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Suppose A is a $3\times n$ matrix whose columns span $\mathbb R^{3}$. Explain how to construct a $3\times n$ matrix $D$ such that $AD = I_{3}$. I want to say that $AD = 0$ must have only the trivial solution, for $I_{3}$ is linearly independent. Is this all I need to say? The final dimensions do match up and so I do not think a restriction on n is needed.

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Please see here for how to typeset common math expressions with MathJax, and see here for how to use Markdown formatting. – Zev Chonoles Jul 6 '13 at 18:57
$AD$ is not defined for all $n$. – Ink Jul 6 '13 at 19:20
If you want both $\;A,D\,$ to be $\,3\times n\,$ matrices, and also that $\,AD\,$ is defined, it will have to be $\,n=3\,$ this really what you want? – DonAntonio Jul 6 '13 at 21:37
up vote 0 down vote accepted

The columns of $A_{3\times n}$ span $\Bbb R^3$ so you have $i,j,k \in \{1,n\}$ so that the columns $i,j$ and $k$ of $A_{3\times n}$ span $\Bbb R^3$.

Then you need a matrix $B_{n\times 3}$ with zeroes everywhere except at $(i,1),(j,2)$ and $(k,3)$.

$A_{3\times n}B_{n\times 3}$ is a $3\times 3$ matrix whose columns are a basis of $\Bbb R^3$ so it is invertible.

You can easily check that $D=B_{n\times 3}\left(A_{3\times n}B_{n\times 3}\right)^{-1}$ works.

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