# Jordan normal form of powers of Jordan normal form

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?

• What exactly do you mean by "order"? What is the difference between $\lambda_2(0)_1$ and $\lambda_2(0)_2$? Jul 19 '20 at 12:24
• 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 $$St = \left[\sum_{r=1}^{i-1} \binom nr k^{n-r} t^r\right]S.$$ Jul 19 '20 at 12:50
• @BenGrossmann, the index is just to represent where in the matrix the block is found. Jul 19 '20 at 14:05
• 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. Jul 19 '20 at 14:09
• And, multiplying by $t$ corresponds to adding a column of 0’s on the left and shifting the entries to the right. Jul 19 '20 at 14:14

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.)

• In the first case, what will be the similarity(change of basis) matrix? Jul 20 '20 at 8:56
• @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. Jul 20 '20 at 15:20
• @razivo That last comment is wrong. Let me correct it in the next comment. Jul 21 '20 at 0:47
• 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. Jul 21 '20 at 0:49