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Recalling that the matrices are rectangular
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Louis 2026-02-09 10:18:57 +01:00
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suivi/2026-6/2026-6.qmd
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@ -122,7 +122,7 @@ On veut qu'il existe $B^{\prime}$ et $B$ avec $B_{:,R} = \vec 0_p$, par les prop
& \iff \exists C \in \mathbb{R}^{n_2}, B^{\top} X - \pmb{1}_R C^{\top} = {B^{\prime}}^{\top} X\\ & \iff \exists C \in \mathbb{R}^{n_2}, B^{\top} X - \pmb{1}_R C^{\top} = {B^{\prime}}^{\top} X\\
& \iff \exists C \in \mathbb{R}^{n_2}, (B^{\top} X - \pmb{1}_R C^{\top}) X^{\top} = {B^{\prime}}^{\top} X X^{\top}\\ & \iff \exists C \in \mathbb{R}^{n_2}, (B^{\top} X - \pmb{1}_R C^{\top}) X^{\top} = {B^{\prime}}^{\top} X X^{\top}\\
& \iff \exists C \in \mathbb{R}^{n_2}, (B^{\top} X - \pmb{1}_R C^{\top}) X^{\top}(X X^{\top})^{-1} = {B^{\prime}}^{\top}\\ & \iff \exists C \in \mathbb{R}^{n_2}, (B^{\top} X - \pmb{1}_R C^{\top}) X^{\top}(X X^{\top})^{-1} = {B^{\prime}}^{\top}\\
& \iff \exists C \in \mathbb{R}^{n_2}, (B^{\top} X - \pmb{1}_R C^{\top})X^{-1} = {B^{\prime}}^{\top}\\
\end{align*} \end{align*}