## Compute the Jordan canonical form of a given matrix over QQ

Compute the Jordan canonical form over $\mathbb{Q}$ of the $n \times n$ matrix whose diagonal entries are 0 and whose off-diagonal entries are 1.

Let $A = [1 - \delta_{i,j}]$. Note the following.

 $(A - (n-1)I)(A + I)$ = $[1 - \delta_{i,j} - \delta_{i,j}(n-1)][1 - \delta_{i,j} + \delta_{i,j}]$ = $[1-\delta_{i,j}n][1]$ = $[\sum_k (1 - \delta_{i,k}n)1]$ = $[\sum_k 1 - \sum_k \delta_{i,k}n]$ = $[n-n]$ = $[0]$

So the minimal polynomial of $A$ divides $(x-(n-1))(x+1)$. Since $A \neq -I$ and $A \neq (n-1)I$, in fact this is the minimal polynomial of $A$.

Suppose now that $v = [v_1\ \ldots\ v_n]$ is an eigenvector with eigenvalue $n-1$. Evidently then for each $i$, we have $\sum_{k \neq i} v_k = (n-1)v_i$, and thus $\sum_k v_k = nv_i$. Since $n \neq 0$, we have $v_i = v_j$ for all $i,j$, so that $v = v_1[1\ \ldots\ 1]^\mathsf{T}$. In particular, the eigenspace with eigenvalue $n-1$ has dimension 1. By the lemma we proved in this previous exercise, $n-1$ has multiplicity 1 in the characteristic polynomial of $A$. So the elementary divisors of $A$ are $x+1$ $n-1$ times and $x-(n-1)$. The Jordan canonical form of $A$ is thus $J = [b_{i,j}]$, where $b_{i,j} = n-1$ if $i=j=n$, $-1$ if $i = j \neq n$, and 0 otherwise.

• newsun157  On December 13, 2011 at 8:28 pm

I need the solutions of some problems in Section 12.3 and 13.3. Can you post is?

Thanks in andvance.

Newsun.

• newsun157  On December 13, 2011 at 8:33 pm

It is eaxactly

Problem 22,23,24,25 in Section 12.2

Problem 4,5,8,12,13, 17,18,19 in Section 12.3

• nbloomf  On December 14, 2011 at 4:53 pm

Did you try any of them?