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

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up vote 2 down vote accepted

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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Thanks a lot! I hope the J C F might then have $1$ as first element and rest are $0$. – sum410 Nov 26 '12 at 7:09
How is $1$ a eigenvalue?! As I have stated, the eigenvalues of $A$ are $0$ (multiplicity is $n-1$) and $n$. – Isomorphism Nov 26 '12 at 7:22
Oh yes! the characteristic polynomial turns out to be $c_A (x) = x^{n - 1} (x - n)$. Thanks a lot! – sum410 Nov 27 '12 at 4:33

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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What about the possibility of $1$s on the super-diagonal? – robjohn Nov 30 '12 at 19:08

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.

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