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Homework 10

  1. (20 pts) Using properties of determinants, prove that \(\det \begin{pmatrix}1 & x_1 & x_1^2 & x_1^3 \\1 & x_2 & x_2^2 & x_2^3 \\1 & x_3 & x_3^2 & x_3^3 \\1 & x_4 & x_4^2 & x_4^3\end{pmatrix} = (x_4 - x_1)(x_4 - x_2)(x_4 - x_3)(x_3 - x_1)(x_3 - x_2)(x_2 - x_1).\)
    ​\(\det \begin{pmatrix}1 & x_1 & x_1^2 & x_1^3 \\1 & x_2 & x_2^2 & x_2^3 \\1 & x_3 & x_3^2 & x_3^3 \\1 & x_4 & x_4^2 & x_4^3\end{pmatrix} = \det \begin{pmatrix}1 & x_1 & x_1^2 & x_1^3 \\0 & x_2-x_1 & x_2^2-x_1^2 & x_2^3-x_1^3 \\0 & x_3-x_1 & x_3^2-x_1^2 & x_3^3-x_1^3 \\0 & x_4-x_1 & x_4^2-x_1^2 & x_4^3-x_1^3\end{pmatrix}\)
    ​\(= (x_2 - x_1)(x_3 - x_1)(x_4 - x_1) \det \begin{pmatrix}1 & x_1 & x_1^2 & x_1^3 \\0 & 1 & x_2+x_1 & x_2^2+x_2x_1+x_1^2 \\0 & 1 & x_3+x_1 & x_3^2+x_3x_1+x_1^2 \\0 & 1 & x_4+x_1 & x_4^2+x_4x_1+x_1^2\end{pmatrix}\)
    ​\(= (x_2 - x_1)(x_3 - x_1)(x_4 - x_1) \det \begin{pmatrix}1 & x_1 & x_1^2 & x_1^3 \\0 & 1 & x_2+x_1 & x_2^2+x_2x_1+x_1^2 \\0 & 0 & x_3-x_2 & (x_3-x_2)(x_3+x_2+x_1) \\0 & 0 & x_4-x_2 & (x_4-x_2)(x_4+x_2+x_1)\end{pmatrix}\)
    ​\(= (x_2 - x_1)(x_3 - x_1)(x_4 - x_1)(x_3 - x_2)(x_4 - x_2) \det \begin{pmatrix}1 & x_1 & x_1^2 & x_1^3 \\0 & 1 & x_2+x_1 & x_2^2+x_2x_1+x_1^2 \\0 & 0 & 1 & x_3+x_2+x_1 \\0 & 0 & 1 & x_4+x_2+x_1\end{pmatrix}\)
    ​\(= (x_2 - x_1)(x_3 - x_1)(x_4 - x_1)(x_3 - x_2)(x_4 - x_2) \det \begin{pmatrix}1 & x_1 & x_1^2 & x_1^3 \\0 & 1 & x_2+x_1 & x_2^2+x_2x_1+x_1^2 \\0 & 0 & 1 & x_3+x_2+x_1 \\0 & 0 & 0 & x_4 - x_3\end{pmatrix}\)
    ​\(= (x_2 - x_1)(x_3 - x_1)(x_4 - x_1)(x_3 - x_2)(x_4 - x_2)(x_4 - x_3)\)

    1. (30 pts) Let \(A = \begin{pmatrix} a & 1 & 0 & 1 \\ 1 & a & 1 & 0 \\ 0 & 1 & a & 1 \\ 1 & 0 & 1 & a \end{pmatrix} \in M_4(\mathbb{R})\).

    (a) Find all \(a \in \mathbb{R}\) such that \(A\) is invertible.
    (b) Find \(\operatorname{adj}(A)\).
    (c) Find the inverse of \(A\) for the values of \(a\) found in (a).

    (a) Let \(\det A \neq 0\). Then \(\det A = a \cdot (-1)^{1+1}\det \begin{pmatrix} a & 1 & 0 \\ 1 & a & 1 \\ 0 & 1 & a \end{pmatrix} + 1 \cdot (-1)^{2+1}\det \begin{pmatrix} 1 & 0 & 1 \\ 1 & a & 1 \\ 0 & 1 & a \end{pmatrix} + 1 \cdot (-1)^{4+1}\det \begin{pmatrix} 1 & 0 & 1 \\ a & 1 & 0 \\ 1 & a & 1 \end{pmatrix} = a(a^{3}- 2a) - (a^{2}+ 1 - 1) - (1 + a^{2}- 1) = a^{4}- 2a^{2}- a^{2}- a^{2}= a ^{4}- 4a^{2}\neq 0\).
    ​\(a^2(a^2 - 4) \neq 0 \implies a^2 \neq 0, 4 \implies a \neq 0, \pm 2\). Thus, when \(a \neq 0, \pm 2\), \(A\) is invertible.

    (b) Let's find the cofactor matrix of \(A\) and we know \(\operatorname{adj}(A) = C^T\).
    ​\(C = \begin{pmatrix} (-1)^{1+1}\det(1|1) & (-1)^{1+2}\det(1|2) & (-1)^{1+3}\det(1|3) & (-1)^{1+4}\det(1|4) \\ (-1)^{2+1}\det(2|1) & (-1)^{2+2}\det(2|2) & (-1)^{2+3}\det(2|3) & (-1)^{2+4}\det(2|4) \\ (-1)^{3+1}\det(3|1) & (-1)^{3+2}\det(3|2) & (-1)^{3+3}\det(3|3) & (-1)^{3+4}\det(3|4) \\ (-1)^{4+1}\det(4|1) & (-1)^{4+2}\det(4|2) & (-1)^{4+3}\det(4|3) & (-1)^{4+4}\det(4|4) \end{pmatrix}\)

    \(\det(1|1) = a^{3} - 2a\), \(\det(1|2) = a^{2}\), \(\det(1|3) = 2a\), \(\det(1|4) = a^{2}\)

    \(\det(2|1) = a^{2}\), \(\det(2|2) = a^{3} - 2a\), \(\det(2|3) = a^{2}\), \(\det(2|4) = 2a\)

    \(\det(3|1) = 2a\), \(\det(3|2) = a^{2}\), \(\det(3|3) = a^{3} - 2a\), \(\det(3|4) = a^{2}\)

    \(\det(4|1) = a^{2}\), \(\det(4|2) = 2a\), \(\det(4|3) = a^{2}\), \(\det(4|4) = a^{3} - 2a\)

    \(C = \begin{pmatrix} a^3 - 2a & -a^2 & 2a & -a^2 \\ -a^2 & a^3 - 2a & -a^2 & 2a \\ 2a & -a^2 & a^3 - 2a & -a^2 \\ -a^2 & 2a & -a^2 & a^3 - 2a \end{pmatrix}\)
    Then \(\operatorname{adj}(A) = \begin{pmatrix} a^3 - 2a & -a^2 & 2a & -a^2 \\ -a^2 & a^3 - 2a & -a^2 & 2a \\ 2a & -a^2 & a^3 - 2a & -a^2 \\ -a^2 & 2a & -a^2 & a^3 - 2a \end{pmatrix}\)​

    (c) we know \(A^{-1} = \frac{1}{\det A} \cdot \text{adj} A\)

    \(\det A = a^4 - 4a^2 \neq 0\)

    Then \(A^{-1} = \frac{1}{a^2(a+2)(a-2)} \begin{pmatrix} a^3 - 2a & -a^2 & 2a & -a^2 \\ -a^2 & a^3 - 2a & -a^2 & 2a \\ 2a & -a^2 & a^3 - 2a & -a^2 \\ -a^2 & 2a & -a^2 & a^3 - 2a \end{pmatrix}\)

