Statements about linear maps | Linear Algebra

[JD_PM]

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!

 

[FactChecker]

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

Why?
If this is essential to the rest of the proof then I have my doubts about the proof. I have not looked at the rest of it.

 

[fresh_42]

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!

You cannot use $(L')^2=L'$ or $L^2=L$. What we have is
$$
\operatorname{ker}L \hookrightarrow V \twoheadrightarrow \operatorname{im}L \stackrel{L'}{\cong} \operatorname{im}L
$$
and the last isomorphism has to be shown, which isn't necessarily an identity.

The second part b) is true for all linear transformations on finite-dimensional vector spaces. You can prove it e.g. by choosing a basis for $\operatorname{ker} L$ and extending it to a basis of $L$ etc.

Once you have b), you can use it to prove a).

An example of such a linear transformation is $V=\mathbb{R}^4\, , \,L=\begin{bmatrix} 0&0&0&0\\0&0&0&0\\0&0&0&1\\0&0&1&0 \end{bmatrix}$

 

[JD_PM]

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

Why?

Given $L: V \to V:v\mapsto L(v)$ and the definition of product of linear maps (for instance, as found in enlightening Axler's Algebra done right book)

One finds that $L \circ L: V \to V:v\mapsto L(v)$, so I would say this is a sufficient condition to assert that these two are equal.

You cannot use $(L')^2=L'$ or $L^2=L$. What we have is
$$
\operatorname{ker}L \hookrightarrow V \twoheadrightarrow \operatorname{im}L \stackrel{L'}{\cong} \operatorname{im}L
$$
and the last isomorphism has to be shown, which isn't necessarily an identity.

Hi @fresh_42, it is nice to discuss with you! You seem to disagree in this step

$$L^2(v) = L(v)$$

So you seem to suggest that $L$ and $L \circ L$ are not the same maps. Might you please explain why in more detail? (I do not follow your argument above).

 

[fresh_42]

Given $L: V \to V:v\mapsto L(v)$ and the definition of product of linear maps (for instance, as found in enlightening Axler's Algebra done right book)

View attachment 287372

Applied to $S=T=L$ we get $L^2(v)=L(L(v))$. Nowhere is said that this equals $L(v)$ except in your proof. Look at my example matrix. It has rank $2$, as has its square, a $2-$dimensional kernel, and $e_3=L^2(e_3)\neq L(e_3)=e_4$ if $e_3,e_4$ denote the third and fourth basis vector. Yet, it is injective on $\operatorname{im}(L).$

Try my approach: Solve b) for any linear transformation $L\, : \,V\longrightarrow V$ and use it for a). Are you allowed to use the rank-nullity theorem, $\dim V =\dim \operatorname{im(L)}+\dim \operatorname{ker}(L),$ or is exercise b) used to prove it?

 

[FactChecker]

One finds that $L \circ L: V \to V:v\mapsto L(v)$,

This is definitely not true in general. How are you proving this? Are you thinking that having the same range implies that they are identical? What about the linear function f(x) = 2x and f(f(x)) = 4x?

 

[pasmith]

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.

This is false. Consider, for example, $L: \mathbb{R}^2 \to \mathbb{R}^2$ where $L(e_1) = 2e_1$ and $L(e_2) = 0$. Here $\ker L = \ker L^2 = \langle e_2 \rangle$, but $$L^2(e_1) = 4e_1 \neq 2e_1 = L(e_1)$$.

 

[JD_PM]

Sorry for replying a bit late, needed to study.

Ohhh, big yikes! I understand my mistake now! :)

Try my approach: Solve b) for any linear transformation $L\, : \,V\longrightarrow V$ and use it for a). Are you allowed to use the rank-nullity theorem, $\dim V =\dim \operatorname{im(L)}+\dim \operatorname{ker}(L),$ or is exercise b) used to prove it?

I am indeed allowed to use the fundamental theorem of linear maps but I am supposed to solve a) before b), so please let me show my new reasoning.

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

Now a) should be correct

b) Show that $V =\ker(L) \oplus \text{Im}(L)$

What I presented was essentially wrong so let's start from scratch.

We need to show that $\ker(L) \cap \text{Im}(L) = \{0\}$ and $V=\ker(L) + \text{Im}(L)$. Let's start by $\ker(L) \cap \text{Im}(L) = \{0\}$. My mistake was again on assuming that $L^2(v) = L(v)$ holds generally.

Let $x \in \ker(L) \cap \text{Im}(L)$. Then $x \in \ker(L)$ and $x \in \text{Im}(L)$, which implies $L(x) = 0$ and $L(v) = x$. Apply $L$ to the latter equation to get

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

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

We are left to show that $V = \ker(L) + \Im(L)$. 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! :)

OK, what I presented is NOT a proof of $V =\ker(L) + \text{Im}(L)$ but that given the sum of any two subspaces of $V$, with their intersection equal to $\{0\}$, their sum must be direct.

But let's tackle it once again (it is actually quite fun because it made me think a lot! :D).

