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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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They have +1,-1 as eigenvalue, but how do I make the connection to the rank?

Thank you

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- #4

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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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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?

They have +1,-1 as eigenvalue, but how do I make the connection to the rank?

Thank you

- #6

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Using the Nullity, I can find the rank of A by the dimension theorem. n-r.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?

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.

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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.

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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