A property of 3×3 orthogonal matrices with determinant -1

Posted: May 22, 2018 in Linear Algebra, Uncategorized
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Problem. Let $A \in M_3(\mathbb{R})$ be orthogonal and suppose that $\det(A)=-1.$ Find $\det(A-I).$

Solution. Since $A$ is orthogonal, its eigenvalues have absolute value $1$ and it can be be diagonalized. Let $D$ be a diagonal matrix such that $PDP^{-1}=A$ for some invertible matrix $P \in M_3(\mathbb{C}).$ Then

$\det(D)=\det(A)=-1, \ \ \det(D-I)=\det(A-I).$

We claim that the eigenvalues of $A$ are $\{-1,e^{i\theta},e^{-i\theta}\}$ for some $\theta.$ Well, the characteristic polynomial of $A$ has degree three and so it has either three real roots or only one real root. Also, the complex conjugate of a root of a polynomial with real coefficients is also a root. So, since $\det(A)=-1,$ the eigenvalues of $A$ are either all $-1,$ which is the case $\theta=\pi,$ or two of them are $1$ and one is $-1,$ which is the case $\theta = 0,$ or one is $-1$ and the other two are in the form $\{e^{i\theta}, e^{-i\theta}\}$ for some $\theta.$ So

\displaystyle \begin{aligned} \det(D-I)=\det(A-I)=(-2)(e^{i\theta}-1)(e^{-i\theta}-1)=-4(1-\cos \theta). \ \Box \end{aligned}

Note that given $\theta,$ the matrix

$\displaystyle A=\begin{pmatrix} -1 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta \end{pmatrix}$

is orthogonal, $\det(A)=-1$ and $\det(A-I)=-4(1-\cos \theta).$

Reduced norm and reduced trace (1)

Posted: October 11, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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Throughout this post, $k$ is a field and $A$ is a finite dimensional central simple $k$-algebra of degree $n.$

Let $a \in M_m(k)$ and suppose that $p(x) = x^m + \alpha_{m-1}x^{m-1} + \ldots + \alpha_1x + \alpha_0 \in k[x]$ is the characteristic polynomial of $a.$ We know from linear algebra that $\text{Tr}(a)= - \alpha_{m-1}$ and $\det(a)=(-1)^m \alpha_0.$ We also know that $\text{Tr}$ is $k$-linear, $\det$ is multiplicative and $\text{Tr}(bc)=\text{Tr}(cb)$ for all $b,c \in M_m(k).$ We’d like to extend the concepts of trace and determinant to any finite dimensional central simple algebra.

Definition. Let $\text{Prd}_A(a,x)=x^n + \alpha_{n-1}x^{n-1} + \ldots + \alpha_1 x + \alpha_0 \in k[x]$ be the reduced characteristic polynomial of $a \in A$ (see the definition of reduced characteristic polynomials in here). The reduced trace and the reduced norm of $a$ are defined, respectively, by $\text{Trd}_A(a) = - \alpha_{n-1}$ and $\text{Nrd}_A(a) = (-1)^n \alpha_0.$

Remark. Let $K$ beĀ  a splitting field of $A$ with a $K$-algebra isomorphism $f : A \otimes_k K \longrightarrow M_n(K).$ So $\text{Trd}_A(a)=\text{Tr}(f(a \otimes_k 1))$ and $\text{Nrd}_A(a)=\det(f(a \otimes_k 1))$ because $\text{Prd}_A(a,x)$ is the characteristic polynomial of $f(a \otimes_k 1).$

Theorem. 1) The map $\text{Trd}_A: A \longrightarrow k$ is $k$-linear and the map $\text{Nrd}_A: A \longrightarrow k$ is multiplicative.

2) $\text{Trd}_A(ab)=\text{Trd}_A(ba)$ for all $a,b \in A.$

3) If $a \in k,$ then $\text{Tr}_A(a)=na$ and $\text{Nrd}_A(a)=a^n.$

4) $\text{Nrd}_A(a) \neq 0$ if and only if $a$ is a unit of $A.$ So $\text{Nrd}_A : A^{\times} \longrightarrow k^{\times}$ is a group homomorphism.

5) $\text{Trd}_A$ and $\text{Nrd}_A$ are invariant under isomorphism of algebras and extension of scalars.

Proof. Fix a splitting field $K$ of $A$ and a $K$-algebra isomorphism $f: A \otimes_k K \longrightarrow M_n(K).$

1) We have already proved that the values of $\text{Trd}_A$ and $\text{Nrd}_A$ are in $k.$ Now, let $a_1,a_2 \in A$ and $\alpha \in k.$ Then

$\text{Trd}_A(\alpha a_1 + a_2) = \text{Tr}(f((\alpha a_1 + a_2) \otimes_k 1)) = \text{Tr}(\alpha f(a_1 \otimes_k 1) + f(a_2 \otimes_k 1)) =$

$\alpha \text{Tr}(f(a_1 \otimes_k 1)) + \text{Tr}(f(a_2 \otimes_k 1)) = \alpha \text{Trd}_A(a_1) + \text{Trd}_A(a_2)$

and

$\text{Nrd}_A(a_1a_2) = \det(f(a_1a_2 \otimes_k 1)) = \det(f(a_1 \otimes_k 1)f(a_2 \otimes_k 1))=$

$\det(f(a_1 \otimes_k 1)) \det (f(a_2 \otimes_k 1)) = \text{Nrd}_A(a_1) \text{Nrd}_A(a_2).$

2) This part is easy too:

$\text{Trd}_A(ab)=\text{Tr}(f(ab \otimes_k 1))=\text{Tr}(f((a \otimes_k 1)(b \otimes_k 1)))=\text{Tr}(f(a \otimes_k 1)f(b \otimes_k 1))=$

$\text{Tr}(f(b \otimes_k 1)f(a \otimes_k 1))= \text{Tr}(f(ba \otimes_k 1))=\text{Trd}_A(ba).$

3) Let $I$ be the identity element of $M_n(K).$ Then

$\text{Trd}_A(a)=a \text{Trd}_A(1) = a \text{Trd}_A(f(1 \otimes_k 1)) = a \text{Tr}(I)=na$

and $\text{Nrd}_A(a)=\det(f(a \otimes_k 1))=\det(a f(1 \otimes_k 1)) = \det(aI) \det(I) = a^n.$

4) If $ab=1$ for some $b \in A,$ then $1 = \text{Nrd}_A(ab)=\text{Nrd}_A(a) \text{Nrd}_A(b)$ and thus $\text{Nrd}_A(a) \neq 0.$ Conversely, if $\text{Nrd}_A(a) \neq 0,$ then $\det(f(a \otimes_k 1)) \neq 0$ and so $f(a \otimes_k 1)$ is invertible in $M_n(K).$ Let $U$ be the inverse of $f(a \otimes_k 1).$ Then $U = f(u),$ for some $u \in A \otimes_k K$ because $f$ is surjective. Since $f$ is injective, it follows that $(a \otimes_k 1)u=1.$ Now if $a$ is not a unit of $A,$ then it is a zero divisor because $A$ is artinian. So $ba = 0$ for some $0 \neq b \in A.$ But then $b \otimes_k 1 = (b \otimes_k 1)(a \otimes_k 1)u = (ba \otimes_k 1)u=0,$ contradiction!

5) By Prp 3 and Prp 4 in this post, reduced characteristic polynomials are invariant under those things. $\Box$