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Let $A$ be an $n\times m$ matrix with rank $n$. Then I would like to conclude that there is a matrix $B$ such that $AB=I_n$.

This is easy, on the intuitive level. The image of $A$ in $\mathbb{R}^m$ is an $n$-dimensional subspace, spanned by the pivotal columns of $A$. Taking $B$ to be the $m\times n$ matrix that sends each pivotal column to the standard basis in $\mathbb{R}^n$ should yield the answer.

I feel like my solution lacks rigor. Any tips for how to make this more precise?

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3 Answers 3

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Let $A:\mathbb{R}^m\to\mathbb{R}^n$ be rank $n$. Then, you know that $A(\mathbb{R}^m)=\mathbb{R}^n$. Pick a basis for $\mathbb{R}^n$, say $\{b_1,\cdots,b_n\}$ and choose $a_i\in A^{-1}(b_i)$. Then, define $B(b_i)=a_i$, and extend by linearity. When you make the matrix for $B$ you will have your right inverse!

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We can use one of the handy formulations of matrix multiplication: if $A,B$ are matrices and $B$ can be written as a matrix of columns $[b_1|b_2|\ldots|b_k]$, then $AB=[Ab_1|Ab_2|\ldots|Ab_k]$. Now since we want $AB=I$, we would like $Ab_i=e_i$, $i=1,...,k$, $e_i$ the $i$-th standard basis vector. So the problem reduces to finding these vectors $b_i$. Can you do this?

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Let $e_k$ be the vector with zeros except for a one at the $k$th position. I will abuse notation slightly by using $e_k$ regardless of the dimension $n$ of the space it is in (as long as $k\leq n$).

Since $\text{rk} A = n$, we can find $i_1,...,i_n$ such that $\{A e_{i_1},...,A e_{i_n}\}$ are linearly independent. If we define $\Pi : \mathbb{R}^n \to \mathbb{R}^m$ by $\Pi = \sum_{k=1}^n e_{i_k} e_k^T$, then we have $A \Pi = \begin{bmatrix} A e_{i_1} & \cdots & A e_{i_n}\end{bmatrix}$, which is invertible. Hence we have $A \Pi (A \Pi)^{-1} = I$. Letting $B=\Pi (A \Pi)^{-1}$ gives the desired result.

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