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Math Help - Product of two non-square matrices is identity

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    Member Last_Singularity's Avatar
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    Product of two non-square matrices is identity

    Question: Suppose A is m \times n with rank m. Prove there exists n \times m matrix B such that AB = I_m.

    I am kind of stuck on this. Any hints on which properties or theorems on rank of matrices that I should appeal to? Thanks!
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    Quote Originally Posted by Last_Singularity View Post
    Question: Suppose A is m \times n with rank m. Prove there exists n \times m matrix B such that AB = I_m.

    I am kind of stuck on this. Any hints on which properties or theorems on rank of matrices that I should appeal to? Thanks!
    let \mathcal{B}=\{\bold{e}_i, \ 1 \leq i \leq m \} be the standard basis for \mathbb{R}^m and let T: \mathbb{R}^n \longrightarrow \mathbb{R}^m be the linear map with [T]_{\mathcal{B}}=A. this map is onto because \text{rank}(A)=m.

    so for any 1 \leq i \leq m, there exists \bold{x}_i \in \mathbb{R}^n such that T(\bold{x}_i)=\bold{e}_i. now define S: \mathbb{R}^m \longrightarrow \mathbb{R}^n by S(r_1 \bold{e}_1 + \cdots + r_m \bold{e}_m)=r_1 \bold{x}_1 + \cdots + r_m \bold{x}_m. see that

    S is a linear map and TS = \text{id}. let B=[S]_{\mathcal{B}}. we have AB=I_m because TS = \text{id}.
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