How to compute the Jordan canonical form for the $n \times n$ matrix over $\mathbb{Q}$ whose entries equals to $1$.
Jordan Canonical form of a matrix over rationals whose all entries are 1.
3 Answers
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?
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0Oh yes! the characteristic polynomial turns out to be $c_A (x) = x^{n - 1} (x - n)$. Thanks a lot! – 2012-11-27
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] $
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1What about the possibility of $1$s on the super-diagonal? – 2012-11-30
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.