0
$\begingroup$

How to compute the Jordan canonical form for the $n \times n$ matrix over $\mathbb{Q}$ whose entries equals to $1$.

3 Answers 3

2

I will call your matrix $A$.

Observe that the dimension of the null space of $A$ is $n-1$(Why?). So you know $n-1$ linearly independent eigenvectors (whose associated eigenvalue is zero). Further, the vector which has all co-ordinates equal to $1$ is clearly an eigenvector for $A$ (associated eigenvalue being $n$).

Can you fill in the gaps and guess the Jordan canonical form?

  • 0
    Oh yes! the characteristic polynomial turns out to be $c_A (x) = x^{n - 1} (x - n)$. Thanks a lot!2012-11-27
1

The characteristic equation is $x^{n-1}(x-n)$. There are $n-1$ Jordan blocks for eigenvalue 0 and only one for $n$.

hence Jordan canonical form is: $ [n,0,0,...0;0 0 0 ,...,0;...;0 0 0 ,...0] $

  • 1
    What about the possibility of $1$s on the super-diagonal?2012-11-30
0

As everyone says above, the characteristic polynomial is $x^{n-1}(x-n)$, but the key to the answer is to show that the minimal polynomial is $x(x-n)$ (You can do this by showing that $A(A-n\cdot I_n)=0$). Conclude that

coker$(xI-A)\cong \frac{\mathbb{Q}[x]}{(x)}\oplus \cdots \oplus \frac{\mathbb{Q}[x]}{(x)}\oplus \frac{\mathbb{Q}[x]}{(x-n)}$

as a $\mathbb{Q}[x]$-module. This ensures that the Jordan blocks are of size $1$ and there are no $1$'s on the superdiagonal.