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This is a nice question I came across in Linear Algebra but I cant figure out how to tackle it. I need some help.

Given two linear transformations, $E$ and $F$ such that $E^2=E$ and $F^2=F$, I am supposed to determine if it is true that $E$ and $F$ are similar if and only if $rank(E)=rank(F)$.

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Have you encountered such transformations (projections) before? Do you know some of their properties? – Olivier Bégassat Oct 27 '11 at 22:13
Hint: What are the possible eigenvalues of $E$ and $F?$ And if you put $E$ and $F$ into Jordan normal form can $1$ occur on their upper diagonals? – jspecter Oct 27 '11 at 22:13
Is there a possibility of avoiding eigenvalues in the solution to this question? – smanoos Oct 27 '11 at 22:43
up vote 2 down vote accepted

Since $E^2=E$ and $F^2=F$, their minimal polynomials must divide $x^2-x=x(x-1)$. Thus their minimal polynomials cannot have repeated factors and so they are both diagonalizable.

Next, by nature of the minimal polynomials dividing $x(x-1)$, the eigenvalues of $E$ and $F$ must be $1$'s and $0$'s. Thus your answer is "Yes." If their rank is the same, the same number of $1$'s will appear in both diagonalizations. If their rank differs, they must have a different number of $1$'s in their diagonalizations and so must not be similar.

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Could this question be solved in another way without the use of eigenvalues? – smanoos Oct 27 '11 at 22:42
Probably by following Olivier's suggestion in the comments above...using projections. But I'm not sure quite how to go about it that way. Problems involving similarity most easily solved by appealing to some canonical form. Any problem like this immediately leads me to recast the problem in terms of Jordan forms. – Bill Cook Oct 27 '11 at 22:52
Half of this is easy - it's an exercise to show that if $E$ and $F$ don't have the same rank then they can't be similar. The other half, maybe this is a good start: let $X$ be the range of $E$, let $Y$ be the range of $F$. If $E$ and $F$ have the same rank then $X$ and $Y$ have the same dimension, so there's an isomorphism $T$ such that $TX=Y$. Maybe there's a way to combine that with $E^2=E$ and $F^2=F$ to get a proof without eigenvalues (though I don't see it just yet). – Gerry Myerson Oct 27 '11 at 23:28
@GerryMyerson. Can you please provide a sketch of the proof of the part you are saying is easy? – smanoos Oct 28 '11 at 0:13
@GerryMyerson Thanks for the clue. The isomorphism thing worked very well. Maybe i will post the answer. – smanoos Oct 28 '11 at 4:27

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