<s>[INST] Eigenvalues in orthogonal matrices Let $A \in M_n(\Bbb R)$.
How can I prove, that

1) if $ \forall {b \in \Bbb R^n}, b^{t}Ab>0$, then all eigenvalues $>0$.
2) if $A$ is orthogonal, then all eigenvalues are equal to $-1$ or $1$

 [/INST] The first part of the problem is well solved above, so I want to emphasize on the second part, which was partially solved. 
An orthogonal transformation is either a rotation or a reflection. I will focus on 3D which has  lots of practical use. Let us then assume that $A$ is an orthonormal matrix in $\mathbb{R}^3 \times \mathbb{R}^3$. 
We find that its eigenvalues are either $1, \text{e}^{\pm i \theta}$ for a rotation or $\pm 1$ for a reflection. 


*

*For a rotation: 
We have the following sequence of equalities (since $\det A = 1$)
\begin{eqnarray*}    \det(I -A) &=& \det(A) \det (I-A)  \\   &=&
\det(A^T) \det(I-A)  \\   &=&  \det(A^T - I)  \\   &=&  \det(A - I) 
\\   &=& -\det(I-A)   \quad , \text{since 3 is odd}, \end{eqnarray*}
So $\det(I-A)=0$ and $\lambda_1=1$ is an eigenvalue of $A$.
Now, let $u_1$ the unit eigenvector of $\lambda_1$, so $A u_1 =
u_1$.  We show that the matrix $A$ is a rotation of an angle
$\theta$  around this axis $u_1$. Let us form a new coordinate
system using $u_1, u_2, u_1 \times u_2$, where $u_2$ is a vector
orthogonal to $u_1$, so the new system is right handed (has
determinant = 1). The transformation between the old system $\{e_1,
e_2, e_3\}$ and the new system is given by a matrix  $B$ with column
vectors $u_1, u_2$, and $u_3$.  So we have that $B e_i = u_i$. Let
us call the coordinate transformation (similarity) matrix $C =
B^{-1}AB$. Then 
\begin{eqnarray*} 
    C e_1 = B^{-1} A B e_1 = B^{-1} A u_1 = B^{-1} u_1 = e_1.  \end{eqnarray*}  The fact that $C e_1 = e_1$ means two things:


*

*The first column of $C$ is  $(1,0,0)^T$.

*The first raw of $C$ is $(1,0,0)$.
Then $C$ is a matrix of the type 
\begin{eqnarray*}  
C = \left (   \begin{array}{ccc}
    1 & 0 & 0 \\
    0 & a & b \\
    0 & c & d \\   \end{array}   \right ) \end{eqnarray*}  Since $A$ is orthogonal $C$ is orthogonal and so the vectors $(a,c)^T$ and
$(b,d)^T$  are orthogonal and since  \begin{eqnarray*}
    1 = \theta A = \det C = ad - bc  \end{eqnarray*} we have that the minor matrix with entries $a,b,c,d$ is a rotation (orthogonal 
with determinat 1). Rotations in 2D are of the form
\begin{eqnarray*}
    \left (
    \begin{array}{cc}
      a  & b \\
      c  & d 
    \end{array}
    \right )
=
    \left (
    \begin{array}{cc}
      \cos \theta  & -\sin \theta \\
      \sin \theta  & \cos \theta 
    \end{array}
    \right ) \end{eqnarray*} then   \begin{eqnarray*}   B^{-1}A B = \left (   \begin{array}{ccc}
    1 & 0 & 0 \\
    0 & \cos \theta & -\sin \theta \\
    0 & \sin \theta & \cos \theta    \end{array}   \right ) \end{eqnarray*} But since the eigenvalues of $B^{-1}AB$ are the same
as those of $A$ we find the eigenvalues of this matrix and those
would be the eigenvalues of $A$.
For the eigenvalues of this matrix we have that
\begin{eqnarray*}   \det ( C - \lambda I)= 0 \implies   (\cos \theta
- \lambda)^2 + \sin^2 \theta = 0. \end{eqnarray*} That is,
\begin{eqnarray*} 1 -2 \lambda \cos \theta + \lambda^2 = 0
\end{eqnarray*} Then
\begin{eqnarray*}   \lambda = \frac{2 \cos \theta \pm \sqrt{4 \cos^2
\theta - 4}}{2} =  \cos \theta \pm i \sin \theta = \text{e}^{\pm i
\theta}. 
\end{eqnarray*}
This is the interesting result by Euler where he claimed that all 
rigid body rotations can be written as rotation about one vector
(here $u_1$ by some angle $\theta$). 


*Eigenvalues of a reflection. If we repeat all the steps above for a
reflection we find that now
\begin{eqnarray*}   B^{-1}A B = \left (   \begin{array}{ccc}
    1 & 0 & 0 \\
    0 & \cos 2 \theta & \sin 2 \theta \\
    0 & \sin 2 \theta & -\cos 2 \theta    \end{array}   \right ) \end{eqnarray*}
Now
\begin{eqnarray*}  
 \det ( C - \lambda I)= 0 \implies   -(\cos
2 \theta  -  \lambda)(\cos 2 \theta + \lambda )  - \sin^2 2 \theta =
0. 
\end{eqnarray*} 
and then 
\begin{eqnarray*} -\sin^2 2 \theta - \cos^2 2 \theta + \lambda^2 =0
\end{eqnarray*} and from here
\begin{eqnarray*}   \lambda =  \pm 1. \end{eqnarray*}
.
Please see the following reference which I used for this answer:
Orthogonal Matrices
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