Let \(A\) be an \(n\times n\) complex matrix with characteristic polynomial \(\chi\) and minimal polynomial \(\mu\). Suppose \[ \chi(x)=\mu(x)(x-i), \qquad \mu(x)^2=\chi(x)(x^2+1). \] Determine the Jordan canonical form of \(A\).
Solution
1. The minimal polynomial is\[ \mu(x)=(x-i)^2(x+i). \]
Substitute \[ \chi(x)=\mu(x)(x-i) \] into the second hypothesis: \[ \mu(x)^2 = \mu(x)(x-i)(x^2+1). \] The minimal polynomial is nonzero, and \(\CC[x]\) is an integral domain, so cancellation gives \[ \mu(x) = (x-i)(x^2+1). \] Since \[ x^2+1=(x-i)(x+i), \] we obtain \[ \mu(x)=(x-i)^2(x+i). \]
2. The characteristic polynomial is\[ \chi(x)=(x-i)^3(x+i). \]
Using the first hypothesis and step 1, \[ \begin{aligned} \chi(x) &= \mu(x)(x-i)\\ &= (x-i)^3(x+i). \end{aligned} \]
3. The Jordan blocks for the eigenvalue \(i\) have sizes \(2\) and \(1\).
For an eigenvalue \(\lambda\), its exponent in the characteristic polynomial is the sum of the sizes of its Jordan blocks, while its exponent in the minimal polynomial is the size of its largest Jordan block. By step 2, the total size of the \(i\)-blocks is \(3\). By step 1, their largest size is \(2\). The only partition of \(3\) with largest part exactly \(2\) is \[ 3=2+1. \] Thus the \(i\)-blocks have sizes \(2\) and \(1\).
4. The eigenvalue \(-i\) has one Jordan block of size \(1\).
Step 2 shows that \(-i\) has algebraic multiplicity \(1\). Hence its Jordan blocks have total size \(1\), so there is exactly one such block and it has size \(1\).
5. Therefore the Jordan canonical form is\[ \boxed{ \begin{pmatrix} i&1&0&0\\ 0&i&0&0\\ 0&0&i&0\\ 0&0&0&-i \end{pmatrix} }. \]
Step 3 gives one block \[ J_2(i)= \begin{pmatrix} i&1\\ 0&i \end{pmatrix} \] and one \(1\times1\) block \((i)\). Step 4 gives one \(1\times1\) block \((-i)\). Their direct sum is the displayed matrix, up to permutation of Jordan blocks.
6. Q.E.D.
Step 5 is the requested Jordan canonical form.