Let \(A,B\) be nonsimilar complex \(n\times n\) matrices with the same minimal polynomial and the same characteristic polynomial. Show that \[ n\ge4 \] and that the minimal polynomial is not equal to the characteristic polynomial.
Solution
For an eigenvalue \(\lambda\), let \[ m_\lambda \] denote its algebraic multiplicity and let \[ s_\lambda \] denote the size of the largest Jordan block for \(\lambda\).
1. The characteristic polynomial determines every\(m_\lambda\), and the minimal polynomial determines every\(s_\lambda\).
In Jordan normal form, the exponent of \[ t-\lambda \] in the characteristic polynomial is the sum of the sizes of all Jordan blocks for \(\lambda\), which is \(m_\lambda\). The exponent of \(t-\lambda\) in the minimal polynomial is the largest such block size, which is \(s_\lambda\).
2. If\[ m_\lambda\leq3, \]then the pair \((m_\lambda,s_\lambda)\) uniquely determines the multiset ofJordan-block sizes for \(\lambda\).
The possible partitions are \[ \begin{array}{c|c} m_\lambda&\text{partitions by block sizes}\\ \hline 1&(1)\\ 2&(2),\ (1,1)\\ 3&(3),\ (2,1),\ (1,1,1). \end{array} \] For each fixed value of \(m_\lambda\) in this table, the possible partitions have distinct largest parts. Thus specifying \(s_\lambda\) selects exactly one partition.
3. If \(n\leq3\), then two complex \(n\times n\) matrices with the samecharacteristic polynomial and the same minimal polynomial are similar.
For every eigenvalue, \[ m_\lambda\leq n\leq3. \] By step 1, the common characteristic and minimal polynomials give the same pair \((m_\lambda,s_\lambda)\) for the two matrices. Step 2 then gives the same Jordan-block partition for every eigenvalue. Hence the two Jordan normal forms are identical up to reordering blocks, so the matrices are similar.
4. Under the hypotheses of the problem,\[ \boxed{n\geq4}. \]
The matrices \(A\) and \(B\) are assumed nonsimilar. Step 3 rules out \(n\leq3\).
5. If a complex matrix has equal minimal and characteristicpolynomials, then for each eigenvalue it has exactly one Jordan block.
Let \(\lambda\) be an eigenvalue. Equality of the two polynomials makes the exponent of \(t-\lambda\) the same in both. By step 1, this says \[ s_\lambda=m_\lambda. \] The largest Jordan block for \(\lambda\) therefore already has size equal to the total size of all Jordan blocks for \(\lambda\). No second nonempty block can occur.
6. Two complex matrices with the same characteristic polynomial, eachof which also has minimal polynomial equal to that characteristicpolynomial, are similar.
The common characteristic polynomial fixes every eigenvalue and its algebraic multiplicity \(m_\lambda\). By step 5, each matrix has exactly one Jordan block \[ J_{m_\lambda}(\lambda) \] for each eigenvalue. Thus their Jordan normal forms are the same up to block order, so the matrices are similar.
7. Under the hypotheses of the problem, the common minimal polynomialis not equal to the common characteristic polynomial.
If they were equal, step 6 would imply that \(A\) and \(B\) are similar, contrary to hypothesis.
8. Q.E.D.
Step 4 proves \(n\geq4\), and step 7 proves that the minimal and characteristic polynomials are unequal.