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10

Homework 10.pdf

  1. (30 points) Let \(A \in M_3(\mathbb{C})\). Consider the matrix \(xI_d - A\) with polynomial entries.

    1. Show that \(\det(xI_d - A)\) is a monic polynomial of degree 3.

      Since \(xI_{d}-A=x\begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}-\begin{pmatrix}a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\end{pmatrix}=\begin{pmatrix}x-a_{11} & -a_{12} & -a_{13}\\ -a_{21} & x-a_{22} & -a_{23}\\ -a_{31} & -a_{32} & x-a_{33}\end{pmatrix}\)

      Then \(\det(xI_{d}-A)=\left(x-a_{11}\right)\left(x-a_{22}\right)\left(x-a_{33}\right)-a_{21}a_{32}a_{13}-a_{31}a_{12}a_{23}-a_{31}\left(x-a_{22}\right)a_{13}-\left(x-a_{11}\right)a_{32}a_{23}-\left(x-a_{33}\right)a_{12}a_{21}\)​

      From above we can see the item of polynomial degree of 3 is \(x^3\) which is a monic polynomial of degree 3. 2. If \(\det(xI_d - A) = (x - c_1)(x - c_2)(x - c_3)\) with \(c_1, c_2, c_3 \in \mathbb{C}\) the roots of \(\det(xI_d - A)\), prove that \(c_1 + c_2 + c_3 = \text{trace}(A)\) and \(c_1c_2c_3 = \det(A)\).

      First, let's focus on the constant item

      From left hand, the constant item is \(-a_{11}a_{22}a_{33}-a_{13}a_{21}a_{32}-a_{12}a_{23}a_{31}+a_{13}a_{22}a_{31}+a_{11}a_{23}a_{32}+a_{12}a_{21}a_{33}\)​

      From the right hand, the constant item is \(-c_1c_2c_3\), then we have \(c_1c_2c_3=a_{11}a_{22}a_{33}+a_{13}a_{21}a_{32}+a_{12}a_{23}a_{31}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}=\det(A)\)​

      Then let's focus on the item of degree 2

      From the left hand, \(-x^2\left(a_{22}+a_{11}+a_{33}\right)\)​

      From the right hand, \(-x^2(c_1+c_2+c_3)\)​

      Then we have \(c_1+c_2+c_3=a_{11}+a_{22}+a_{33}=\text{trace}(A)\)​

  2. (20 points) Find the inverse of the following matrix using the adjoint: \(A = \begin{pmatrix} -2 & 3 & 2 & -6 \\ 0 & 4 & 4 & -5 \\ 5 & -6 & -3 & 2 \\ -3 & 7 & 0 & 0 \end{pmatrix}.\)

    First we calculate \(\det A=\sum_{j=1}^{n}\left(-1\right)^{4+j}a_{4j}\det A\left(4\left|j\right.\right)\) we choose row 4 here

    For \(j=1\), \((-1)^5a_{41}\det A(4|1)=3\cdot\left(3\cdot4\cdot2+4\cdot3\cdot6+6\cdot2\cdot5-6\cdot6\cdot4-3\cdot5\cdot3-2\cdot2\cdot4\right)=-147\)​

    For \(j=2\), \((-1)^6a_{42}\det A(4|2)=7\cdot\left(-2\cdot4\cdot2+0\cdot3\cdot6+5\cdot2\cdot-5+6\cdot5\cdot4+2\cdot5\cdot3-2\cdot2\cdot0\right)=588\)

    Thus \(\det(A)=441\)​

    Then we need the matrix of cofactors \(c_{ij} = (-1)^{i+j} \det(A(i,j)) \rightarrow C = \begin{pmatrix} c_{11} & c_{12} & c_{13} & c_{14} \\ c_{21} & c_{22} & c_{23} & c_{24}\\ c_{31} & c_{32} & c_{33} & c_{34}\\ c_{41} & c_{42} & c_{43} & c_{44} \end{pmatrix}\)

    After calculating, we get \(C=\begin{pmatrix}-49 & -21 & -109 & -104\\ 98 & 42 & 92 & 19\\ 98 & 42 & -97 & -44\\ 49 & 84 & -89 & -4\end{pmatrix}\)​

