{\a - \Diag{a}}^{\bf+}\LR{\phi\o-a} \\ }$$ where the pseudoinverse is used since the matrix in parentheses can be singular.

lynn
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    This is neat! and the $\phi 1 - a$ term appears as for the other answer. But why is $(a I - \Diag{a})^{+}$ necessarily diagonal? – Colin Sep 26 '22 at 13:45
  • You're right. In my initial post, I transposed ${\tt1}x^T$ on the first line. Propagating that through the rest of the calculation, it turns out that the matrix is not diagonal. – lynn Sep 26 '22 at 15:07
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If I understand you correctly, your matrix is

$$ 0 \psi_1 + a_2 \psi_2 + a_3 \psi_3 = \phi -a_1\\ a_1\psi_1 + 0 \psi_2 + a_3 \psi_3 = \phi -a_2\\ a_1 \psi_1 + a_2 \psi_2 + 0 \psi_3 = \phi -a_3$$

Using Gaussian Elimination, we arrive at

$$\begin{bmatrix} 1 & 0 & 0 & \dfrac{a_1-a_2-a_3+\phi}{2 a_1}\\0 & 1 & 0 & \dfrac{-a_1+a_2-a_3+\phi}{2 a_2} \\ 0 & 0 & 1 & \dfrac{-a_1-a_2+a_3+\phi}{2 a_3} \end{bmatrix}$$

Note that any $a_i$ equal to zero causes issues.

Moo
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