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    #1
    Hey!!

    Let $\mathbb{K}$ be a field.

    1. Find a matrix $M\in \mathbb{K}^{2\times 2}$ such that for the linear mapping $f:\mathbb{K}^2\rightarrow \mathbb{K}^2, x\mapsto Mx$ it holds that $f\neq 0$ and $f^2:=f\circ f=0$.
    2. Let $V$ be a $\mathbb{K}$-vector space and $\psi:V\rightarrow V$ be a linear mapping, with $\psi^k\neq 0$ and $\psi^{k+1}=0$ for some $k>0$. Show that there is an element $x\in V$ such that the set $\{x, \psi (x), \ldots , \psi^k(x)\}$ is linearly independent.



    I have done the following:

    1. Since $f\neq 0$ it holds that $M$ is not the zero matrix.
      We have that $f^2(x):=f(f(x))=f(Mx)=MMx=M^2x$.
      So, we have to find a matrix $M$ such that $M^2$ is the zero matrix.
      Let $M=\begin{pmatrix}m_1&m_2 \\m_3&m_4\end{pmatrix}$.
      Then we have the following:
      $$M^2=0 \Rightarrow \begin{pmatrix}m_1&m_2 \\m_3&m_4\end{pmatrix}\begin{pmatrix}m_1&m_2 \\m_3&m_4\end{pmatrix}=\begin{pmatrix}0&0 \\0&0\end{pmatrix} \Rightarrow\begin{pmatrix}m_1^2+m_2m_3&m_1m_2+m_2m_4 \\m_3m_1+m_4m_3&m_3m_2+m_4^2\end{pmatrix}=\begin{pmatrix}0&0 \\0&0\end{pmatrix}$$
      So, we are looking for a solution for the following system:
      $$m_1^2+m_2m_3=0 \\ m_1m_2+m_2m_4=0 \Rightarrow m_2(m_1+m_4)=0\\m_3m_1+m_4m_3=0\Rightarrow m_3(m_1+m_4)=0 \\ m_3m_2+m_4^2=0$$
      When we take from the second and third equation that $m_1+m_4=0 \Rightarrow m_1=-m_4$, let $m_1=-m_4=1$, then we get from the first and fourth equation that $1+m_2m_3=0\Rightarrow n_2m_3=-1$. Let $m_2=-m_3=1$.
      So, we get the matrix: $$M=\begin{pmatrix}1&1 \\-1&-1\end{pmatrix}$$

      Is this correct?

    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 $c_0x+ c_1\psi (x)+ \ldots + c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$.
      $$\psi (c_0x+ c_1\psi (x)+ \ldots + c_{k-1}\psi^{k-1}(x)+ c_k\psi^k(x))=\psi (0)=0 \\ (\text{ it holds that } \psi (0)=0 \text{ since it is a linear mapping, or not? }) \\ \Rightarrow c_0\psi (x)+ c_1\psi^2 (x)+ \ldots + c_{k-1}\psi^k(x)+c_k\psi^{k+1}(x)=0 \\ \Rightarrow c_0\psi (x)+ c_1\psi^2 (x)+ \ldots + c_{k-1}\psi^k(x)=0$$

      Is this correct so far? How could we continue?
    Last edited by mathmari; January 10th, 2017 at 13:48.

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    #2 Thread Author
    Quote Originally Posted by mathmari View Post
    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?

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    #3
    Hey mathmari!!

    Quote Originally Posted by mathmari View Post
    1. Find a matrix $M\in \mathbb{K}^{2\times 2}$ such that for the linear mapping $f:\mathbb{K}^2\rightarrow \mathbb{K}^2, x\mapsto Mx$ it holds that $f\neq 0$ and $f^2:=f\circ f=0$.
    2. Let $V$ be a $\mathbb{K}$-vector space and $\psi:V\rightarrow V$ be a linear mapping, with $\psi^k\neq 0$ and $\psi^{k+1}=0$ for some $k>0$. Show that there is an element $x\in V$ such that the set $\{x, \psi (x), \ldots , \psi^k(x)\}$ is linearly independent.


    ...

    $$M=\begin{pmatrix}1&1 \\-1&-1\end{pmatrix}$$
    Is this correct?
    Yep.

    Alternatively, we can observe that $M$ must be similar to a Jordan normal form.
    Suppose for $v\ne 0$ we have $Mv=\lambda v$. Then $M^2v=\lambda^2 v$. Therefore $\lambda = 0$.
    Since $M \ne 0$, it follows that $M$ is similar to the Jordan normal form:
    $$M \sim \begin{pmatrix}0&1 \\ 0&0\end{pmatrix}$$

    Your matrix $M$ is indeed similar to that form.

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

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    #4 Thread Author
    Quote Originally Posted by I like Serena View Post
    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?

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    #5
    It looks all correct to me.

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

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    #6 Thread Author
    Quote Originally Posted by I like Serena View Post
    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?

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    #7
    Quote Originally Posted by mathmari View Post
    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 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?

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    #8 Thread Author
    Quote Originally Posted by HallsofIvy View Post
    "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?

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    #9 Thread Author
    Quote Originally Posted by mathmari View Post
    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.

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    #10
    I think we should proof for an arbitrary $k$ that $c_0 = 0, ..., c_i=0, ..., c_k=0$.

    Then the base case is to proof that $c_0 = 0$.
    And the induction step is to assume that $c_0=...=c_i=0$, and to proof that $c_{i+1}=0$.

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