Let \(n\) be a positive integer and let \(B\) denote the \(n\times n\) matrix over \(\Bbb C\) such that every entry is 1. Find the Jordan normal form of \(B\).
Solution
Let \[ \mathbf 1=(1,\ldots,1)^t\in\CC^n. \]
1. The image of \(B\) is the line \(\CC\mathbf 1\).
For \(v=(v_1,\ldots,v_n)^t\), \[ Bv=\left(\sum_{i=1}^n v_i\right)\mathbf 1. \] Hence every image vector is a multiple of \(\mathbf1\), and \(B\mathbf1=n\mathbf1\ne0\), so the image is exactly that line.
2. The eigenvalues are \(n\) with multiplicity \(1\) and \(0\) with multiplicity \(n-1\).
By 1, \[ B\mathbf1=n\mathbf1, \] so \(n\) is an eigenvalue. Also \[ \ker B=\left\{v:\sum_i v_i=0\right\} \] has dimension \(n-1\), so \(0\) is an eigenvalue with geometric multiplicity \(n-1\).
3. The matrix \(B\) is diagonalizable.
A direct multiplication gives \[ B^2=nB. \] Thus the minimal polynomial divides \[ x(x-n). \] Since \(n\ne0\) in \(\CC\), these two roots are distinct. Therefore the minimal polynomial is squarefree, so \(B\) is diagonalizable.
4. Its Jordan normal form is\[ J=\operatorname{diag}(n,0,\ldots,0). \]
By 2 the eigenvalue multiplicities are \(1\) and \(n-1\), and by 3 every Jordan block has size \(1\). Hence the displayed diagonal matrix is the Jordan form.