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Rank of adj(A)

Yankel

Active member
Jan 27, 2012
398
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
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.