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Let $F$ be a subfield of the complex numbers (or, a field of characteristic zero). Let $V$ be a finite-dimensional vector space over $F$, Suppose that $E_1,...,E_k$ are projections of $V$ and that $E_1+...+E_k=I$. Prove that $E_iE_j=0$ for $i$ (does not equal) $j$

(Hint: Use the trace function and ask yourself what the trace of the projection is.)

Attempt: If $E$ is projection then $V = R \bigoplus N$ where $R = im E$ and $N$ is nullspace of $E$. If $\{ \alpha_1, ... , \alpha_r \}$ is base of $R$ and $\{ \alpha_{r+1}, ... , \alpha_n \}$ is base of $N$, then the matrix of $E$ is

$\left(\matrix {I&0\\0&0} \right)$,

where I is the identity matrix $r \times r$. Then see that $Trace(E)=dimR$.

Now

$E_1+...+E_k=I \rightarrow E_1^2+...+E_1E_k=E_1\rightarrow E_1E_2+...+E_1E_k=0 $

applying the trace operator and using its linearity

$tr(E_1E_2)+...+tr(E_1E_k)=tr(0)=0 $

now using the observation from the beginning

$dim Im(E_1E_2)+...+dim Im(E_1E_k)=0 $

it is $E_1E_k=0=0 ,\,\ \forall k$. And the other cases would be analogous. Is it right? Any better suggestions?

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    $\begingroup$ Actually, you do not even know that $E_1E_k$ is a projection. $\endgroup$ – Mindlack Feb 11 '19 at 19:34
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Should be everything correct, till $$E_1^2 + \ldots + E_1E_k = E_1 \implies E_1E_2+ \ldots + E_1E_k =0$$

It should be $$E_1^2 + \ldots + E_1E_k = E_1 \implies E_1^2 - E_1 + E_1E_2+ \ldots + E_1E_k=0$$

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    $\begingroup$ But $E_1^2= E_1$ so that's the same. $\endgroup$ – quid Aug 29 '19 at 23:50

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