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.

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Comments

  • 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?

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