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I am wondering about the following matrix equality $$ \begin{pmatrix} 0 \\ 1 & \lambda_{1} \\ & 1 & \lambda_{2} \\ && \ddots & \ddots \\ &&& 1 & \lambda_{k} \\ &&&& 1 & 0 \\ &&&&& \ddots & \ddots \\ &&&&&& 1 & 0 \end{pmatrix} \begin{pmatrix} c_{1} \\ c_{2} \\ c_{3} \\ \vdots \\ c_{k+1} \\ c_{k+2} \\ \vdots \\ c_{k+1 + \ell}\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\0 \\ \vdots \\ 0 \\d_{1} \\ \vdots \\ d_{\ell}\\ \end{pmatrix}. $$ with $\lambda_{i} \neq 0$ for $1 \leq i \leq k$. Is there a way to show that this equation has a solution for the $c_{i}$ and $d_{i}$ such that $c_{1} \neq 0$?

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My own answer:

Let $c_{1} = 1$, then the equation can be rewritten to

$$ \left( \begin{array}{ccccc|cc} \lambda_{1} &&&&&& \\ 1&\lambda_{2}&&&&& \\ &1&\lambda_{2} &&&& \\ && \ddots& \ddots & & & & \\ &&&1 & \lambda_{k} \\ \hline &&&& 1 & 0 \\ &&&&& \ddots & \ddots \\ &&&&&&1 & 0\end{array} \right) \begin{pmatrix} c_{2} \\ c_{3} \\ c_{4} \\ \vdots \\ c_{k+1} \\ \hline c_{k+2} \\ \vdots \\ c_{k+1+\ell} \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 0 \\ \vdots \\ 0 \\ \hline d_{1} \\ \vdots \\ d_{\ell} \end{pmatrix} $$ The matrix on the upper left is invertible on the first $k$ entries. Since in the second part everything is free to choose, there is always a solution. This completes the proof.

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