Trace(matrix) = 0 and the dimension of subspace

Click For Summary
SUMMARY

The discussion centers on the dimension of the subspace of trace-zero matrices within the space of all nxn matrices, denoted as M_n(F). It is established that the dimension of M_n(F) is n^2, and the dimension of the trace function Tr[M_n(F)] is 1, not 0. Consequently, the dimension of the null space of the trace function is confirmed to be n^2 - 1. The confusion arises from the distinction between the entire space of matrices and the subset of trace-zero matrices.

PREREQUISITES
  • Understanding of linear transformations and their properties
  • Familiarity with matrix theory, specifically M_n(F)
  • Knowledge of the trace function and its implications in linear algebra
  • Concept of null space in the context of linear transformations
NEXT STEPS
  • Study the properties of the trace function in linear algebra
  • Explore the concept of null space and its applications in matrix theory
  • Learn about the structure of subspaces in vector spaces
  • Investigate the implications of dimension in linear transformations
USEFUL FOR

Mathematicians, students studying linear algebra, and anyone interested in the properties of matrices and their subspaces.

Ioiô
Messages
3
Reaction score
0

Homework Statement



Claim: Matrices of trace zero for a subspace of M_n (F) of dimension n^2 -1 where M_n (F) is the set of all nxn matrices over some field F.

Homework Equations



Tr(M_n) = sum of diagonal elements

The Attempt at a Solution



I view the trace Tr as a linear transformation Tr: M_n (F) -> F. I find the dimension of the subspace spanned by M_n (F) by using the fact:

dim M_n (F) = dim Tr [M_n (F)] + dim null [M_n (F)]

Since dim M_n (F) = n^2, we have

n^2 = dim Tr [M_n (F)] + dim null [M_n (F)]

I'm confused on what's in the right part of the above equation:

Isn't dim Tr [M_n (F)] = 0 since Tr [M_n (F)] = 0?

Shouldn't dim null [M_n (F)] =n^2 - dim Tr [M_n (F)] = n^2 - 1?

Also, I know that the subspace is generated by the following matrices: A, B, and [A,B] (the commutator of A and B).

Thanks!
 
Physics news on Phys.org
You said

> Isn't dim Tr [M_n (F)] = 0 since Tr [M_n (F)] = 0?

No, dim Tr [M_n (F)] = 1 since Tr [M_n (F)] = F
You have confused M_n (F) (all nxn matrices over F) with its subset of trace zero matrices. And a (non degenerate) field over itself obviously has dimension 1 because it's spanned by its element 1 (multiplicative identity). So you get indeed

dim null [M_n (F)] =n^2 - dim Tr [M_n (F)] = n^2 - 1
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
15
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K