Usually, I am given the explicit vector space $V$ when I am asked to show that $V = U_1 + U_2$. For instance, an easy example: "show that $\Bbb R^2 = U_1 + U_2$, where $U_1 = \{ (x, 0) | x \in \Bbb R\}$ and $U_2 = \{ (0, y) | y \in \Bbb R\}$". We then would proceed to show both inclusions i.e. $V \subset U_1 + U_2$ and $U_1 + U_2 \subset V$. To show the first we of course take any $x \in \Bbb R^2$, write it as a linear combination of basis vectors of $\Bbb R^2$ and notice that $x = \underbrace{a_1(1,0)}_{\in U_1} + \underbrace{a_2(1,0)}_{\in U_2}$ so $x \in U_1 + U_2$. To show $U_1 + U_2 \subset V$, we would take any $x \in U_1 + U_2$ and, by definition of sum of subspaces, notice that $\exists u_1 \in U_1, u_2 \in U_2$ such that $x = u_1 + u_2 \in \Bbb R^2$.

Let's go back to the main problem.

What's tricky here is that I see no way of showing $V \subset \ker(L) + \text{Im}(L)$ because we are not given the explicit form of $V$.

So the idea to show such equality would be showing the inclusion $\ker(L) + \text{Im}(L) \subset V$ (which is also tough because, again, we are not given the explicit form of $V$ but at least I see how to start) and $\dim \left( \ker(L)+\text{Im}(L)\right) = \dim V$

What do you think of this approach? If you agree it is feasible I'll show my attempt.

Please let me know if there is an easier way, thanks!

 

[fresh_42]

Sorry for replying a bit late, needed to study.

Ohhh, big yikes! I understand my mistake now! :)



I am indeed allowed to use the fundamental theorem of linear maps but I am supposed to solve a) before b), so please let me show my new reasoning.

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

Now a) should be correct

Let's see. You apparently use $x=L(v)\in \mathfrak{I}(V)$ and assume $x\in \ker L'$. Then $L^2(x)=0$ so $x\in \ker L^2.$ Clearly $\ker L \subseteq \ker L^2$. But why is $$\ker L^2 \subseteq \ker L\qquad ?$$ This is the crucial point of your argument, but I do not see why this should be the case.

You have to use the condition $\operatorname{rk}L^2=\operatorname{rk}L$ somewhere! One possibility is to prove part b) first. If you do not want to prove b) first, then you have to emulate the reasoning of part b) here in part a).

What I presented was essentially wrong so let's start from scratch.

We need to show that $\ker(L) \cap \text{Im}(L) = \{0\}$ and $V=\ker(L) + \text{Im}(L)$. Let's start by $\ker(L) \cap \text{Im}(L) = \{0\}$. My mistake was again on assuming that $L^2(v) = L(v)$ holds generally.

Let $x \in \ker(L) \cap \text{Im}(L)$. Then $x \in \ker(L)$ and $x \in \text{Im}(L)$, which implies $L(x) = 0$ and $L(v) = x$. Apply $L$ to the latter equation to get

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

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

Same as above!

OK, what I presented is NOT a proof of $V =\ker(L) + \text{Im}(L)$ but that given the sum of any two subspaces of $V$, with their intersection equal to $\{0\}$, their sum must be direct.

But let's tackle it once again (it is actually quite fun because it made me think a lot! :D).

Usually, I am given the explicit vector space $V$ when I am asked to show that $V = U_1 + U_2$. For instance, an easy example: "show that $\Bbb R^2 = U_1 + U_2$, where $U_1 = \{ (x, 0) | x \in \Bbb R\}$ and $U_2 = \{ (0, y) | y \in \Bbb R\}$". We then would proceed to show both inclusions i.e. $V \subset U_1 + U_2$ and $U_1 + U_2 \subset V$. To show the first we of course take any $x \in \Bbb R^2$, write it as a linear combination of basis vectors of $\Bbb R^2$ and notice that $x = \underbrace{a_1(1,0)}_{\in U_1} + \underbrace{a_2(1,0)}_{\in U_2}$ so $x \in U_1 + U_2$. To show $U_1 + U_2 \subset V$, we would take any $x \in U_1 + U_2$ and, by definition of sum of subspaces, notice that $\exists u_1 \in U_1, u_2 \in U_2$ such that $x = u_1 + u_2 \in \Bbb R^2$.

Let's go back to the main problem.

What's tricky here is that I see no way of showing $V \subset \ker(L) + \text{Im}(L)$ because we are not given the explicit form of $V$.

So the idea to show such equality would be showing the inclusion $\ker(L) + \text{Im}(L) \subset V$ (which is also tough because, again, we are not given the explicit form of $V$ but at least I see how to start) and $\dim \left( \ker(L)+\text{Im}(L)\right) = \dim V$

What do you think of this approach? If you agree it is feasible I'll show my attempt.

Watch out for hidden assumptions!

Please let me know if there is an easier way, thanks!

Well, Wikipedia has a, which I find, quite detailed proof. Maybe you want to study it, and you let me know if you have any difficulties with it.
https://en.wikipedia.org/wiki/Rank–nullity_theorem#First_proof