    \(= \begin{pmatrix} \frac{a^2 - 2}{a^3 - 4a} & -\frac{1}{a^2 - 4} & \frac{2}{a^3 - 4a} & -\frac{1}{a^2 - 4} \\ -\frac{1}{a^2 - 4} & \frac{a^2 - 2}{a^3 - 4a} & -\frac{1}{a^2 - 4} & \frac{2}{a^3 - 4a} \\ \frac{2}{a^3 - 4a} & -\frac{1}{a^2 - 4} & \frac{a^2 - 2}{a^3 - 4a} & -\frac{1}{a^2 - 4} \\ -\frac{1}{a^2 - 4} & \frac{2}{a^3 - 4a} & -\frac{1}{a^2 - 4} & \frac{a^2 - 2}{a^3 - 4a} \end{pmatrix}\)

  3. (30 pts) Prove by induction that if \(a_0, a_1 \ldots, a_{n-1} \in \mathbb{F}\), then

    \(\det \begin{pmatrix}t & 0 & 0 & \cdots & 0 & a_0 \\-1 & t & 0 & \cdots & 0 & a_1 \\0 & -1 & t & \cdots & 0 & a_2 \\\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\0 & 0 & 0 & \cdots & -1 & t + a_{n-1}\end{pmatrix} = t^n + a_{n-1}t^{n-1} + \cdots + a_1 t + a_0.\)

    Use induction

    When \(n=1\), \(\det(t + a_0) = t + a_0\).

    Suppose it is true for \(n\), then for \(n+1\), we have \(\det \begin{pmatrix}t & 0 & 0 & \cdots & 0 & a_0 \\-1 & t & 0 & \cdots & 0 & a_1 \\0 & -1 & t & \cdots & 0 & a_2 \\\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\0 & 0 & 0 & \cdots & t & a_{n-1} \\0 & 0 & 0 & \cdots & -1 & t + a_n \end{pmatrix}\)

    \(= \sum_{i=1}^{n+1} (-1)^{i+1} a_{i1} \det A(i|1) = t \det A(1|1) + (-1)(-1) \cdot \det A(2|1)\)

    Since \(A(2|1)=\begin{pmatrix} 0 & 0 & \cdots & 0 & a_{0} \\ -1 & t & \cdots & 0 & a_{2} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & t & a_{n-1} \\ 0 & 0 & \cdots & -1 & t + a_{n} \end{pmatrix},\) then \(\det A(2|1)=a_{0}\left(-1\right)^{n+1}\det A(2|1)(1|n)=a_{0}(-1)^{n+1+n-1}=a_{0}\).

    And for \(\det(1|1)\), use inductive hypothesis, \(\det(1|1) = t^{n}+ a_{n}t^{n-1}+ \cdots + a_{2}t + a_{1}\).

    Then \(\det=t^{n+1}+a_{n}t^{n}+\cdots+a_{2}t^{2}+a_{1}t+a_{0}\)

    Thus the statement is true for all \(n\)​

  4. (20 pts) Let \(A \in M_n(\mathbb{F})\) be any matrix. Prove that \(\det(\operatorname{adj}(A)) = \det(A)^{n-1}\).

    We know \(\det AI_{n}=A\cdot\operatorname{\mathrm{adj}}A\), then \(\det(\det A I_n) = \det(A \cdot \operatorname{adj} A).\)
    Then \((\det A)^n \cdot 1 = \det A \cdot \det(\operatorname{adj} A),\) then if \(\det A \neq 0\), \(\det(\operatorname{adj} A) = (\det A)^{n-1}.\)

    If \(\det A = 0\) and \(n>1\)​\, then \(0_{n}=A\cdot\operatorname{\mathrm{adj}}A\) and \(\text{rank}A\leq n-1\) since \(\det A=0\)\, then \(\dim \ker A\geq 1\)

    Then at least 1 column\row of \(\text{adj}A\) is in \(\ker A\), then at least 1 column\row of \(\text{adj}A\) is zero.

    Then \(\det(\operatorname{\mathrm{adj}}(A))=0\) since it has a zero column\row.

    If \(n=1\), then \(A=(c)\), \(\det A=c,\det A^{0}=c^{0}=1\). And \(\text{adj}A=(1),\det(\text{adj}A)=1\), thus \(\det(\operatorname{adj}(A)) = \det(A)^{n-1}\)

    Thus \(\det(\operatorname{adj}(A)) = \det(A)^{n-1}\) is true in all cases

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