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.

If I have a matrix $A \in R^{(m \times n)}$ with $m \leq n$. All rows in matrix a are linearly independent and therefore $A$ has a full row rank. I can decompose matrix $A$ such that $A = [B|N]$ with $B \in R^{(m \times m)}$ and $N \in R^{(m \times (n-m))}$

How can the following statement in Griffin, C; p70 be justified, "we know that $B$ is invertible since $A$ has a full row rank". $B$ is invertible as comes from having been constructed from $m$ linearly independent columns Griffin, C; p45. Would it be fair to say that, if a $A$ has full row rank, there will therefore always exist some matrix $B$ such that $A=[B|A]$ with $B$ having linearly independent columns and therefore being invertible?

share|improve this question

2 Answers 2

up vote 2 down vote accepted

Yes that is always true. when $m\leq n$ and $A\in\mathbb{R}^{m \times n}$ has linearly independent rows.

To be more precise...

The column rank of a matrix A is the maximum number of linearly independent column vectors of A.

The row rank of a matrix A is the maximum number of linearly independent row vectors of A.

Equivalently, the column rank of A is the dimension of the column space of A, while the row rank of A is the dimension of the row space of A.

A result of fundamental importance in linear algebra is that the column rank and the row rank are always equal

Since your matrix has row rank m, you can always find m linearly independent columns. Those columns will compose matrix B you're looking for.

For a more detailed proof check: wikipedia rank of matrix

share|improve this answer

Read page no:45 of that reference.

The author explains how he forms the matrix $A$. $A$ corresponds to a set of linear equations. $B$ corresponds to the $m$ linearly independent columns in set of linear equations, which are then "re-arranged" (which are equivalent to rearranging the linear equations) so that the first $m$ columns of (new) $A$ are linearly independent.

EDIT: op updated his question which is now equivalent to asking if row rank and column rank of matrix is same.

Yes, it is

User epsilon has already pointed out this. A proof which depends only on basic linear algebra is given here and also here

share|improve this answer
    
Yes Sir, i did get that. Sorry, I still miss the connecting however. I rephrased my question in that regard. –  entropy Jan 2 '13 at 16:39

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.