Originally Posted by mathmari

2. Let $x\in V$. Suppose that the set $\{x, \psi (x), \ldots , \psi^k(x)\}$ is linearly dependent, i.e., there are $c_i$'s not all zero such that $c_0x+ c_1\psi (x)+ \ldots + c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$.
$$\psi (c_0x+ c_1\psi (x)+ \ldots + c_{k-1}\psi^{k-1}(x)+ c_k\psi^k(x))=\psi (0)=0 \\ (\text{ it holds that } \psi (0)=0 \text{ since it is a linear mapping, or not? }) \\ \Rightarrow c_0\psi (x)+ c_1\psi^2 (x)+ \ldots + c_{k-1}\psi^k(x)+c_k\psi^{k+1}(x)=0 \\ \Rightarrow c_0\psi (x)+ c_1\psi^2 (x)+ \ldots + c_{k-1}\psi^k(x)=0$$

$$\psi (c_0x+ c_1\psi (x)+ \ldots + c_{k-1}\psi^{k-1}(x)+ c_k\psi^k(x))=\psi (0)=0 \\ \Rightarrow c_0\psi (x)+ c_1\psi^2 (x)+ \ldots + c_{k-1}\psi^k(x)+c_k\psi^{k+1}(x)=0 \\ \Rightarrow c_0\psi (x)+ c_1\psi^2 (x)+ \ldots + c_{k-1}\psi^k(x)=0 \\ \Rightarrow \psi (c_0\psi (x)+ c_1\psi^2 (x)+ \ldots + c_{k-2}\psi^{k-1}(x)+c_{k-1}\psi^k(x))=\psi (0) \\ \Rightarrow c_0\psi^2 (x)+ c_1\psi^3 (x)+ \ldots + c_{k-2}\psi^k(x)+c_{k-1}\psi^{k+1}(x)=0 \\ \Rightarrow c_0\psi^2 (x)+ c_1\psi^3 (x)+ \ldots + c_{k-2}\psi^k(x)=0 \\ \ldots \\ \Rightarrow c_0\psi^k(x)+c_1\psi^{k+1}(x)=0 \\ \Rightarrow c_0\psi^k(x)=0$$ Since $\psi^k(x)\neq 0$ we conclude that $c_0$.

Do we do the same again $k$ times to cocnlude that $c_i=0$ for all $i=0, \ldots , k$ ?

Or do we show it in an other way?

Originally Posted by I like Serena

Yep.

Now let's add the initial choice for $x$ such that $\psi^k(x) \ne 0$. Such an $x$ must exist, since $\psi^k \ne 0$.
Then it follows that we have a contradiction.
Therefore, with that choice of $x$, it follows that $x, \psi(x), ..., \psi^k(x)$ are linearly independent.

Could we formulate it also as follows?

Let $x\in V$ such that $\psi^k(x) \ne 0$. The set $\{x, \psi (x), \ldots , \psi^k\}$ ist linearly independent if and only if it holds that $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$ for $c_0=c_1=\ldots =c_{k-1}=c_k=0$.

Since $\psi$ is a linear mapping we have that $\psi (0)=0$.

Since $\psi^{k+1}(x)=0$ and $\psi (0)=0$ we have that $\psi^n(x)=0, \ \forall n\geq k+1$.

We have the following:
$$\psi^k (c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x))=\psi^k (0) \\ \Rightarrow c_0\psi^k(x)+c_1\psi^{k+1} (x)+\ldots +c_{k-1}\psi^{2k-1}(x)+c_k\psi^{2k}(x)=0 \\ \Rightarrow c_0\psi^k(x)=0 \\ \Rightarrow c_0=0$$

So, we get $c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$.

We have that
$$\psi^{k-1} (c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x))=\psi^{k-1} (0) \\ \Rightarrow c_1\psi^k (x)+\ldots +c_{k-1}\psi^{2k-2}(x)+c_k\psi^{2k-1}(x)=0 \\ \Rightarrow c_1\psi^k(x)=0 \\ \Rightarrow c_1=0$$

So, we get $c_2\psi^2 (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$.

When we repeat this $(k-2)$-times (or not?) we get that $c_i=0, \ \forall i=1, \ldots k$.


Is everything correct? Could I improve something?

Originally Posted by I like Serena

Formally, I think we're supposed to set up a proof by induction though...

Ah ok. So, is the induction as follows?

Let $x\in V$ such that $\psi^k(x) \ne 0$. We wil show that the set $\{x, \psi (x), \ldots , \psi^k\}$ ist linearly independent, i.e., that it holds it when $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$ then $c_i=0, \forall i$.

