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I am trying to understand the ways to find a rectangular matrix $\mathcal{H}$, of size $m \times n$, in the problem : $$ g = \mathcal{H} f $$

Here, $g$ and $f$ are given and are vectors of sizes $m$ and $n$, respectively.

The matrix notation is given as:

$$\begin{bmatrix} g_{1}\\ \vdots\\ g_{m}\\ \end{bmatrix} = \begin{bmatrix} \mathcal{H}_{11} & \mathcal{H}_{12} & \dots & \mathcal{H}_{1n} \\ \vdots & \vdots &\ddots & \vdots \\ \mathcal{H}_{m1} & \mathcal{H}_{m2} & \dots & \mathcal{H}_{mn} \end{bmatrix} \begin{bmatrix} f_{1}\\ f_{2}\\ \vdots\\ f_{n}\\ \end{bmatrix}$$

Can anyone explain to me the ways to solve this problem and explain the issues of finding a unique solution to this problem?.

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The pseudoinverse of the vector $f$ is
$$f^+ = \frac{f^T}{f^Tf}$$ which can be used to write the general solution as $${H = gf^+ + A(I_n-ff^+)}$$ where $A\in{\mathbb R}^{m\times n}$ is an arbitrary matrix. The least squares solution occurs at $A=0$.

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  • $\begingroup$ Is your second equation of the form: Particular solution + Null space of $f$?. $\endgroup$ – dykes Jun 30 '18 at 17:34
  • $\begingroup$ @dykes Yes, absolutely. $\endgroup$ – greg Jun 30 '18 at 19:16
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If $f$ is equal to $0$ than there exists that matrix $H$ if and only if $g=0$.

If $f\neq 0$ than there exists $f_j\neq 0$ and so you can choose the matrix $H$ such that $H_{s,t}=0$ if $t\neq j$ and $H_{s,j}=\frac{g_s}{f_j}$ if $t=j$

For example for $n=3$ and $m=2$ if $f_2\neq 0$ (for example) you have that $\left[\begin{matrix}g_1\\ g_2\end{matrix}\right]=\left[\begin{matrix}0 & \frac{g_1}{f_2} &0 \\0& \frac{g_2}{f_2} & 0\end{matrix}\right]\left[\begin{matrix}f_1 \\ f_2 & \\f_3\end{matrix}\right] $

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  • $\begingroup$ Could you elaborate your answer?. $\endgroup$ – dykes Jun 25 '18 at 20:33
  • $\begingroup$ If you don’t understand tell me and I will don’t it more esplicitly $\endgroup$ – Federico Fallucca Jun 25 '18 at 20:42
  • $\begingroup$ Yes, I understand now. Got confused with the notations. $\endgroup$ – dykes Jun 26 '18 at 9:58

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