The rank of a block matrix as a function of the rank of its submatrice

Click For Summary
SUMMARY

The discussion centers on determining the rank of a block matrix defined as M = [S1 C; C^T S2], where S1 and S2 are covariance matrices that are symmetric and positive semi-definite, and C is a cross covariance matrix. Participants agree that the rank of M is influenced by the ranks of S1, S2, and the relationship between them, particularly through shared eigenvalues. The conclusion emphasizes that the function f, which expresses rank(M) in terms of rank(S1) and rank(S2), must also account for the rank of C.

PREREQUISITES
  • Understanding of block matrices and their properties
  • Knowledge of covariance matrices and their characteristics
  • Familiarity with eigenvalues and eigenvectors
  • Basic linear algebra concepts, particularly matrix rank
NEXT STEPS
  • Research the properties of block matrices in linear algebra
  • Study the implications of covariance matrix rank on overall matrix rank
  • Explore the relationship between eigenvalues of covariance matrices
  • Investigate existing functions that relate ranks of block matrices
USEFUL FOR

Mathematicians, data scientists, and statisticians interested in matrix theory, particularly those working with covariance matrices and their applications in multivariate statistics.

GoodSpirit
Messages
18
Reaction score
0
Hello everyone,
I would like to post this problem here in this forum.
Having the following block matrix:

<br /> \begin{equation}<br /> M=\begin{bmatrix}<br /> S_1 &amp;C\\<br /> C^T &amp;S_2\\<br /> \end{bmatrix}<br /> \end{equation}<br />

I would like to find the function $f$ that holds rank(M)=f( rank(S1), rank(S2)).
S_1 and S_2 are covariance matrices-> symmetric and positive semi-definite.
C is the cross covariance that may be positive semi-definite.

Can you help me?

I sincerely thank you! :)

All the best

GoodSpirit
 
Last edited:
Physics news on Phys.org


Are you sure that this function exists?

<br /> \begin{equation}<br /> M=\begin{bmatrix}<br /> 1 &amp;1\\<br /> 1 &amp;1\\<br /> \end{bmatrix}<br /> \end{equation}<br />
=> rank(M)=1
<br /> \begin{equation}<br /> M=\begin{bmatrix}<br /> 1 &amp;.5\\<br /> .5 &amp;1\\<br /> \end{bmatrix}<br /> \end{equation}<br />
=> rank(M)=2
 


Hi mfb,

Thank you for answering! :)

True! it depends on something more!

M is also a covariance matrix so C is related to S1 and S2.

It is a good idea to make the rank M dependent of the C rank.

The rank of M may be dependent of the eigen values that are shared by S1 and S2

Thankk you again

All the best

GoodSpirit
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K