A question about vector space manifold

[huyichen]

If k is an integer between 0 and min(m,n),show that the set of
mxn matrices whose rank is at least k is an open submanifold of
M(mxn, R).Show that this is not true if "at least k"is replaced by
"equal to k."
For this problem, I don't understand why the statement is not true if we replace "at least k" by "equal to k", I am thinking that the set of m*n matrices whose rank is equal to k will have some k*k minor nonsingular, thus any such rank k matrices will have a open nbd contained in Mk(m*n, R) (which is the set of m*n matrices whose rank is equal to k) , and thus make Mk(m*n, R) open in M(m*n,R) (which is the set of all matrices with real entries).
Could anyone help me with the second part of this problem?

 

[Office_Shredder]

In general rank is highly unstable. For example given a matrix of rank 0, you can make arbitrarily small changes to a single entry to change the rank to 1.

If you have a matrix of rank k, even if you have a k*k minor which is nonsingular, a small change in the entries won't make that singular but could result in you having a larger non-singular minor

 

[huyichen]

So is it true that Mk1(m*n,R) is contained in Mk2(m*n,R) for any k2>k1?

 

[Office_Shredder]

No. A matrix can have only one rank, so it can't be contained in two such sets. They're mutually exclusive

 

[huyichen]

Could you point to me any reference on that part? I need more detailed formal argument.

 

[Office_Shredder]

The rank can be defined as either the largest k so that a kxk minor exists, the dimension of the image of the matrix, the dimension of the span of the rows of the matrix, or the dimension of the span of the columns of the matrix.

Whichever definition you're using, it's obvious that a matrix can only have one rank. So if a matrix is in both $$M_k(m\times n)$$ AND $$M_r(m\times n)$$, then it has two different ranks which doesn't make any sense

 

[huyichen]

I mean I want to know about how small changes will make rank k matrice into rank r where r>k, for this part I am not clear.

 

[Office_Shredder]

It won't necessarily change the rank, but all we need is that it could change the rank

Let's look at the row span definition. If we have rank of k, then there are k linearly independent rows, and then say a row that is a linear combination. Now let's assume that the k linearly independent rows do not form a basis for R^m (if they do, you have a full rank matrix, which does give us an open submanifold). Then there is some vector that does not lie in the span of our k linearly independent rows - this is not a row of the matrix, but some vector exists.

Do you see how you can modify one of your rows using the existence of this matrix to increase the rank?

 

[huyichen]

You mean change one of the dependent row into a row that is not in the span of k independent rows? In that way you can change the rank?

 

[Office_Shredder]

That's right