Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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 $n$. –  Vectk 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 $\,n=3\,$ ...is this really what you want? –  DonAntonio Jul 6 '13 at 21:37
add comment

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.