MHB Prove These 3 Statements About A Matrix: Rank of Adjoint

  • Thread starter Thread starter Yankel
  • Start date Start date
  • Tags Tags
    Matrix rank
Yankel
Messages
390
Reaction score
0
Hello all

I need to prove these 3 statements, and I don't know how to start...

A is an nxn matrix:

1) if rank(A)=n then rank(adj(A))=n
2) if rank(A)=n-1 then rank(adj(A))=1
2) if rank(A)<n-1 then rank(adj(A))=0

thanks...:confused:
 
Physics news on Phys.org
Yankel said:
Hello all

I need to prove these 3 statements, and I don't know how to start...

A is an nxn matrix:

1) if rank(A)=n then rank(adj(A))=n
2) if rank(A)=n-1 then rank(adj(A))=1
2) if rank(A)<n-1 then rank(adj(A))=0

thanks...:confused:

Write A in Jordan Normal Form:
$$A = P J P^{-1}$$
where P is an invertible matrix and J is an upper triangular matrix with its eigenvalues on its diagonal, and more specifically J consists of Jordan blocks.

If rank(A)=n-1, then J can be written with a row consisting of zeroes, a column consisting of zeroes, and the corresponding minor will be non-zero.

If rank(A)<n-1, then J can be written with at least two rows consisting of zeroes, and at least two columns consisting of zeroes.
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
6K
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K