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 every entry is equal to 1.


Let A = [1] be an n \times n matrix. Note the following.

A(A-nI)  =  [1][1-\delta_{i,j}n]
 =  [\sum_k 1_{i,k}(1_{k,j} - \delta_{k,j}n)]
 =  [\sum_k 1 - \sum_k \delta_{k,j} n]
 =  [n-n]
 =  [0]

So we have A(A-nI) = 0, and thus the minimal polynomial of A divides x(x-n). Since A \neq 0 and A \neq nI, the minimal polynomial of A is x(x-n).

Next we prove a lemma.

Lemma: Let A be an n \times n matrix over a field F containing the eigenvalues of A, and let \lambda be an eigenvalue of A with eigenspace W \subseteq F^n. If the minimal polynomial of A over F is a product of distinct linear factors, then the dimension of W (over F) is precisely the multiplicity of \lambda among the roots of the characteristic polynomial of A. Proof: Let V = F^n. As an F[x] module via A, we have V = \bigoplus_{i=1}^n F[x]/(x-\lambda_i), where \lambda_i are the eigenvalues of A. The eigenspace of \lambda consists precisely of those direct factors with \lambda_i = \lambda, each of which has dimension 1 over F. \square

For our given matrix A, we claim that the eigenspace W of the eigenvalue n has dimension 1 over F. To see this, let v = [v_1\ \ldots\ v_n]^\mathsf{T} and suppose Av = nv. Evidently, every entry of Av is \sum_k v_k, and so we have nv_i = nv_j for all i,j. Since n \neq 0, we have v_i = v_j for all i,j, and so v = v_1[1\ \ldots\ 1]^\mathsf{T}. So W has a basis containing one element, and so has dimension 1 over F as desired.

By the lemma, n has multiplicity 1 among the roots of the characteristic polynomial of A. Since the remaining roots are 0, the invariant factors of A are x n-2 times and x(x-n). The corresponding Jordan canonical form matrix is J = [b_{i,j}], with b_{n,n} = n and b_{i,j} = 0 otherwise.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: