Show that $V = \mbox{ker}(f) \oplus \mbox{im}(f)$ for a straight map with $f \circ f = f$

Question : Let $V$ be a $K$ - Vectorspace and also $f: V \rightarrow V$ be straight. It holds that $f \circ f = f$. Show that $V = \mbox{ker}(f) \oplus \mbox{im}(f)$.

My effort : So i hunch that the $\oplus$ represents a straight amount which suggests i need to show that (i) $V = \mbox{ker}(f) + \mbox{im}(f)$ and also (ii) $\mbox{ker}(f) \cap \mbox{im}(f) = \{0\}$.

I attempted to do (ii) first : Let $v \in \mbox{im}(f) \cap \mbox{ker}(f)$

$\Rightarrow \exists u: f(u)=v \wedge f(v) = 0$

(can i placed a "Rightarrow" below?) $(f \circ f)(u)=f(f(u)) = f(v) = 0$

As for (i) i am having trouble with a strategy to revealing that $V = \mbox{ker}(f) + \mbox{im}(f)$. Should I also be attempting to do this to begin with? if so, any kind of suggestions regarding just how?

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2019-12-02 02:54:24
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Answers: 3

$v = (v - f(v)) + f(v)$

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2019-12-03 04:25:20
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Let is see (ii) first : (Yes, your use the effects icon $\Rightarrow$ is ideal.) You intend to make use of that $f\circ f=f$, which you do not state clearly. The factor is : In (ii) you require to show that the $v$ you are taking into consideration is $0$. You have : $f\circ f=f$, so $v=f(u)=(f\circ f)(u)$ etc, as you did. This offers $v=0$.

For (i), allow $v$ be any kind of component of $V$. What can we claim concerning $v-f(v)$? Allow is call it $w$. We have $f(w)=f(v-f(v))=f(v)-f(f(v))=f(v)-f(v)=0$. Below, I made use of that $f$ is straight and also (once more) that $f\circ f=f$. This reveals that $w\in{\rm ker}(f)$. So, $v=(v-f(v))+f(v)=w+f(v)$ is the amount of a component of the bit (particularly, $w$) and also among the photo (particularly, $f(v)$), which is specifically what you required.

(A meta - inquiry below is just how would certainly one consider taking into consideration $v-f(v)$. Ok, allow is attempt to function in reverse. Intend we currently recognize that $v=a+b$ with $a$ in the bit and also $b$ in the array. After that $f(v)=f(a)+f(b)=f(b)$, given that $a$ remains in the bit. Additionally, $f(v)=f(f(v))$, so it is practical, as a first strategy, to allow $b=f(v)$, and also see if we can straight confirm that $a=v-b$ remains in the bit. And also this is specifically what we did.)

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2019-12-03 04:25:12
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For 1) allow v be an approximate component in the junction of ker f and also photo f. This 2nd declaration, as you have actually mentioned, suggests that v = f (u) for some u, and also the first declaration, suggests that f (v) = 0. So 0 = f (v) = f (f (u)) which is f (u) (or v) by presumption. This reveals v = 0.

For 2) I think you can do this with absolutely nothing greater than set concept. Simply confirm V is a part of ker (f)+im (f) and afterwards confirm the reverse incorporation (this 2nd incorporation is unimportant given that ker (f) and also im (f) are plainly both parts of V)

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2019-12-03 04:25:00
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