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

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The discussion focuses on calculating the basis of the matrix space T, which consists of all rank-n matrices. It confirms that T is indeed a matrix space and discusses the use of standard basis matrices Eij, which are linearly independent and span the space of nxn matrices. The properties of a basis are outlined, emphasizing that a basis must consist of independent vectors that span the space and match the dimension of the space, which for nxn matrices is n^2. Any set of n^2 independent matrices can serve as a basis for this space. Understanding these principles is crucial for effectively calculating the basis of matrix space T.
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 ]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.
 
thank you!
 
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