Rank and Similarity

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let A and B be n x n matrices over a field F. Suppose that A^2 = A and B^2 = B. Prove that A and B are similar if and only if they have the same rank.
 

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  • #2
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Can I use the same eigenvalues argument?
They have +1,-1 as eigenvalue, but how do I make the connection to the rank?

Thank you
 
  • #3
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Perhaps this will help: Every vector x can be written as x=(I-A)x+Ax. If [tex]A^2=A[/tex] then the range of (I-A) and the range of A are disjoint complementary subspaces (needs a proof). A vanishes on the first subspace while the second one consists of vectors of eigenvalues 1. Similarly for B.
 
  • #4
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Perhaps this will help: Every vector x can be written as x=(I-A)x+Ax. If [tex]A^2=A[/tex] then the range of (I-A) and the range of A are disjoint complementary subspaces (needs a proof). A vanishes on the first subspace while the second one consists of vectors of eigenvalues 1. Similarly for B.
So basically to prove similarity of A and B, I am proving they have the same eigenvalues. Am I wrong?
we have
[tex]A^2 -A =0[/tex]
[tex]A( A - I) =0[/tex]
A( A - I) is not invertible.
To say A is not invertible is to say A has eigenvalue of 0.
and A - I is not invertible , means A has eigenvalue of 1. Similarly for B.
What do you think?
 
  • #5
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Can I use the same eigenvalues argument?
They have +1,-1 as eigenvalue, but how do I make the connection to the rank?

Thank you
If you know the eigenvalues, what can you tell about the null space and therefore the nullity? Using the nullity, what can you tell about rank given that they have the same dimensions?
 
  • #6
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If you know the eigenvalues, what can you tell about the null space and therefore the nullity? Using the nullity, what can you tell about rank given that they have the same dimensions?
Using the Nullity, I can find the rank of A by the dimension theorem. n-r.
If I know the eigenvalues, then rank A = [tex]n - rank (A- \lambda*I)[/tex]I) For all eigenvalues.

Note that Null space of A alone equals the eigenspace of its eigenvalue 0.
 
  • #7
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Equality of the ranks should tell you that the two complementary subspaces have the same dimension for A i B. Therefore you can find .... what?
 
  • #8
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Be careful, you have [tex] \begin{pmatrix}1&0\\0&1\end{pmatrix}, \begin{pmatrix}1&1\\0&1\end{pmatrix}[/tex] which are not similar but have the same eigenvalues. Better check for a similarity matrix between two idempotent matrix and use trace(P) = rank(P) for an idempotent matrix P. I did not check it myself but that was the first thing came to my mind.

EDIT : Actually this proves the [itex](\Longrightarrow)[/itex] direction since if they are similar, their trace equals to the number of "1" eigenvalues which is their corresponding matrix rank.

Note: Idempotent matrix eigenvalues are either 1 or zero.
 
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