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Let $X = [\vec{v_1} ... \vec{v_k}]$ and $Y = [\vec{u_1} ... \vec{u_k}]$ be $n \times k$ matrices. Prove that if Span {$\vec{v_1}, ..., \vec{v_k}$} $\subseteq$ Span {$\vec{u_1}, ..., \vec{u_k}$}, then there exists a matrix $A$ such that $X=YA$.

I have a hunch this has something to do with columns spaces but I can't wrap my head around it. Any hints?

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  • $\begingroup$ Hint: $v_i\in\mbox{Span}\left\{u_1,...,u_k\right\}$ so there exist $a_{i,1},..., a_{i, k}\in\mathbb{R}$ such that $$v_i=a_{i,1}u_1+...+a_{i, k}u_{k}$$ $\endgroup$
    – Ixion
    Commented Aug 1, 2017 at 0:52

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Since $\DeclareMathOperator{Span}{Span}\Span\{v_1,\dotsc,v_k\}\subset\Span\{u_1,\dotsc,u_k\}$, each $v_i$ is a linear combination of the vectors $\{u_1,\dotsc,u_k\}$. This means that there are scalars $a_{ij}$ satisfying \begin{array}{rcrcrcrcrc} v_1 &=& a_{11}\cdot u_1 &+& a_{21}\cdot u_2 &+& \dotsb &+& a_{k1}\cdot u_k \\ v_2 &=& a_{12}\cdot u_1 &+& a_{22}\cdot u_2 &+& \dotsb &+& a_{k2}\cdot u_k \\ \vdots & & \vdots & & \vdots & & \ddots & & \vdots \\ v_k &=& a_{1k}\cdot u_1 &+& a_{2k}\cdot u_2 &+& \dotsb &+& a_{kk}\cdot u_k \\ \end{array} Now, let $A=[a_{ij}]$. What is $YA$?

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