    Therefore, \(A^{-1}=\frac{1}{\det(A)}\text{adj}(A)=\frac{1}{441}C^{T}=\frac{1}{441}\begin{pmatrix}-49 & 98 & 98 & 49\\ -21 & 42 & 42 & 84\\ -109 & 92 & -97 & -89\\ -104 & 19 & -44 & -4\end{pmatrix}\)

  3. (30 points) Prove by induction that if \(k_1, \dots, k_n \in F\), then:

    \[ \det \begin{pmatrix} 1 + k_1 & k_2 & k_3 & \cdots & k_n \\ k_1 & 1 + k_2 & k_3 & \cdots & k_n \\ k_1 & k_2 & 1 + k_3 & \cdots & k_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ k_1 & k_2 & k_3 & \cdots & 1 + k_n \end{pmatrix} = 1 + k_1 + \cdots + k_n. \]

    Basic step: \(n=1\), then \(\det (1+k_1)=1+k_1\) which is true

    Suppose it is true for \(n\).

    Which is \(\det\begin{pmatrix}1+k_1 & k_2 & k_3 & \cdots & k_{n}\\ k_1 & 1+k_2 & k_3 & \cdots & k_{n}\\ k_1 & k_2 & 1+k_3 & \cdots & k_{n}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ k_1 & k_2 & k_3 & \cdots & 1+k_{n}\end{pmatrix}=\det\begin{pmatrix}1+k_1 & k_2 & k_3 & \cdots & k_{n}\\ -1 & 1 & 0 & \cdots & 0\\ 0 & -1 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & 1\end{pmatrix}=1+k_1+\cdots+k_{n}.\)

    The determinant is same since we do the row operation \(R_{n-1}-R_{n}\ldots,R_1-R_2\)

    Then for \(n+1\).

    Let \(A=\begin{pmatrix}1+k_1 & k_2 & k_3 & \cdots & k_{n+1}\\ k_1 & 1+k_2 & k_3 & \cdots & k_{n+1}\\ k_1 & k_2 & 1+k_3 & \cdots & k_{n+1}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ k_1 & k_2 & k_3 & \cdots & 1+k_{n+1}\end{pmatrix}=\begin{pmatrix}1+k_1 & k_2 & k_3 & \cdots & k_{n+1}\\ -1 & 1 & 0 & \cdots & 0\\ 0 & -1 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & 1\end{pmatrix}\).

    The determinant is same since we do the row operation \(R_{n}-R_{n+1},R_{n-1}-R_{n}\ldots,R_1-R_2\)

    Then we choose row \(n+1\), \(\det A=\sum_{j=1}^{n}\left(-1\right)^{n+1+j}a_{\left(n+1\right)j}\det A\left(n+1\left|j\right.\right)=\left(-1\right)^{n+1+n}\left(-1\right)\det A\left(n+1\left|n\right.\right)+\left(-1\right)^{n+1+n+1}\det A\left(n+1\left|n+1\right.\right)\)

    Then for \(\det(n+1|n)\) we have image, then we choose column n, \(\det(n+1|n)=(-1)^{n+1}k_{n+1}\det(n+1|n)\left(1\left|n\right.\right)=\left(-1\right)^{n+1}k_{n+1}\left(-1\right)^{n-1}=k_{n+1}\) since image is lower triangular matrix, then \(\det(n+1|n)(1|n)\) is the product of diagonal

    Then for \(\det(n+1|n+1)\) we use hypothesis, \(\det(n+1|n+1)=1+k_1+\cdots+k_{n}\)​

    Then \(\det A=1+k_1+\cdots+k_{n}+k_{n+1}\)​

    Thus it is true for all \(n\)​

  4. (20 points) A matrix \(A \in M_n(\mathbb{R})\) is called skew-symmetric if \(A^T = -A\).

    1. Prove that if \(n\) is odd and \(A \in M_n(\mathbb{R})\) is skew-symmetric, then \(\det(A) = 0\).

      Since we know \(\det A=\det A^{T}=\det\left(-A\right)=\left(-1\right)^{n}\det A\), since \(n\) is odd, then \(\det A=-\det A\)​

      Thus \(\det A=0\)

    2. For every even \(n\), find a skew-symmetric matrix \(A \in M_n(\mathbb{R})\) such that \(\det(A) \neq 0\).

      \(A=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix},\det A=1\), but \(A^{T}=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix},-A=\begin{pmatrix}-1 & 0\\ 0 & -1\end{pmatrix}\)​

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