[Question 2014a13] Answer

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[Question 2014a13] Answer

Let's introduce two simple conclusions.

conclusion 1 set \ (\varphi\) is a linear transformation on \ (n\) dimension linear space \ (v\), if there is a positive integer \ (k\), so that \ (\mathrm{r} (\varphi^k) =\mathrm{r} (\varphi^{k+1} ) \), then \[\mathrm{im\,}\varphi^k=\mathrm{im\,}\varphi^{k+1}=\mathrm{im\,}\varphi^{k+2}=\cdots.\]

Conclusion 1 of the proof of the arbitrary positive integer \ (k\), obviously have \ (\mathrm{im\,}\varphi^k\supseteq\mathrm{im\,}\varphi^{k+1}\), so by \ (\mathrm{r} (\ varphi^k) =\mathrm{r} (\varphi^{k+1}) \ (\mathrm{im\,}\varphi^k=\mathrm{im\,}\varphi^{k+1}\). So we just have to prove to the arbitrary positive integers \ (l\geq k\), \ (\mathrm{im\,}\varphi^l=\mathrm{im\,}\varphi^{l+1}\). On the one hand the inclusion is obvious, the present certificate \ (\mathrm{im\,}\varphi^l\subseteq\mathrm{im\,}\varphi^{l+1}\). either Fetch \ (\varphi^l (\alpha) \in\mathrm{im\,}\varphi^l\), then \ (\varphi^l (\alpha) =\varphi^{l-k} (\varphi^k (\alpha)) \). There is also \ (\beta\in v\), which makes \ (\varphi^k (\alpha) =\varphi^{k+1} (\beta) \), thus \ (\varphi^l (\alpha) =\varphi^{l-k} (\varphi^{k+1 } (\beta)) =\varphi^{l+1} (\beta) \in\mathrm{im\,}\varphi^{l+1}\).

conclusion 2 set \ (\varphi,\psi\) is a linear transformation on \ (n\) dimension linear space \ (v\), then \ (\mathrm{r} (\VARPHI\PSI) \geq \mathrm{r} (\varphi) +\mathrm{ R} (\PSI)-n\).

Conclusion 2 proves that this is the Sylvester inequality of rank, can refer to the No. 201 page of Fudan Gaodai Exercise 7.

return to the proof of the subject . By assuming that there are sufficiently large positive integers \ (n\), make \ (\varphi^n=0\). The following is an inductive method to prove \ (\mathrm{r} (\varphi^k) =n-k\), \ (1\leq K\leq n\). \ (k=1\) is the hypothesis in the question. Set \ (\mathrm{r} (\varphi^k) =n-k\), where \ (1\leq k<n\), certificate \ (\mathrm{r} (\varphi^{k+1}) =n-(k+1) \). By Conclusion 2 \[\mathrm{r} (\varphi^{k+1}) =\mathrm{r} (\varphi^k \varphi) \geq \mathrm{r} (\varphi^k) +\mathrm{r} (\ Varphi)-n=n-(k+1). \ (\mathrm{r} (\varphi^{k+1}) \leq \mathrm{r} (\varphi^k) =n-k\), if  \ (\mathrm{r} (\ Varphi^{k+1}) =\mathrm{r} (\varphi^k) =n-k\), then by the conclusion 1  \ (0 \neq N-k=\mathrm{r} (\varphi^k) =\mathrm{r} (\varphi^N) =0 \), contradiction. Therefore, only   \ (\mathrm{r} (\varphi^{k+1}) =n-(k+1) \) can be used to conclude the evidence. In particular, we know \ (R (\varphi^{n-1}) =1\) and \ (\varphi^n=0\). Fetch \ (\alpha\in v\) makes \ (\varphi^{n-1} (\alpha) \neq 0\), but \ (\varphi^n (\alpha) =0\). Finally by the Fudan Gaodai No. 206 page review question 13 know \ (\alpha,\varphi (\alpha), \cdots,\varphi^{n-1} (\alpha) \) is a set of \ (v\), thus \[v=l (\alpha,\varphi (\ Alpha), \cdots,\varphi^{n-1} (\alpha), \varphi^n (\alpha), \cdots). \,\,\,\box\]

note after learning the matrix of the Jordan standard form theory, we can give a very concise algebraic proof of [problem 2014a13].

[Question 2014a13] Answer

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