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Previous related question: Jordan normal form powers

Let $A$ be a $n\times n$ Matrix such that $A=PBP^{-1}$ where $B$ is in Jordan normal form with $\lambda_i(k)_j$ Where $i$ is the size, $k$ is the eigenvalue and $j$ the order.

From the previous question I know that each Jordan block $\lambda_i(k)_j$ when the matrix is raised to the $n$-th power is an upper triangular matrix $$\sum_{r=0}^{i-1} {n \choose r} k^{n-r}t^r$$ Where $t$ is the matrix with 1’s on it’s super diagonal and 0’s everywhere else. How can I get this matrix to Jordan normal form?

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  • $\begingroup$ What exactly do you mean by "order"? What is the difference between $\lambda_2(0)_1$ and $\lambda_2(0)_2$? $\endgroup$ Commented Jul 19, 2020 at 12:24
  • $\begingroup$ Some thoughts: I assume that the index $j$ plays no role. Suppose that $k \neq 0$ and $M = \lambda_i(k)$. What we want is an invertible matrix $S$ such that $M^n = S\lambda_{i}(k^n)S^{-1}$. Subtracting $k^n I$ from both sides, what we want is a similarity $S$ such that $$ St S^{-1} = \sum_{r=1}^{i-1} \binom nr k^{n-r} t^r. $$ In other words, we want an invertible $S$ that satisfies the linear equation $$ St = \left[\sum_{r=1}^{i-1} \binom nr k^{n-r} t^r\right]S. $$ $\endgroup$ Commented Jul 19, 2020 at 12:50
  • $\begingroup$ @BenGrossmann, the index is just to represent where in the matrix the block is found. $\endgroup$
    – razivo
    Commented Jul 19, 2020 at 14:05
  • $\begingroup$ Your equation seems to simplify it, and making a diagonal matrix $S*$ Composed out of the $S$’s of each block will get as a change of basis matrix for the entire matrix. $\endgroup$
    – razivo
    Commented Jul 19, 2020 at 14:09
  • $\begingroup$ And, multiplying by $t$ corresponds to adding a column of 0’s on the left and shifting the entries to the right. $\endgroup$
    – razivo
    Commented Jul 19, 2020 at 14:14

1 Answer 1

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So you want to know the Jordan canonical form of the $i \times i$ matrix $$ A = \sum_{r=0}^{i-1} \left( n \atop r \right) k^{n-r} t^r .$$ Since $A$ has $k^n$ as an $i$-fold repeated eigenvalue, it is sufficient to find the Jordan form for $$ A - k^n I = \sum_{r=1}^{i-1} \left( n \atop r \right) k^{n-r} t^r .$$ First consider the case $k \ne 0$. Then $$ (A- k^n I)^{i-1} = n^{i-1} k^{(n-1)(i-1)} t^{i-1} \ne 0$$ since $t^r = 0$ for $r \ge i$. Similarly $(A- k^n I)^i = 0$. Therefore the minimal polynomial for $A$ is $p(x) = (x - k^n)^i$, and its Jordan canonical form must be $k^n I + t$, that is, a single block of size $i$.

Next, consider the case $k = 0$, when $A = t^n$. Denote the unit vectors by $e_r$ with $1 \le r \le i$. Then the unit vectors split into groups:

  • $e_1, e_{n+1}, e_{2n+1}, \dots$ of size $[(i+n-1)/n]$;
  • $e_2, e_{n+2}, e_{2n+2}, \dots$ of size $[(i+n-2)/n]$;
  • $e_3, e_{n+3}, e_{2n+3}, \dots$ of size $[(i+n-3)/n]$;
  • $\vdots$
  • $e_n, e_{2n}, e_{3n}, \dots$ of size $[i/n]$;

where $[x]$ denotes the integer part of $x$. On each group, $A$ acts as a Jordan block. So its Jordan canonical form is a collection of blocks of size $[(i+n-1)/n], [(i+n-2)/n], \dots, [i/n]$. And if you think about it, this is $n - i + n[i/n]$ blocks of size $[i/n]$ and $i - n[i/n]$ blocks of size $[i/n]+1$. (In particular, if $n \ge i$, then it is $i$ blocks of size $1$, that is, $A = 0$ is diagonal.)

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  • $\begingroup$ In the first case, what will be the similarity(change of basis) matrix? $\endgroup$
    – razivo
    Commented Jul 20, 2020 at 8:56
  • $\begingroup$ @razivo If $f_1, f_2,\dots,f_i$ is the basis for the Jordan canonical form, then by computing the range of $A^{r-1}$, you can see that the span of $f_r,\dots,f_i$ is the same as the span of $e_r,\dots,e_i$ for all $1 \le r \le i$. In this manner, you could probably write an explicit similarity transform. I suspect the formula will be quite complicated. But I haven't tried to do it. $\endgroup$ Commented Jul 20, 2020 at 15:20
  • $\begingroup$ @razivo That last comment is wrong. Let me correct it in the next comment. $\endgroup$ Commented Jul 21, 2020 at 0:47
  • $\begingroup$ If $f_1, f_2,\dots,f_i$ is the basis for the Jordan canonical form, then by computing the kernel of $A^r$, we see that the span of $f_1,\dots,f_r$ is the same as the span of $e_1,\dots,e_r$ for all $1 \le r \le i$. In this manner, you could probably write an explicit similarity transform. I suspect the formula will be quite complicated. But I haven't tried to do it. $\endgroup$ Commented Jul 21, 2020 at 0:49

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