Calculating Basis of Matrix Space T: All Rank-n Matrices | Tips & Tricks

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Homework Help Overview

The discussion revolves around the calculation of the basis for the space of all rank-n matrices, specifically focusing on the properties and characteristics of such a matrix space.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the definition of the matrix space T and question how to calculate its basis. There is a discussion on the properties of a basis, including independence and spanning, as well as the dimensionality of the space of n by n matrices.

Discussion Status

Some participants have provided insights into the properties of bases and suggested the use of standard basis matrices. Others are seeking alternative methods for calculating the basis of the space T.

Contextual Notes

There is an assumption that the matrices discussed are of size n by n, and the conversation touches on the constraints of matrix addition and the implications for dimensionality.

xuej1112
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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!
 
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xuej1112 said:
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 Eij 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.
 
thank you.that's right. is there any other method to calculate the basis of space T?
 
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 [itex]]n^2[/itex] matrices, it follows that the dimension of the set of all n by n matrices is [itex]n^2[/itex].

So any set of [itex]n^2[/itex] matrices that is independent is a basis and any set of [itex]n^2[/itex] matrices that span the space is a basis.
 
thank you!
 

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