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 \(E_1, \ldots, E_k\) are projections of \(V\) and that \(E_1 + \cdots + E_k = I\). Prove that \(E_iE_j = 0\) for \(i \neq j\) (Hint: Use the trace function and ask yourself what the trace of a projection is.)
Solution
1. For any projection \(E\) on a finite-dimensional space \(V\), \(\operatorname{tr}(E) = \dim \operatorname{im}(E)\):
1.1. A projection satisfies \(E^2 = E\), so its minimal polynomial divides \(x(x - 1)\).
definition of projection operator.
1.2. Hence \(E\) is diagonalizable with eigenvalues in \(\{0, 1\}\), and \(V = \operatorname{im}(E) \oplus \ker(E)\).
direct sum decomposition into eigenspaces.
1.3. In an eigenbasis, the diagonal entries are \(1\) on \(\operatorname{im}(E)\) and \(0\) on \(\ker(E)\), so \(\operatorname{tr}(E) = \dim \operatorname{im}(E)\).
trace is the sum of eigenvalues with multiplicity.
2. Show that \(V = \operatorname{im}(E_1) \oplus \cdots \oplus \operatorname{im}(E_k)\) is an internal direct sum:
2.1. Taking the trace of the identity \(I = \sum_{i=1}^k E_i\):\[ \dim V = \operatorname{tr}(I) = \operatorname{tr}\left(\sum_{i=1}^k E_i\right) = \sum_{i=1}^k \operatorname{tr}(E_i) = \sum_{i=1}^k \dim \operatorname{im}(E_i). \]
1.
1.2. For any \(v \in V\), \(v = I(v) = E_1(v) + \cdots + E_k(v) \in \operatorname{im}(E_1) + \cdots + \operatorname{im}(E_k)\), so \(V = \sum_{i=1}^k \operatorname{im}(E_i)\).
range of linear maps.
1.3. For any collection of subspaces \(\{W_i\}\), \(\dim(\sum W_i) \le \sum \dim W_i\), with equality if and only if the sum is direct.
standard dimension formula for subspace sums.
1.4. Since \(\dim V = \sum_{i=1}^k \dim \operatorname{im}(E_i)\), the sum is a direct sum:\[ V = \operatorname{im}(E_1) \oplus \operatorname{im}(E_2) \oplus \cdots \oplus \operatorname{im}(E_k). \]
1, 2, and 3.
3. Prove \(E_i E_j = 0\) for \(i \neq j\):
3.1. Let \(j \in \{1, \dots, k\}\) and let \(w \in \operatorname{im}(E_j)\).
setup.
3.2. Since \(w \in \operatorname{im}(E_j)\), the unique representation of \(w\) in the direct sum \(V = \bigoplus_{r=1}^k \operatorname{im}(E_r)\) is \(w = 0 + \cdots + w + \cdots + 0\) (with \(w\) in the \(j\)-th slot and \(0\) elsewhere).
definition of direct sum.
3.3. On the other hand, \(w = I(w) = E_1(w) + \cdots + E_k(w)\), where \(E_r(w) \in \operatorname{im}(E_r)\) for each \(r\).
definition of image.
3.4. By uniqueness of the direct sum decomposition, comparing 2 and 3 yields:\[ E_i(w) = 0 \quad \text{for all } i \neq j. \]
uniqueness of coordinates in direct sum.
3.5. For any \(v \in V\), the vector \(w = E_j(v) \in \operatorname{im}(E_j)\).
definition of image.
3.6. Applying 4 to \(w = E_j(v)\) gives:\[ (E_i E_j)(v) = E_i(E_j(v)) = 0 \quad \text{for all } i \neq j \text{ and all } v \in V. \]
4 applied to \(E_j(v)\).
3.7. Thus \(E_i E_j = 0\) for all \(i \neq j\).
6.
4. Conclusion: \(E_i E_j = 0\) for all \(i \neq j\).
2 and 3.
Q.E.D.