Let \(A\) be the complex matrix \[ \begin{bmatrix} 2 & 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 & 0 & 0 \\ -1 & 0 & 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 2 & 0 & 0 \\ 1 & 1 & 1 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 1 & -1 \end{bmatrix}. \] Find the Jordan form for \(A\).
Solution
A direct determinant computation gives \[ \chi_A(x)=(x-2)^5(x+1). \] Moreover \[ m_A(x)=(x-2)^4(x+1). \] Thus the eigenvalue \(2\) has algebraic multiplicity five and largest Jordan block size four, while \(-1\) has multiplicity one [@HK71].
To determine the remaining \(2\)-block structure, compute \[ \dim\ker(A-2I)=2. \] The dimension of this eigenspace equals the number of Jordan blocks for the eigenvalue \(2\). Hence there are exactly two such blocks. Their sizes add to five and the largest has size four, so the sizes are \(4\) and \(1\).
Therefore, up to permutation of blocks, \[ \boxed{J(A)=J_4(2)\oplus J_1(2)\oplus J_1(-1).} \] Explicitly, one possible Jordan form is \[ \begin{pmatrix} 2&1&0&0&0&0\\ 0&2&1&0&0&0\\ 0&0&2&1&0&0\\ 0&0&0&2&0&0\\ 0&0&0&0&2&0\\ 0&0&0&0&0&-1 \end{pmatrix}. \]