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$$\phi : {F^{N_0}}\rightarrow{F^{N_0}}$$

$$ x \mapsto {({x_{k+1}})}_{k \in {N_0}}$$ \

$\phi $ is called the left shift of F-vectorspace ${F^{N_0}}$

and let ${F^{N_0}_{<n}} = \{x \in{F^{N_0}} \mid x_k=0 , $for $ k\in {N_0} $ and $ k>=n\}$

$\textbf{Questions }$

a) Prove that ${F^{N_0}_{<n}}$ is a $\phi$-invariant F-subvecotorsapce which means $\phi({F^{N_0}_{<n}}) \subseteq {F^{N_0}_{<n}}$

b)find the Transformation matrix and Eigenvalues and eigenvectors of the restricted $\phi$ $$\phi : {F^{N_0}_{<4}}\rightarrow {F^{N_0}_{<4}}$$

c) find the Eigenvalues and eigenvectors of $ \phi$

$\textbf{My Solution }$

a) first we have to show that ${F^{N_0}_{<n}}$ is a subvector-space ( easy and no problem)

the let $ x \in {F^{N_0}_{<n}}$ which means that $$x = \begin{bmatrix} x_0&\\ x_1&\\ \vdots \\ x_{n-1}\\ 0\\ \vdots\\ 0\\ \end{bmatrix}$$ then $$\phi(x)= \begin{bmatrix} x_1&\\ x_2&\\ \vdots \\ x_{n-1}\\ 0\\ \vdots\\ 0\\ \end{bmatrix}$$

let then $y_k=x_{k+1}$ then $$\phi(x)= \begin{bmatrix} x_1&\\ x_2&\\ \vdots \\ x_{n-1}\\ 0\\ \vdots\\ 0\\ \end{bmatrix}= \begin{bmatrix} y_0&\\ y_1&\\ \vdots \\ y_{n-2}\\ 0\\ \vdots\\ 0\\ \end{bmatrix}$$

then we that for $ K>= n-1 $ ist $x_k=0 \rightarrow k>=n $ ist $ x_k=0 \rightarrow \phi(x) \in {F^{N_0}_{<n}}$

b) for ${F^{N_0}_{<4}}$ we see that $${F^{N_0}_{<4}}= \langle e_0,e_1,e_2,e_3\rangle $$

with
$$e_0=\begin{bmatrix} 1\\ 0\\ 0\\ \vdots\\ 0\\ \end{bmatrix},e_1=\begin{bmatrix} 0\\ 1\\ 0\\ \vdots\\ 0\\ \end{bmatrix},e_2=\begin{bmatrix} 0\\ 0\\ 1\\ \vdots\\ 0\\ \end{bmatrix}e_3=\begin{bmatrix} 0\\ 0\\ 0\\ 1\\ \vdots\\ 0\\ \end{bmatrix}$$

with $\phi(e_0)=0$ ,$\phi(e_1)=e_0$,$\phi(e_2)=e_1$,,$\phi(e_3)=e_2$

then the transformation Matris is : $$M_{e,e}(\phi_{\mid{F^{N_0}_{<4}}})=\begin{bmatrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0\\ \end{bmatrix}$$

in order to find the Eigenvalues and eigenvectors, we have to find the characteristic polynomial

$\chi_{M_{e,e}(\phi_{\mid{F^{N_0}_{<4}}})}= det(XE_4-M_{e,e}(\phi_{\mid{F^{N_0}_{<4}}}))= X^n$ $\rightarrow$ 0 is Zero of $\chi \rightarrow$ 0 is Eigenvalue of $\phi_{\mid{F^{N_0}_{<4}}}$ then : $$Eig_0(M_{e,e}(\phi_{\mid{F^{N_0}_{<4}}}))= Sol(M_{e,e}(\phi_{\mid{F^{N_0}_{<4}}})-0E_4,0)=\langle \begin{bmatrix} 1\\ 0\\ 0\\ 0\\ \end{bmatrix} \rangle $$

then is : $$Eig_0(\phi_{\phi_{\mid{F^{N_0}_{<4}}}}) = \langle e_0 \rangle= \langle \begin{bmatrix} 1\\ 0\\ 0\\ \vdots\\ 0\\ \end{bmatrix}\rangle$$

for c , it is the same idea and the result will be the same the only is eigenvalue vor $\phi$ is 0

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  • $\begingroup$ Is $N_0$ a subset of the natural numbers $\mathbb{N}$? and is $F$ an arbitrary field like $(\{0,1\},\cdot,+,-,^{-1})$ or do you demand $\operatorname{char} F = 0$? $\endgroup$ – Nathanael Skrepek Jul 11 '17 at 11:13
  • $\begingroup$ ${N_0}={0,1,2,3,4,..............}=\{0\} \cup N$ and F is Filed yeah $\endgroup$ – Mohbenay Jul 11 '17 at 11:17
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All your ideas are technically probably correct. However, the presentation is lacking in a lot of places, making it unclear what exactly it is you are trying to say.


