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    Determinant, characteristic polynomial, and minimal polynomial of an invariant direct sum

    E-HK-67-4

    Area
    Algebra
    Topics
    Determinants, Invariant Direct Sums, Linear Algebra
    Status
    draft

    Let \(T\) be a linear operator on \(V\). Suppose \(V = W_1 \oplus \cdots \oplus W_k\), where each \(W_i\) is invariant under \(T\). Let \(T_i\) be the induced (restriction) operator on \(W_i\).

    1. Prove that \(\det(T) = \det(T_1) \cdots \det(T_k)\).

    2. Prove that the characteristic polynomial for \(T\) is the product of the characteristic polynomials for \(T_1, \ldots, T_k\).

    3. Prove that the minimal polynomial for \(T\) is the least common multiple of the minimal polynomials for \(T_1, \ldots, T_k\). (Hint: Prove and then use the corresponding facts about direct sums of matrices.)

    Solution

    Choose an ordered basis \(\mathfrak B_i\) of each \(W_i\) and concatenate them to an ordered basis \(\mathfrak B\) of \(V\). Since every \(W_i\) is \(T\)-invariant, the matrix of \(T\) in this basis is block diagonal: \[ [T]_{\mathfrak B} =\operatorname{diag}([T_1]_{\mathfrak B_1},\ldots,[T_k]_{\mathfrak B_k}). \]

    1. Determinants multiply across the invariant direct sum.

    The determinant of a block diagonal matrix is the product of the determinants of its diagonal blocks [@HK71]. Therefore \[ \boxed{\det T=\prod_{i=1}^k\det T_i}. \]

    2. Characteristic polynomials multiply across the invariant direct sum.

    For an indeterminate \(x\), \[ xI-[T]_{\mathfrak B} =\operatorname{diag} \bigl(xI-[T_1]_{\mathfrak B_1},\ldots, xI-[T_k]_{\mathfrak B_k}\bigr). \] Taking determinants and using step 1 for the block matrix gives \[ \chi_T(x) =\prod_{i=1}^k\det(xI-[T_i]) =\boxed{\prod_{i=1}^k\chi_{T_i}(x)}. \]

    3. The minimal polynomial is the least common multiple of the block minimal polynomials.

    Let \(m_i\) be the minimal polynomial of \(T_i\), and let \[ \ell=\operatorname{lcm}(m_1,\ldots,m_k), \] taken monic. Since each \(m_i\) divides \(\ell\), \[ \ell(T_i)=0 \] for every \(i\). Polynomial evaluation preserves the block diagonal form, so \[ \ell(T) =\operatorname{diag}(\ell(T_1),\ldots,\ell(T_k))=0. \] Hence the minimal polynomial \(m_T\) divides \(\ell\).

    Conversely, if a polynomial \(p\) satisfies \(p(T)=0\), then restricting this identity to the invariant subspace \(W_i\) gives \[ p(T_i)=0. \] By the defining property of the minimal polynomial, \(m_i\) divides \(p\) for every \(i\). Therefore their least common multiple \(\ell\) divides \(p\). Applying this to \(p=m_T\) gives \(\ell\mid m_T\).

    Both \(m_T\) and \(\ell\) are monic, and each divides the other, so \[ \boxed{m_T=\operatorname{lcm}(m_1,\ldots,m_k)}. \]

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