Linear map $L \neq O$ having trivial image only at $L^2=L \circ L$

Question

From S.L Linear Algebra:

Let $L:ℝ^2 \rightarrow ℝ^2$ be a linear map such that $L \neq O$ but $L^2=L \circ L=O$. Show that there exists a basis $\{A, B\}$ of $ℝ^2$ such that, $L(A)=B$ and $L(B)=O$.

---

According to my logical standpoint, I find such problem slightly confusing. Let's say we have some arbitrary vector $v=\{v_1, v_2\} \in ℝ^2$, then $L(v) \neq O$, but why does $(L \circ L)(v)=L(L(v))=O$? Isn't this contradiction of previous logical assertion that $L(v) \neq O$?

What I currently know about the linear map:

1. $L$ is a linear map that is surjective and injective (bijective).
2. $L$ has trivial kernel of $\{0\}$ due to being injective.
3. $L$ is invertible, due to being bijective.
4. $L$ is an isomorphism.
5. For $v, u \in \mathbb{R}^2$, $L(v+u)=L(v)+L(u)$.
6. For $c \in \mathbb{R}$ and $v \in \mathbb{R}^2$, $cL(v)=L(cv)$.

But this knowledge doesn't seem to be sufficient, since I'm unable to understand logical validity of the statement given above from S.L Linear Algebra.

Is there any mistake in the book? If not, why are both assertions mutually true? (why is $L^2=L \circ L=O$ when $L \neq O$?)

Answer 1

In fact, $L$ is neither surjective nor injective. And $L\neq 0$ does not contradict $L^2=0$. You can see this by an example: $$ L=\begin{bmatrix} 0 & 1 \\ 0 &0 \end{bmatrix}. $$ The solution is quite simple and straightforward once you get it. We can see that $$ (0)<\ker L <\ker L^2=\mathbb{R}^2\quad\cdots(*). $$ So choose $w\in \ker L^2\setminus \ker L$ and let $v=Lw$. We immediately see that $Lv = L^2w =0$. To show that they form a basis, it suffices to show they are linearly independent. Suppose $$\alpha v +\beta w =0.$$ Applying $L$ yields $$ \beta Lw =0. $$ By the assumption that $w\not\in\ker L$, we have $\beta = 0$ and $\alpha =0$. This shows $\{v,w\}$ is a basis.

$\textbf{EDIT:}$ I guess you are not very familiar with linear algebra, so I'll add some details. First, any $n\times m$ matrix $A$ can be seen as a linear map from $\mathbb{F}^m$ into $\mathbb{F}^n$ acting by $$ A\begin{bmatrix} x_1 \\ x_{2} \\ \vdots\\ x_m \end{bmatrix} = \begin{bmatrix} a_{11}x_1 + a_{12}x_2 +\cdots +a_{1n}x_n \\ a_{21}x_1 + a_{22}x_2 +\cdots +a_{2n}x_n \\ \vdots\\ a_{m1}x_1 + a_{m2}x_2 +\cdots +a_{mn}x_n \end{bmatrix}. $$ Hence, any matrix represents a linear map. Moreover, given a linear map $L:V\to W$, we can find a matrix representation of $L$ using fixed ordered bases of $V$ and $W$. In this sense, linear map and matrix are equivalent objects. Thus the matrix example is legitimate.

If we have $\ker L = (0)$, observe that $$L^2v = L(Lv) =0$$ for all $v\in \mathbb{R}^2$ implies that $$Lv =0,$$ and in turn that $$v=0.$$ (because if $Lx =0$, then $x\in \ker L$ and $x=0$.) This leads to contradiction, so $\ker L$ is not the trivial space $(0)$. This means that $L$ is not injective since it implies that there is $x\neq 0$ such that $Lx=0$. On the other hand, we can see that $\ker L \neq \ker L^2 =\mathbb{R}^2$ since if it were, then $L$ must be $O$. This proves $(*)$ holds.

To see that $L$ is not surjective, you should be aware of argument using dimension theorem. If $L:V\to V$ is a linear map, then a version of pigeonhole principle holds for $L$ which says that $L$ is surjective if and only if it is injective (just like any function defined on a finite set $F$.)

Answer 2

According to my logical standpoint, I find such problem slightly confusing. Let's say we have some arbitrary vector $v=\{v1,v2\}∈\Bbb R^2$, then $L(v)≠O\ ...$

That is incorrect. The problem states $L\ne O$, which means the matrix of the linear map, $L$, is not the zero matrix, or that the linear map is not the zero-transformation. This, however, does not mean that $L(v)=Lv\ne\vec0\ \forall v\in\Bbb R^2$. For example, let $L=\begin{bmatrix}0&0\\1&0\end{bmatrix}, v=\begin{bmatrix}0&1\end{bmatrix}^T;Lv=\vec0, L\ne O$.

Now, note that the matrix of $L\circ L$ is $L\times L=L^2=O$, which is given.

$\implies \det(L^2)=[\det(L)]^2=0\\\implies \det(L)=0$

Since $L$ is a $2\times2$ matrix, this means the rank of $L<2$. It cannot be $0$ since $L\ne O$, so the rank of $L=1$. By the rank-nullity theorem, the nullity of $L$ is also $1$, which indicates that the basis of the null-space of $L$ is singleton, and all vectors belonging to the null-space of $L$ are of the form $kv, k\in\Bbb R$, for some $v\in\Bbb R^2-\{(0,0)\}$.

This fact invalidates your first $4$ observations:

1. $L$ is a linear map that is surjective and injective (bijective).
2. $L$ has trivial kernel of $\{0\}$ due to being injective.
3. $L$ is invertible, due to being bijective.
4. $L$ is an isomorphism.

Now, take vector $A$ linearly independent of $v$, and the basis of $\Bbb R^2$ as $\{A, kv\}, k\ne0$, which is the answer. Can you argue why such an $A$ exists?