Let $v_i \in \mathbb{R}^n$ and $u_i \in \mathbb{R}^m$, where $n \ge m$. We have $m+1$ pairs $(v_i, u_i), i=1,...,m+1$, where only $m$ many $v_i$ are lineary independent (i.e., $\mathrm{dim}\,\mathrm{span}\{v_1, ..., v_{m+1}\} = m$), and likewise for $u_i$. We wish to find a linear mapping $A$ which sends $v_i$ to $u_i$ up to scaling. That is, $$ Av_i = \lambda_i u_i, \quad i=1,...,m+1 $$ for some $\lambda_i \in \mathbb{R}^*$.

An equivalent formulation

Let $n \ge m$, and let $V \in \mathbb{R}^{n\times(m+1)}$ and $U \in \mathbb{R}^{m\times(m+1)}$ be rank $m$ matrices. Let $A \in \mathbb{R}^{m\times n}$ and let $D \in \mathbb{R}^{(m+1)\times(m+1)}$ be a nonsingular diagonal matrix such that $$ AV = UD. $$ Given $U$ and $V$, how can one find $A$ (and hence $D$) to satisfy the above?

Solutions are not unique. I am interested in a method that yields any solution.

A method for a special case

A simple method presents itself when $U$ is in reduced row echelon form, so that $AV = UD$ looks like: $$ A \underbrace{\begin{bmatrix} \vec{v}_1 \cdots \vec{v}_{m+1} \end{bmatrix} }_V = \underbrace{ \begin{bmatrix} \hat{\mathbf{e}}_1 & \cdots & \hat{\mathbf{e}}_m & \vec{c} \end{bmatrix} }_U D = \underbrace{ \begin{bmatrix} 1 & 0 & \cdots & 0 & c_1 \\ 0 & 1 & & 0 & c_2 \\ \vdots & & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & c_m \\ \end{bmatrix} }_U \underbrace{ \begin{bmatrix} \lambda_1 & & \\ & \ddots & \\ & & \lambda_{m+1} \end{bmatrix} }_D $$

Let $B := \begin{bmatrix} \vec{v}_1 \cdots \vec{v}_{m} \end{bmatrix}^{-1}_L$ be the left inverse, so that $B\begin{bmatrix} \vec{v}_1 \cdots \vec{v}_{m} \end{bmatrix} = I$.

Now, $Bv_i = \hat{\mathbf{e}}_i$ for $i=1,...,m$. Define $d_i$ by $$ Bv_{m+1} =: \begin{bmatrix} d_1 \\ \vdots \\ d_m \end{bmatrix} $$ and now define $$ A = \begin{bmatrix} c_1/d_1 & & \\ & \ddots & \\ & & c_m/d_m \end{bmatrix}B .$$

It's easy to verify that $Av_i = \lambda_i \hat{\mathbf{e}}_i$ with $\lambda_i = c_i/d_i$ for $i=1,...,m$ and $Av_{m+1} = \lambda_{m+1}\vec{c}$ with $\lambda_{m+1} = 1$ as required.

However, I am not sure how to generalise this method to the case where $U$ is not in reduced form. Is there a way of transforming $U$ into $\mathrm{rref}(U)$ without changing the system?

  • $\begingroup$ "I am interested in a method that yields any solution." In that case, you can take $A$ and $D$ to be zero matrices. $\endgroup$ – Gerry Myerson Apr 23 at 2:48

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