# Let $V$ be a finite-dimensional vector space over $F$, Suppose that $E_1,…,E_k$ are projections of $V$ [duplicate]

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?

• Actually, you do not even know that $E_1E_k$ is a projection. – Mindlack Feb 11 '19 at 19:34

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 $$E_1^2 + \ldots + E_1E_k = E_1 \implies E_1^2 - E_1 + E_1E_2+ \ldots + E_1E_k=0$$
• But $E_1^2= E_1$ so that's the same. – quid Aug 29 '19 at 23:50