Let \(A\) be a \(4\times4\) matrix over a field of characteristic not equal to \(2\) with \[ \chi_A(x)=(x-1)^2(x+1)^2 \] and minimal polynomial \[ m_A(x)=(x-1)(x+1)^2. \] Determine the Jordan canonical form and rational canonical form of \(A\).
Solution
1. The Jordan blocks for the eigenvalue \(1\) are two \(1\times1\) blocks.
The algebraic multiplicity of \(1\) is \(2\). The exponent of \(x-1\) in the minimal polynomial is \(1\), so the largest Jordan block for \(1\) has size \(1\). Therefore both units of algebraic multiplicity occur as separate \(1\times1\) blocks.
2. The Jordan blocks for the eigenvalue \(-1\) consist of one \(2\times2\) block.
The algebraic multiplicity of \(-1\) is \(2\), while the exponent of \(x+1\) in the minimal polynomial is \(2\). Hence there must be a block of size \(2\), which exhausts the full algebraic multiplicity.
3. Thus\[ J(A)=J_2(-1)\oplus[1]\oplus[1]. \]
Combine 1 and 2.
4. The invariant factors are\[ d_1=x-1, \qquad d_2=(x-1)(x+1)^2. \]
The largest invariant factor is the minimal polynomial. Their product is the characteristic polynomial, so \[ d_1=\frac{\chi_A}{d_2}=x-1. \] The divisibility condition \(d_1\mid d_2\) holds.
5. Therefore the rational canonical form is\[ C(x-1)\oplus C((x-1)(x+1)^2). \]
Rational canonical form is the direct sum of the companion matrices of the invariant factors. Since \[ (x-1)(x+1)^2=x^3+x^2-x-1, \] one standard companion-matrix convention gives \[ \begin{pmatrix} 0&0&1\\ 1&0&1\\ 0&1&-1 \end{pmatrix} \] for the cubic factor, together with the \(1\times1\) block \([1]\).