- #1

JD_PM

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- Homework Statement
- Let ##V## be a finite-dimensional vector space and let ##L: V \to V:v\mapsto L(v)## be a linear transformation such that the rank of ##L## (which, by definition, equals ##\dim(\Im L)##) equals the rank of ##L^2 = L \circ L## and ##\ker(L) = \ker(L^2)##.

a) Show that the linear map ##L' : \Im(L) \to V : x \mapsto L(x)## is injective.

b) Show that ##V =\ker(L) \oplus \Im(L)##

- Relevant Equations
- N/A

First thing to notice is that ##L## and ##L \circ L## are precisely equal linear maps.

What we know

$$L \ \text{is injective} \iff \ker(L)=\{0\}$$

$$\ker L' = \{ x \in \Im(L) \ | \ L'(x)=0\}$$

$$\Im(L)=\{ x \in V \ | \ \exists \ v \in V \ \text{such that} \ L(v)=x\}$$

Besides, we notice that ##x \in \Im(L) \subseteq V \Rightarrow x\in V##.

To prove: $$\ker(L') = \{0\}$$

Let ##x\in \ker(L')##. Then

$$0=L'(x) \Rightarrow 0=L(x) \Rightarrow 0 = L (L(v)) \Rightarrow 0 = L^2(v) = L(v) \Rightarrow 0 = L(v) =x$$

So ##x=0## and hence ##\ker L' = \{ 0\}##.

b) I saw in class the following proposition: given two subspaces ##U_1, U_2## of ##V## we have $$W= U_1 \oplus U_2 \iff W=U_1 + U_2 \ \& \ U_1 \cap U_2 = \{0\}$$

TP: $$V=\ker(L) + \Im(L) \ \& \ \ker(L) \cap \Im(L) = \{0\}$$

Let ##x \in \ker(L) \cap \Im(L)##. Then ##x \in \ker(L)## and ##x \in \Im(L)##, which implies ##L(x) = 0## and ##L(v) = x##. Apply ##L## to the former equation to get

$$L(L(v)) = L(x) \Rightarrow L(L(v)) = L(v) = 0 \Rightarrow x=0$$

So ##\ker(L) \cap \Im(L) = \{0\}##.

We are left to show that ##V = \ker(L) + \Im(L)## is unique. Take ##x_1, x_1' \in \ker(L)## and ##x_2, x_2' \in \Im(L)##. Suppose it is not unique i.e. there's ##v\in V## such that ##v=x_1+x_2## and ##v=x_1'+x_2'##. Then ##x_1+x_2=x_1'+x_2' \Rightarrow x_1-x_1'=x_2'-x_2##. But ##x_1-x_1' \in \ker(L)## and ##x_2'-x_2 \in \Im(L)## and we just proved that ##\ker(L) \cap \Im(L) = \{0\}## so ##x_1-x_1'=0=x_2'-x_2 \Rightarrow x_1=x_1', x_2=x_2'##, a contradiction! :)

I think I got this one mainly right! I'd like to get some feedback/slash correction from you guys, thanks!

What we know

$$L \ \text{is injective} \iff \ker(L)=\{0\}$$

$$\ker L' = \{ x \in \Im(L) \ | \ L'(x)=0\}$$

$$\Im(L)=\{ x \in V \ | \ \exists \ v \in V \ \text{such that} \ L(v)=x\}$$

Besides, we notice that ##x \in \Im(L) \subseteq V \Rightarrow x\in V##.

To prove: $$\ker(L') = \{0\}$$

Let ##x\in \ker(L')##. Then

$$0=L'(x) \Rightarrow 0=L(x) \Rightarrow 0 = L (L(v)) \Rightarrow 0 = L^2(v) = L(v) \Rightarrow 0 = L(v) =x$$

So ##x=0## and hence ##\ker L' = \{ 0\}##.

b) I saw in class the following proposition: given two subspaces ##U_1, U_2## of ##V## we have $$W= U_1 \oplus U_2 \iff W=U_1 + U_2 \ \& \ U_1 \cap U_2 = \{0\}$$

TP: $$V=\ker(L) + \Im(L) \ \& \ \ker(L) \cap \Im(L) = \{0\}$$

Let ##x \in \ker(L) \cap \Im(L)##. Then ##x \in \ker(L)## and ##x \in \Im(L)##, which implies ##L(x) = 0## and ##L(v) = x##. Apply ##L## to the former equation to get

$$L(L(v)) = L(x) \Rightarrow L(L(v)) = L(v) = 0 \Rightarrow x=0$$

So ##\ker(L) \cap \Im(L) = \{0\}##.

We are left to show that ##V = \ker(L) + \Im(L)## is unique. Take ##x_1, x_1' \in \ker(L)## and ##x_2, x_2' \in \Im(L)##. Suppose it is not unique i.e. there's ##v\in V## such that ##v=x_1+x_2## and ##v=x_1'+x_2'##. Then ##x_1+x_2=x_1'+x_2' \Rightarrow x_1-x_1'=x_2'-x_2##. But ##x_1-x_1' \in \ker(L)## and ##x_2'-x_2 \in \Im(L)## and we just proved that ##\ker(L) \cap \Im(L) = \{0\}## so ##x_1-x_1'=0=x_2'-x_2 \Rightarrow x_1=x_1', x_2=x_2'##, a contradiction! :)

I think I got this one mainly right! I'd like to get some feedback/slash correction from you guys, thanks!