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$E$ is finite dimensional space and $P_1,\ldots,P_k$ are projections i.e. $P_i^2=P_i$ such that $$P_1+\ldots+P_k=\operatorname{Id_E}\qquad (*)$$ I want to prove that $$E=\bigoplus_{i=1}^k \operatorname{Im}(P_i)$$ and then $P_iP_j=0$ if $i\ne j$.

My Try: To prove the equality of direct sum I tried to prove that if $$0_E=y_1+\cdots+y_k$$ where every $y_i\in \operatorname{Im}(P_i)$ then $y_i=0_E$. The problem that for every $y_i$ exists $x_i$ not necessary unique such that $y_i=P_i(x_i)$. If it was unique then from $(*)$ we get that this unique $x_i=0_E$ and then $y_i=0_E$. For the second question I haven't any idea how to prove it.

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Let $P : E \to E$ be a projection. $\DeclareMathOperator{\Tr}{Tr}$$\DeclareMathOperator{\Ima}{Im}$$\DeclareMathOperator{\rank}{rank}$ Notice that $P(P-1) = P^2 - P = 0$ so the only eigenvalues of $P$ can be $0$ and $1$. Furthermore, if $P\ne 0, I$ then $\mu_P(x) = x(x-1)$ is precisely the minimal polynomial of $P$, so $P$ is diagonalizable since $\mu_P$ consists only of linear factors.

This implies that $\Tr P = \rank(P)$, since both will count the number of ones on the diagonal of the diagonalized $P$.

Back to the problem, from $P_1 + \ldots + P_k = I$ we get that $$E = \sum_{i=1}^k \Ima P_i$$

It suffices to prove that the sum is direct:

$$\dim E = \Tr I = \Tr (P_1 + \ldots + P_n) = \sum_{i=1}^k \Tr P_i = \sum_{i=1}^k \rank(P_i) = \sum_{i=1}^k \dim(\Ima P_i)$$

Now assume $\Ima P_{j_1} \cap \Ima P_{j_2} \ne \{0\}$ for some $j_1 \ne j_2$:

\begin{align}\dim E &= \dim\left(\sum_{i=1}^k \Ima P_i\right) \\ &\le \sum_{i\ne j_1,j_2} \dim\Ima P_i + \dim(\Ima P_{j_1} + \Ima P_{j_2}) \\ &= \sum_{i\ne j_1,j_2} \dim\Ima P_i + \dim\Ima P_{j_1} + \dim\Ima P_{j_2} - \underbrace{\dim(\Ima P_{j_1} \cap \Ima P_{j_2})}_{>0} \\ &< \sum_{i=1}^k \dim\Ima P_i\\ &= \dim E \end{align}

This is a contradiction. Thus, the sum is indeed direct.

Now, take $i \ne j$. We have

$$P_i(\underbrace{P_jx}_{\in\Ima P_j}) = 0$$

since $\Ima P_i \cap \Ima P_j = \{0\}$. Therefore, $P_iP_j = 0$.

Note: This question was very helpful.

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