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Here's the prompt:

Given $\rho$ and $\sigma$ are commuting projections, prove that $\rho \sigma$ is a projection and show that $\operatorname{ker}(\rho \sigma) = \operatorname{ker}(\rho) + \operatorname{ker} (\sigma)$.

I've already proven the first part, but am having trouble with the second. I know that I need to show that given $v$ in $\operatorname{ker} (\rho \sigma)$ that one must write $v$ as $v = l + k$, where $l$ is in $\operatorname{ker}(\rho)$ and $k$ is in $\operatorname{ker}(\sigma)$. I've seen where people start that proof with letting $x$ be in $\operatorname{ker}(\rho \sigma)$ and then using $x = \sigma(x) + (x-\sigma(x))$ to show the two parts, but I'm not seeing it. Any suggestions on where to start?

Thanks in advance.

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  • $\begingroup$ Hint : try to use the same decomposition trick a second time on $\sigma(x)$. What do you get ? $\endgroup$
    – yago
    Apr 21 '14 at 22:49
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Let $x\in\ker(\rho\sigma)$. As you noticed, $x=\sigma(x)+(x-\sigma(x))$. Notice that $\rho(\sigma(x))=0$ and $\sigma(x-\sigma(x))=\sigma(x)-\sigma^2(x)=\sigma(x)-\sigma(x)=0$, so $\sigma(x)\in\ker(\rho)$ and $(x-\sigma(x))\in\ker(\sigma)$. Hence, $$x=\underbrace{\sigma(x)}_{\in\ker(\sigma)}+\underbrace{(x-\sigma(x))}_{\in\ker(\rho)}\in\ker(\rho)+\ker(\sigma).$$ Therefore, $\ker(\rho\sigma)\subseteq\ker(\rho)+\ker(\sigma)$.

Conversely, if $y\in\ker(\rho)$ and $z\in\ker(\sigma)$, then $\rho\sigma(y+z)=\rho\sigma(y)+\rho\sigma(z)=\sigma\rho(y)+\rho\sigma(z)=\sigma(0)+\rho(0)=0+0=0$, so $\ker(\rho)+\ker(\sigma)\subseteq\ker(\rho\sigma)$.

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