Decomposition of Vector Spaces

Let $E_1, E_2,\dots, E_s$ be linear transformations on an $n$ -dimensional vector space such that $I=E_1+E_2+\dots+E_s$ and $E_iE_j=0$ for $i\neq j$ . Then $V=E_1V\oplus E_2V\oplus\dots\oplus E_sV$ .

How does this happen? Take the expression $I=E_1+E_2+\dots+E_s$ and multiply by any $v\in V$ on both sides. We see that $v=E_1v+E_2v+\dots+E_sv$ . Hence any vector $v$ can be expressed as a sum of elements in $E_iV$ for $i\in\{1,2,\dots,s\}$ .

Why do we have a direct sum decomposition? Let $v_1+v_2+\dots+v_s=0$ for $v_i\in E_iV$ . Then consider $E_k(v_1+v_2+\dots+v_s)=E_k(0)=0$ . For any $v_i$ where $i\neq k$ , $v_i=E_i v$ for some $v\in V$ . Hence $E_kv_i=E_kE_iv=0.v=0$ . Hence, we have $E_kv_k=0$ . Now $v_k=E_kv'$ for some $v'\in V$ . Hence $E_kv_k=E_k^2v'$ . Note that $E_k^2=E_k$ (just multiply the expression $I=E_1+E_2+\dots+E_s$ by $E_k$ on both sides). Hence, we have $E_k v'=0$ . Now $E_kv'=v_k$ . Hence, we have $v_k=0$ . This is true for all $k\in\{1,2,\dots,s\}$ . Hence, all the $v_i's=0$ , which proves that we have a direct sum decomposition of $V=E_1V\oplus E_2V\oplus\dots\oplus E_sV$ .

Why is all this relevant? Because using the minimal polynomial $f(x)=p_1(x)^{e_1}\dots p_s(s)^{e_s}$ of any transformation $T\in L(V,V)$ , we can construct such $E_i$ ‘s which satisfy the above two conditions, and can hence decompose the vector space as a direct sum of $s$ subspaces. Moreover, these subspaces have the additional property that they’re $T$ -invariant. Each $E_i=f(x)/p_i(x)^{e_i}$