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I'm working on a problem out of Roman's Advanced Linear Algebra.

Let $V$ be a vector space over a field $F$ ($\operatorname{char}(F)\neq 2$), with $\rho$ and $\sigma$ projections. Prove:

The difference $\rho-\sigma$ is a projection iff $\rho\sigma=\sigma\rho=\sigma$, in which case $$ \operatorname{im}(\rho-\sigma)=\operatorname{im}(\rho)\cap\ker(\sigma),\quad \ker(\rho-\sigma)=\ker(\rho)\oplus \operatorname{im}(\sigma). $$

I've essentially proven everything except that $\ker(\rho-\sigma)\subseteq\ker(\rho)\oplus\operatorname{im}(\sigma)$. From the relations $\rho\sigma=\sigma\rho=\sigma$ I see $\ker(\rho)\subseteq\ker(\sigma)$ and $\operatorname{im}(\sigma)\subseteq\operatorname{im}(\rho)$. I take $v\in\ker(\rho-\sigma)$, and conclude that $\rho(v)=\sigma(v)$, but I don't see way to write $v$ as a sum of elements in $\ker(\rho)$ and $\operatorname{im}(\sigma)$. Does anyone know how to show this containment? Thanks.

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How about $v=(v-\sigma(v))+\sigma(v)$?


To arrive at this, one could guess at $\sigma(v)$ for the second summand simply because it is something available in the image of $\sigma$, and then check that it works. A more systematic approach would be as follows.

Suppose that $v= x+ y$ with $x\in\ker\rho$ and $y\in\mathrm{im}\ \sigma$. Then $\sigma(v)=\sigma(x)+\sigma(y)=0+y=y$, hence $y=\sigma(v)$ and $x=v-\sigma(v)$. (We used $x\in\ker\rho\subseteq\ker \sigma$ and $\sigma\sigma=\sigma$.) Then, because $\sigma(v)=\rho(v)$ it follows that $v-\sigma(v)$ is in fact in $\ker\rho$.

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  • $\begingroup$ Damn! Thanks. ${}{}$ $\endgroup$ – Noomi Holloway Dec 15 '12 at 8:23

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