Finding a basis for the set of all rank-n matrices

[xuej1112]

If T={Bi} Bi are the all matrix of rank n. So,T is a matrix space(right?). How to calculate the basis of T? are the basis of T also some matrix?

Thank you!

 

[LCKurtz]

If T={Bi} Bi are the all matrix of rank n. So,T is a matrix space(right?). How to calculate the basis of T? are the basis of T also some matrix?

Thank you!

Of course you can't add matrices that don't have the same dimensions, so I'm assuming you are talking about nxn matrices. Let E_ij be the matrix with 1 in the ij'th position and 0 everywhere else. Should be pretty easy to show they are linearly independent and span the space of nxn matrices.

 

[xuej1112]

thank you.that's right. is there any other method to calculate the basis of space T?

 

[HallsofIvy]

As I am sure you have learned about vector spaces in general, a basis has three properties:

a) The vectors are independent.
b) They span the space
c) The number of vectors in the space is equal to the dimension of the space.

Further, if any two of those is true, the third is true.

Since the "standard basis" LCKurtz gave has $]n^2$ matrices, it follows that the dimension of the set of all n by n matrices is $n^2$.

So any set of $n^2$ matrices that is independent is a basis and any set of $n^2$ matrices that span the space is a basis.

 

[xuej1112]

thank you!