Let \(M\) be the \(n\times n\) matrix over a field \(F\) all of whose entries are \(1\).
Find the characteristic polynomial of \(M\).
Is \(M\) diagonalizable?
Find the Jordan canonical form of \(M\), and discuss the extent to which the Jordan form depends on the characteristic of \(F\).
Solution
Let \[ u= \begin{pmatrix} 1\\ \vdots\\ 1 \end{pmatrix} \in F^n, \] and write \[ \overline{n}=n\cdot1_F\in F. \]
1. For every \(v=(v_1,\ldots,v_n)^T\in F^n\),\[ Mv=(v_1+\cdots+v_n)u. \]Consequently,\[ \operatorname{rank}M=1 \]and\[ M^2=\overline{n}M. \]
Every row of \(M\) consists entirely of \(1\)’s, so each coordinate of \(Mv\) is \(v_1+\cdots+v_n\), giving the first formula. The image is therefore the line \(Fu\), and \(u\neq0\), so the rank is \(1\).
Also \[ Mu=\overline{n}u. \] Applying \(M\) to the first formula gives \[ M^2v =(v_1+\cdots+v_n)Mu = \overline{n}Mv. \]
2. The kernel of \(M\) is the hyperplane\[ H = \left\{ v\in F^n: v_1+\cdots+v_n=0 \right\}, \]which has dimension \(n-1\).
By step 1, \(Mv=0\) exactly when \(v_1+\cdots+v_n=0\). The defining linear functional is nonzero, so its kernel has codimension \(1\).
3. The characteristic polynomial of \(M\) is\[ \boxed{ \chi_M(t)=t^{n-1}(t-\overline{n}) }. \]
Choose a basis \(h_1,\ldots,h_{n-1}\) of \(H\) and append a vector \(w\notin H\). Step 2 shows that \(M\) vanishes on all \(h_i\), so in this basis the first \(n-1\) columns have only zeros.
Modulo \(H\), the action of \(M\) is multiplication by \(\overline{n}\). Indeed, if \(s(v)=v_1+\cdots+v_n\), then \[ s(Mv)=\overline{n}s(v). \] Thus the matrix of \(M\) in this basis is upper triangular with diagonal entries \[ 0,\ldots,0,\overline{n}. \] Its characteristic polynomial is therefore the displayed one.
4. If \(\overline{n}\neq0\), then \(M\) is diagonalizable and itsJordan form is\[ \boxed{ \operatorname{diag} (\overline{n},0,\ldots,0) }. \]
By step 2, \(M\) vanishes on the \((n-1)\)-dimensional space \(H\). Step 1 gives \[ Mu=\overline{n}u. \] Since \(\overline{n}\neq0\), \[ u\notin H, \] because the sum of the coordinates of \(u\) is \(\overline{n}\). Therefore \[ F^n=H\oplus Fu. \] Relative to a basis of \(H\) followed by \(u\), the matrix is diagonal with eigenvalues \(0,\ldots,0,\overline{n}\).
5. If \(\overline{n}=0\), then \(M\) is not diagonalizable.
Step 1 gives \[ M^2=0, \] while \(M\neq0\) because all of its entries are \(1\). Thus \(M\) is a nonzero nilpotent matrix. A diagonalizable nilpotent matrix must be the zero matrix, so \(M\) is not diagonalizable.
6. If \(\overline{n}=0\), the Jordan form of \(M\) is\[ \boxed{ J_2(0)\oplus 0\oplus\cdots\oplus0 }, \]with one block of size \(2\) and \(n-2\) blocks of size \(1\).
In this case step 5 gives \(M^2=0\), so every nilpotent Jordan block has size at most \(2\). Step 1 gives \[ \operatorname{rank}M=1. \] Each size-\(2\) nilpotent Jordan block contributes rank \(1\), while each size-\(1\) zero block contributes rank \(0\). Hence there is exactly one size-\(2\) block. The remaining \(n-2\) dimensions are size-\(1\) zero blocks.
7. The dependence on the characteristic is exactly whether\(\overline{n}\) vanishes:\[ \boxed{ \overline{n}=0 \quad\Longleftrightarrow\quad \operatorname{char}F=p>0\text{ for some prime }p\mid n }. \]
The scalar \(\overline{n}\) vanishes precisely when \(F\) has positive characteristic \(p\) dividing the integer \(n\). If it does not vanish, step 4 gives the diagonal Jordan form. If it vanishes, steps 5 and 6 give the nontrivial nilpotent Jordan form.
8. Q.E.D.
Step 3 answers part 1, steps 4 and 5 answer part 2, and steps 4, 6, and 7 answer part 3.