General Canonical Forms (12.3.25-28)

Dummit and Foote Abstract Algebra, section 12.3, exercises 25-28:

25. Determine the Jordan canonical form of the $n \times n$ matrix over $\mathbb{Q}$ whose entries are all equal to $1$.
26. Determine the Jordan canonical form of the $n \times n$ matrix over $\mathbb{F}_p$ whose entries are all equal to $1$.
27. Determine the Jordan canonical form of the $n \times n$ matrix over $\mathbb{Q}$ whose diagonal entries are all equal to $0$ and other entries are all $1$.
28. Determine the Jordan canonical form of the $n \times n$ matrix over $\mathbb{F}_p$ whose diagonal entries are all equal to $0$ and other entries are all $1$.

Proof: When $n=1$ all these matrices are already in Jordan canonical form, so assume $n > 1$.

(25) Let $B=A-n$ and $C=AB$. We see$$c_{ij}=\sum_{k=1}^n a_{ik}b_{kj}=\sum_{k=1}^nb_{kj}=b_{jj}+\sum_{k=1,k \neq j}b_{kj}=1-n+n-1=0$$so now $m_A(x) \mid x(x-n)$ and since $A,B \neq 0$ we have $m_A(x)=x(x-n)$ and $A$ is diagonalizable with $n$s and $0$s.

Assume there are multiple $n$s down the diagonal of the Jordan form of $A$, so there are multiple $x-n$ factors in the invariant decomposition. Letting $1_a,1_b∈\mathbb{Q}^n$ be the $\mathbb{Q}[x]$ generators of these summands we see $x(1_a)$ and $x(1_b)$ are linearly independent over $\mathbb{Q}$. But we see $A \begin{bmatrix}f_1 \\ ... \\ f_n \end{bmatrix}=\begin{bmatrix}\sum f_i \\ ... \\ \sum f_i \end{bmatrix}$ so that all elements under the image of $A$ are either zero or associate to $\begin{bmatrix}1 \\ ... \\ 1 \end{bmatrix}$, a contradiction. Therefore the Jordan canonical form of this matrix is the $n \times n$ matrix with $a_{1,1}=n$ and all other entries $0$.

(26) Assume $p \not \mid n$. Then as before we can conclude $m_A(x)=x(x-n)$ and that there is only one $x-n$ invariant factor and the Jordan canonical form is as above. Assume $p \mid n$. Then $x=x-n$ and since $A \neq 0$ we have $m_A(x)=x^2$. Once again we see that assuming there is more than invariant factor not dividing $x$ leads to linearly independent elements in the image of $A$, so that we may conclude the invariant decomposition is $x,...,x,x^2$ and the Jordan canonical form is simply the matrix with $a_{1,2}=1$ and all other entries $0$.

(27) Letting $A$ be this matrix and $A'$ the matrix of (25), we see $A+1=A'$ and thus $((x+1)-n)(x+1)=(x-n+1)(x+1)=0$. As before we note there is one associativity class in the image of $A'=A+1$ so that there is exactly one invariant factor of $x-n+1$ in the decomposition, and so the Jordan canonical form is the matrix with $a_{1,1}=n-1$, the other diagonal entries $-1$ and the other entries $0$.

(28) When $p \not \mid n$ we have $n \not ≡ 0$ and so $x-n+1 \neq x+1$ and the Jordan canonical form is as above. When $p \mid n$ we again notice there is one invariant factor of $(x+1)^2$ and thus the Jordan canonical form is the matrix with $a_{1,2}=1$, the diagonal entries all $-1$ and the other entries $0$.$~\square$