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I have a matrix $A \in \mathbb{R}^{n \times p}$ where $p>n$ and column rank of $A=n$. Consider a matrix $M=[e \quad A^T]$ where $e \in \mathbb{R}^p$ is a vector of $1$. What is the column rank of $M$ ?

If $M$ doesn't have full column rank, then can we find certain columns of $A$ deleting which, will lead to a new full column rank matrix $\hat{M}=[e \quad \hat{A}^T]$, where $\hat{A}$ is obtained by deleting certain columns of $A$. However, I want to delete such potentials columns, as few as possible.

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  • $\begingroup$ We will always be able to delete $p-n$ columns of $A$ - no matter what the rank of $M$ is. $\endgroup$ – amsmath Oct 10 '17 at 15:23
  • $\begingroup$ Thanks. I want to delete as few as possible, number of columns from $A$. I edited the question. $\endgroup$ – user402940 Oct 10 '17 at 15:30
  • $\begingroup$ The rank of $M$ is either $n$ or $n+1$. In the second case, it is already of full (column) rank. Actually, I don't understand your objective? $\endgroup$ – amsmath Oct 10 '17 at 15:42
  • $\begingroup$ Thanks for the answer. I was confused about its rank. I actually had few vectors (whose collection is $A$), over which $M$ has to be chosen/computed such that column rank of $M$ must be full. I have the liberty to drop few vectors from $A$ in case something goes wrong. Thats why the second part. $\endgroup$ – user402940 Oct 10 '17 at 15:57

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