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Going through many articles, questions on MSE and my book; I now know how by looking at a given matrix we can tell whether it is of full rank or not: its columns must be independent to be of full rank.

My question is, suppose matrix $A$ is not full rank, then what can be said about its rank in general? Thank you.

Gigili
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  • It depends on what you know about the matrix I guess. There's a generalization of your statement that the columns must be independent for the matrix to be full rank. The rank is the dimension of the columnspace, i.e. it's the size of the largest independent subset of column vectors. – EuYu Dec 03 '13 at 10:02
  • @EuYu: Thank you for your comment. So nothing can be said about the rank of matrix $A$? Then how in this question one can find rank of the given matrix, I don't get it. Is there any special readings you recommend me to follow to better understand the concept? – Gigili Dec 03 '13 at 10:05
  • It really depends on what you mean by "what can be said". If you give me a matrix and let me explicitly compute its rank, then I know everything about the rank of the matrix. But I'm sure that's not what you mean. – EuYu Dec 03 '13 at 10:07
  • @EuYu: Right. I mean are there other properties that when a matrix does fulfill them its rank must be a specific number, one or two for instance, something like what we do about full rank matrices. – Gigili Dec 03 '13 at 10:09
  • That question uses what I said. The rows are spanned by the $2$ vectors, so the dimension of the rowspace is $2$. That means the rank of the matrix is $2$. In case it's not clear, we phrased rank in terms of the columns, but the same results apply for the rows as well. For example, the rows being linearly independent means the matrix is full rank and the dimension of the rowspace is the rank. All of this follows from the fact that the rank of a matrix is the same as the rank of its transpose. – EuYu Dec 03 '13 at 10:10
  • @EuYu: Great, thank you. When is the rank of an $m \times n$ matrix $min(m,n)$? And in case the rows are spanned by, say, $n$ vectors, the rank is $n$, right? Are we supposed to find all of those vectors to collect information about the rank? – Gigili Dec 03 '13 at 10:17
  • If the rank of an $m \times n$ matrix is $\min(m,n)$ then the matrix either has linearly independent columns or linearly independent rows depending on whether $n \le m$ or $m\le n$. If the rowspace is spanned by $n$ linearly independent vectors (and no others) then the rank is $n$. We really need the vectors to form a basis and not simply be spanning or linearly independent. The traditional way to calculate rank is to row reduce the matrix and read the information off the echelon form. In this case finding the rank will simultaneously allow you to find the corresponding basis vectors. – EuYu Dec 03 '13 at 10:24
  • @EuYu: I now know a lot more and can find more information on Google. Thank you for your detailed responses. – Gigili Dec 03 '13 at 10:26

2 Answers2

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One thing that can be said is that a matrix is of rank $r$ if and only if the largest invertible square submatrix has dimensions $r\times r$. Actually some people use this as the definition of rank.

Casteels
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I didn't see it in the above statements or comments but by the Rank-Nullity theorem, $$\text{Rank}(A)=n-\text{nullity}(A)$$ where $n$ is the number of columns of $A$ and nullity is the dimension of the null-space of $A$. If nullity$(A)$=0, then we have an invertible matrix which means of course full Rank.

Eleven-Eleven
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