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1.3

  1. Let \(A\in M_{n}(\mathbb{C})\).

    (a) Let \(\chi_A(x)\) be the characteristic polynomial of \(A\). Prove that the constant coefficient of \(\chi_A(x)\) is given by \((-1)^n \det(A)\) and the coefficient of the \((n-1)\)-th term is given by \(-\text{trace}(A)\).

    Proof

    \(\chi_{A}(0)=\det(0\text{Id}-A)=\det\left(-A\right)=\left(-1\right)^{n}\det A\).

    For the second one, use induction.

    Base case: \(n=2\)

    Then \(\chi_{A}(x)=\det\left(x\text{Id}-A\right)=\left(x-a\right)\left(x-c\right)-bc=x^2-\left(a+c\right)x+ac-bc\)

    Then it's true

    Inductive step: Suppose it's true for \(n\), that is \(c_{n-1}=-\left(a_{11}+...+a_{nn}\right)\), then for \(n+1\)

    Then \(\chi_{A}(x)=\det\left(x\text{Id}-A\right)=\det\begin{pmatrix}x-a_{11} & a_{12} & \cdots & a_{1n} & a_{1\left(n+1\right)}\\ a_{21} & x-a_{22} & \cdots & a_{2n} & a_{2\left(n+1\right)}\\ a_{31} & a_{32} & \cdots & a_{3n} & a_{3\left(n+1\right)}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & x-a_{nn} & a_{n\left(n+1\right)}\\ a_{\left(n+1\right)1} & a_{\left(n+1\right)2} & \cdots & a_{\left(n+1\right)n} & x-a_{\left(n+1\right)\left(n+1\right)}\end{pmatrix}\)

    Then we choose the first row, \(\det_{n+1}=\left(-1\right)^{n+1+n+1}\left(x-a_{\left(n+1\right)\left(n+1\right)}\right)\det_{n}=\left(x-a_{\left(n+1\right)\left(n+1\right)}\right)\det_{n}\)

    Then the n-th term of \(n+1\) comes from the sum of \(1\cdot\)(n-1 term coefficient of \(n\)) and \(a_{(n+1)(n+1)}\)

    Thus it's equal to \(-\text{trace}(A)\)

    (b) Prove that if \(c_1, \ldots, c_n\) are the eigenvalues of \(A\) (where some may be repeated), then \(\det(A) = c_1 \cdots c_n\) and \(\text{trace}(A) = c_1 + \cdots + c_n\).

    \(\chi_{A}(x)=\det\left(x\text{Id}-A\right)=\left(x-c_1\right)\cdot\ldots\cdot\left(x-c_{n}\right)\)

    When \(x=0\), \(\det-A=(-1)^{n}\det A=\left(-1\right)^{n}c_1\cdot\ldots\cdot c_{n}\)​

    Then \(\det A=c_1\cdot ...\cdot c_n\)

    Since we have \(n\) eigenvalues, then we have corresponding \(n\) eigenvectors, then the matrix can be changed into diagonal by \(P\)​

    Then \(A\) is similar to diagonal matrix, then trace\(A\) is equal to trace diagonal matrix.

  1. Let \(A \in M_n(\mathbb{F})\). Prove that \(A\) and \(A^T\) have the same eigenvalues.

    Consider \(\chi_{A}(x)=\det\left(x\text{Id}-A\right)=0\) and \(\chi_{A^{T}}(x)=\det\left(x\text{Id}-A^{T}\right)=\det\left(x\text{Id}^{T}-A^{T}\right)=\det\left(x\text{Id}-A\right)^{T}=0\)​

    Give an example where the eigenvectors are different.

  1. Suppose \(R, T \in L(\mathbb{F}^3, \mathbb{F}^3)\) each have \(2, 6, 7\) as eigenvalues. Prove that there exists an invertible operator \(S \in L(\mathbb{F}^3, \mathbb{F}^3)\) such that \(R = S^{-1} T S\).

    Since both \(R\) and \(T\) have 3 distinct eigenvalues, they are diagonalizable i.e. there exists invertible matrices \(P, Q\) such that \(R = PDP^{-1}\) and \(T = QDQ^{-1}\), where \(D = \text{diag}(2, 3, 5)\). Substituting one equation into the other, we get that

    \(R = P(Q^{-1} T Q)P^{-1} = (PQ^{-1})T(PQ^{-1})^{-1}.\)Thus, \(S = PQ^{-1}\) is the invertible matrix you're after.