For vector V, if we apply linear transformation T to V n times, what will happen? i.e. what is c(T^n V) ? Can anyone provide some hints about this problem set?
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In the future, please take the time to enter important parts of your question as text instead of pasting an image. Images are neither searchable nor accessible to people using screen readers, nor do they show up in summaries. You can find a quick reference to formatting mathematical expressions with MathJax here. – amd Mar 30 '18 at 00:25
2 Answers
Write the powers of the Jordan matrix explicitly. This may help.
Let $w_1, \ldots, w_4$ be the basis corresponding to the Jordan matrix (so that $T w_1 = \alpha w_1$, $T w_2 = \beta w_2$, $T w_3 = \beta w_3 + w_2$, and $T w_4 = \beta w_4 + w_3$).
Then for $v=a_1 w_1 + \cdots + a_4 w_4$ you can write $$T^n(v) = T^n(a_1 w_1 + \cdots + a_4 w_4) = a_{n,1} a_1 w_1 + \cdots + a_{n,4} a_4 w_4,$$ where $a_{n,j}$ are coefficients you can compute from your explicit Jordan matrix power computation. They will depend on $n$, $\alpha$, and $\beta$. For instance, $a_{n,1} = \alpha^n$.
Then $$\frac{c(T^n v)}{c(T^{n-1} v)} = \frac{c(a_{n,1} a_1 w_1 + \cdots + a_{n,4} a_4 w_4)}{c(a_{n-1,1} a_1 w_1 + \cdots + a_{n-1,4} a_4 w_4)} = \frac{a_{n,1} a_1 c(w_1) + \cdots + a_{n,4} a_4 c(w_4)}{a_{n-1,1} a_1 c(w_1) + \cdots + a_{n-1,4} a_4 c(w_4)}.$$ The numerator will be a polynomial that has terms like $\alpha^n$, $\beta^n$, $\beta^{n-1}$, and $\beta^{n-2}$. The denominator will have polynomial terms like $\alpha^{n-1}$, $\beta^{n-1}$, $\beta^{n-2}$, and $\beta^{n-3}$. Then you can compute the limit of this ratio, which will depend on the "coefficients" $a_j c(w_j)$.
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HINT:
It is not that hard to find the $n$-th power of a matrix given in standard Jordan form. In this case $$T^n = \left[\begin{matrix} \alpha^n & 0 & 0 & 0 \\ 0 & \beta^n & n \beta^{n-1} & \binom{n}{2} \beta^{n-2} \\ 0& 0 & \beta^n & n \beta^{n-1}\\ 0& 0& 0& \beta^{n-1} \end{matrix} \right ] $$
Now observe that $T^n = \alpha^n \cdot S_n$ where the sequence of matrices $(S_n)$ has limit $$\lim S_n = \left[\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{matrix} \right ] $$
The expression for which you consider the limit is defined only for vectors with first component $\ne 0$. For such vectors, the limit equals $\alpha$.
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Note that the matrix for $T$ is only similar to, not necessarily equal to, the given Jordan matrix. – amd Mar 30 '18 at 00:34
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@amd: Thank you, you are right! I have to think how to fix the proof. – orangeskid Mar 30 '18 at 08:25