First mistake:

Writing $$x = \begin{bmatrix} x_0&\\ x_1&\\ \vdots \\ x_{n-1}\\ 0\\ \vdots\\ 0\\ \end{bmatrix}$$

is very misleading, because it implies that $x$ is a finite vector (because you wrote the last zero). Instead, you should write

$$x = \begin{bmatrix} x_0&\\ x_1&\\ \vdots \\ x_{n-1}\\ 0\\ \vdots \end{bmatrix}$$


Second mistake:

then we that for $ K>= n-1 $ ist $x_k=0 \rightarrow k>=n $ ist $ x_k=0 \rightarrow \phi(x) \in {F^{N_0}_{<n}}$

This is very unclearly written. Since you already defined the $k$-th element of $\phi(x)$ as $y_k$, we know that $\phi(x)\in F_{<n}^{N_0}$ is $y_k=0$ for $k\geq n$. So you have to prove it. I don't see any $y_k$ in your argument. This should be improved.

******************************EDIT************************************

Here's an example of how I would write the proof to be as clear as possible:

Claim:

$\phi(F_{\leq n}^{N_0})\subseteq F_{\leq n}^{N_0}$

Proof:

We want to prove that $\forall x\in F_{\leq n}^{N_0}: \phi(x)\in F_{\leq n}^{N_0}$.

Let $x\in F_{\leq n}^{N_0}$ and let the $i$-th component of $x$ be $x_i$ (in other words, $$x=\begin{bmatrix}x_0\\x_1\\x_2\\\vdots\end{bmatrix}$$

Because $x\in F_{\leq n}^{N_0}$, we know that $x_n=x_{n+1}=\cdots =0$.

We want to prove that $\phi(x)\in F_{\leq n}^{N_0}$. Let $y_k$ be the $k$-th component of $\phi(x)$. By definition of $F_{\leq n}^{N_0}$, we know that $$\phi(x)\in F_{\leq n}^{N_0}\iff y_k=0\text{ for }k\geq n$$

by the definition of $\phi$, we know that $y_k = x_{k+1}$. Now let $k\geq n$. Then, $k+1\geq n$ and, therefore, $x_{k+1} = 0$. Since $y_k=x_{k+1}$, we conclude that $y_k=0$. Because $k$ was an arbitrary number $\geq n$, we have proven the statement $$\forall k\geq n: y_k=0$$ which means that $\phi(x)\in F_{\leq n}^{N_0}$.

QED.

************************END OF EDIT ****************************************


Third thing to improve:

$\chi_{M_{e,e}(\phi_{\mid{F^{N_0}_{<4}}})}= det(XE_4-M_{e,e}(\phi_{\mid{F^{N_0}_{<4}}}))= X^n$ $\rightarrow$ 0 is Zero of $\chi \rightarrow$ 0 is Eigenvalue of $\phi_{\mid{F^{N_0}_{<4}}}$ then : $$Eig_0(M_{e,e}(\phi_{\mid{F^{N_0}_{<4}}}))= Sol(M_{e,e}(\phi_{\mid{F^{N_0}_{<4}}})-0E_4,0)=\langle \begin{bmatrix} 1\\ 0\\ 0\\ 0\\ \end{bmatrix} \rangle $$

This is all horribly written, even though the original idea is probably correct.

  1. The first mistake is that this is not a sentence. All good mathematical arguments are, grammatically, still sentences They are formed usually by saying "because _____, we know ______, and therefore ______" and so on. You write what you plan on proving, and then you prove it. Here, you just throw out a bunch of equations that are clear to you, sure, but not to anyone reading what you wrote.
  2. you say $\chi_{M_{e,e}(\phi_{\mid{F^{N_0}_{<4}}})}= det(XE_4-M_{e,e}(\phi_{\mid{F^{N_0}_{<4}}}))$ But since the expression $\chi_{M_{e,e}(\phi_{\mid{F^{N_0}_{<4}}})}$ has no variable $X$, I have no idea wherethe $X$ came from on the right side of the equation. Furthermore, what is $E_4$? If you mean the identity matrix, you should write $I_4$ or just $I$, because that's the standard notation for the identity.

Conclusion:

My advice is to first use $A$ or $A_n$ as the notation for the matrix instead of the incredibly long-winded $M_{e,e}(\phi_{\mid{F^{N_0}_{<4}}})$. Then, define $p$ as the characteristic polynomial of $A$ and write down what $p(\lambda)$ is. Be as clear as possible, and read what you wrote and ask yourself "if this is the first time reading this, would I understand anything"?

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  • $\begingroup$ Oh you are totally right , those notation which I wrote , are written in our lessen , that s why I did use thosé rather than what you have advised .. $\endgroup$ – Mohbenay Jul 11 '17 at 10:50
  • $\begingroup$ @Mohbenay Well, you can keep the notation. Just be careful with other things, like $X$ appearing out of nowhere. Because usually, we use $\lambda$ for the free variable, and also, writing $p=\lambda^n$ is wrong, while $p(\lambda)=\lambda^n$ is correct. The difference, mathematically, is huge. $\endgroup$ – 5xum Jul 11 '17 at 10:51
  • $\begingroup$ Yeah I totally know , so can you tell me if (b) correct is ? $\endgroup$ – Mohbenay Jul 11 '17 at 10:53
  • $\begingroup$ @Mohbenay I don't know because I don't know what your answer is. The matrix is correct, but then you really need to rewrite the part with the eigenvalues, because I don't really know what your answer even is. $\endgroup$ – 5xum Jul 11 '17 at 10:55
  • $\begingroup$ can you just tell me then , how would i answer (a) in more understandable way , just some hints $\endgroup$ – Mohbenay Jul 11 '17 at 11:12

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