Base case: For i=0 it holds the following:

Since $\psi$ is a linear mapping we have that $\psi (0)=0$.
$$\psi^k (c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x))=\psi^k (0) \\ \Rightarrow c_0\psi^k(x)+c_1\psi^{k+1} (x)+\ldots +c_{k-1}\psi^{2k-1}(x)+c_k\psi^{2k}(x)=0 \\ \Rightarrow c_0\psi^k(x)=0 \\ \Rightarrow c_0=0 \ \ \checkmark$$


Inductive hypothesis: We suppose that it holds that for $i\leq m$ : $$c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0 \Rightarrow c_i=0, \forall 1\leq i\leq m$$


Inductive step: We want to shw that it holds for $i=m+1$:
From the inductive hypothesis we get that $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0 \Rightarrow c_i=0, \forall 1\leq i\leq m$. So $c_{m+1}\psi^{m+1}(x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$.
Then we have the following:
$$\psi^{k-m-1} (c_{m+1}\psi^{m+1}(x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x))=\psi^{k-1} (0) \\ \Rightarrow c_{m+1}\psi^k (x)+\ldots +c_{k-1}\psi^{2k-m-2}(x)+c_k\psi^{2k-m-1}(x)=0 \\ \Rightarrow c_{m+1}\psi^k(x)=0 \\ \Rightarrow c_{m+1}=0$$



So, we have that when $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$ then $c_i=0, \forall i$, i.e., the set $\{x, \psi (x), \ldots , \psi^k\}$ ist linearly independent.





This is the only way to show that the set is inearly independent, right?

Originally Posted by mathmari

Ah ok. So, is the induction as follows?

Let $x\in V$ such that $\psi^k(x) \ne 0$. We wil show that the set $\{x, \psi (x), \ldots , \psi^k\}$ ist linearly independent, i.e., that it holds it when $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$ then $c_i=0, \forall i$.

Base case: For i=0 it holds the following:

"i" is not the index here, k is.

Quote:

Since $\psi$ is a linear mapping we have that $\psi (0)=0$.
$$\psi^k (c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x))=\psi^k (0) \\ \Rightarrow c_0\psi^k(x)+c_1\psi^{k+1} (x)+\ldots +c_{k-1}\psi^{2k-1}(x)+c_k\psi^{2k}(x)=0 \\ \Rightarrow c_0\psi^k(x)=0 \\ \Rightarrow c_0=0 \ \ \checkmark$$


Inductive hypothesis: We suppose that it holds that for $i\leq m$ : $$c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0 \Rightarrow c_i=0, \forall 1\leq i\leq m$$


Inductive step: We want to shw that it holds for $i=m+1$:
From the inductive hypothesis we get that $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0 \Rightarrow c_i=0, \forall 1\leq i\leq m$. So $c_{m+1}\psi^{m+1}(x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$.
Then we have the following:
$$\psi^{k-m-1} (c_{m+1}\psi^{m+1}(x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x))=\psi^{k-1} (0) \\ \Rightarrow c_{m+1}\psi^k (x)+\ldots +c_{k-1}\psi^{2k-m-2}(x)+c_k\psi^{2k-m-1}(x)=0 \\ \Rightarrow c_{m+1}\psi^k(x)=0 \\ \Rightarrow c_{m+1}=0$$



So, we have that when $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$ then $c_i=0, \forall i$, i.e., the set $\{x, \psi (x), \ldots , \psi^k\}$ ist linearly independent.





This is the only way to show that the set is inearly independent, right?

Originally Posted by HallsofIvy

"i" is not the index here, k is.

So, do we have to show that for each $k\geq 0$ it holds that $$c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0\Rightarrow c_0=\ldots =c_k=0$$ ?

Base case: For $k=0$ we have that $\psi (x)\neq 0$ and then $$c_0x=0\Rightarrow \psi(c_0x)=\psi(0)\Rightarrow c_0\psi(x)=0\Rightarrow c_0\ \checkmark$$

Inductive hypothesis: We suppose that it holds for $k=n$:
$$c_0x+c_1\psi (x)+\ldots +c_{n-1}\psi^{n-1}(x)+c_n\psi^n(x)=0\Rightarrow c_0=\ldots =c_n=0$$

Inductive step: We want to show that it holds for $k=n+1$:
$$c_0x+c_1\psi (x)+\ldots +c_{n}\psi^{n}(x)+c_{n+1}\psi^{n+1}(x)=0\Rightarrow c_0=\ldots =c_{n+1}=0$$
What can we do here? How can we use the inductive hypothesis? We could apply it only if we would have $c_0x+c_1\psi (x)+\ldots +c_{n-1}\psi^{n-1}(x)+c_n\psi^n(x)=0$, or not?

Originally Posted by mathmari

Inductive step: We want to show that it holds for $k=n+1$:
$$c_0x+c_1\psi (x)+\ldots +c_{n}\psi^{n}(x)+c_{n+1}\psi^{n+1}(x)=0\Rightarrow c_0=\ldots =c_{n+1}=0$$

Tp apply here the inductive hypothesis, we have to show first that $c_{n+1}=0$, or not? But how? When we apply any map $\psi^i$ the term $c_{n+1}\psi^{n+1}(x)$ will get zero.