Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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 $\ker\rho$.

share|improve this answer
    
Damn! Thanks. ${}{}$ –  Noomi Holloway Dec 15 '12 at 8:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.