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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?

Edwards
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2 Answers2

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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" $a_j c(w_j)$.

angryavian
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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 $\alpha$.

orangeskid
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