12.20

1. A matrix \(A \in M_n(F)\) is called orthogonal if \(A^T A = I_d\), where \(I_d\) is the identity matrix. If A is orthogonal, prove that \(\text{det}(A) = \pm 1\)​

   \(\det(A)\cdot\det(A^{T})=1\Rightarrow\det(A)^2=1\)​

2. A matrix \(A \in M_n(\mathbb{C})\) is called unitary if \(A A^* = I_d\) (where \(A^*\) denotes the conjugate transpose of \(A\)). If \(A\) is unitary, prove that \(|\text{det}(A)| = 1\).

   \(\det AA^{*}=1=\det A\det A=1\)​

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

   \[ \det\left(\begin{array}{cccccc}t & 0 & 0 & \cdots & 0 & a_0\\ -1 & t & 0 & \cdots & 0 & a_1\\ 0 & -1 & t & \cdots & 0 & a_2\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & -1 & t+a_{n-1}\end{array}\right)=t^{n}+a_{n-1}t^{n-1}+\cdots+a_1t+a_0. \]

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

   Suppose it is true for \(n\), then for \(n+1\), we have \(\det\left(\begin{array}{cccccc}t & 0 & 0 & \cdots & 0 & a_0\\ -1 & t & 0 & \cdots & 0 & a_1\\ 0 & -1 & t & \cdots & 0 & a_2\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & t & a_{n-1}\\ 0 & 0 & 0 & \ldots & -1 & t+a_{n}\end{array}\right)=\sum_{i=1}^{n+1}\left(-1\right)^{i+1}a_{i1}\det A\left(i\left|1\right.\right)=t\det A\left(1\left|1\right.\right)+\left(-1\right)\left(-1\right)\cdot\det A\left(2\left|1\right.\right)\)

   Since \(\left(\begin{array}{cccccc} & 0 & 0 & \cdots & 0 & a_0\\ & -1 & t & \cdots & 0 & a_2\\ & \vdots & \vdots & \ddots & \vdots & \vdots\\ & 0 & 0 & \cdots & t & a_{n-1}\\ & 0 & 0 & \ldots & -1 & t+a_{n}\end{array}\right)\), then \(\det A(2|1)=a_0\det A(2|1)(1|n)=a_0\left(-1\right)^{n-1}\)​

   Then \(\det=t^{n+1}+a_{n-1}t^{n}+\cdots+a_1t^2+a_0t+a_0\left(-1\right)^{n-1}=t^{n+1}+a_{n}t^{n}+\cdots+a_2t^2+a_1t+a_0\)​

4. Let \(A, B, C, D \in M_n(F)\) where \(A\) is invertible.

   (a) Prove that

   \[ \text{det} \begin{pmatrix} A & 0 \\ 0 & D \end{pmatrix} = \text{det}(A) \text{det}(D). \]

   

   (b) Prove that

   \[ \text{det} \begin{pmatrix} A & B \\ 0 & D \end{pmatrix} = \text{det}(A) \text{det}(D). \]

   

   (c) Prove that if \(AC = CA\), then

   \[ \text{det} \begin{pmatrix} A & B \\ C & D \end{pmatrix} = \text{det}(AD - BC). \]

   

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