91Ó°ÊÓ

In an inner-product space V, a linear operator T is called normal if it commutes with its adjoint: \(TT^* = T^*T\). The minimal polynomial of T is the monic polynomial of least degree such that when you plug in T, you get the zero operator. Repeated roots in the minimal polynomial mean that a root (eigenvalue) appears more than once in the polynomial.

Let the minimal polynomial of T be given by \(m_T(\lambda) = (\lambda - \lambda_0)^2 p(\lambda)\), for some polynomial p, where λ0 is a repeated root. Note that in this case, both \((\lambda - \lambda_0)\) and \((\lambda - \lambda_0)^2\) must annihilate T (meaning when plugged into T, the resulting operator is the zero operator).

Since \((\lambda - \lambda_0)\) annihilates T, there exists a vector v such that \(Tv - \lambda_0v = 0\). We also know that \((\lambda - \lambda_0)^2\) annihilates T, so \((T - \lambda_0I)^2v = 0\). Now, let's consider the inner product \(\langle (T - \lambda_0I)v, (T - \lambda_0I)v \rangle\). Since T is normal, we can apply properties of the adjoint and get: \[ \langle (T - \lambda_0I)v, (T - \lambda_0I)v \rangle = \langle v, (T^* - \overline{\lambda_0}I)(T - \lambda_0I)v \rangle \] Using \((T - \lambda_0I)v = 0\), we have: \[ \langle 0, (T^* - \overline{\lambda_0}I)(0) \rangle = 0 \] However, this implies that \((T - \lambda_0I)v = 0\) for any vector v, which means the polynomial \((\lambda - \lambda_0)\) would also annihilate T, leading to a contradiction since the minimal polynomial is unique and of the least degree. Thus, the assumption that the minimal polynomial of T has a repeated root is false. Therefore, the minimal polynomial of a normal operator T in an inner-product space V has no repeated roots.