# Similar Linear Transformations

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

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.
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