Eigenvalues in orthogonal matrices

Question

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 $1$

Answer 1

(1):

Let $Av=\lambda v$ with $v\not=0$, i.e. $\lambda$ is an eigenvalue of $A$. Then $$0<v^t Av=\lambda v^t v=\lambda \|v\|^2.$$

Since $\|v\|^2>0$, we get $\lambda>0$.

(2):

You certainly mean that the determinant of $A$ is $\pm 1$, since the statement about the eigenvalues is not true, for consider the orthogonal matrix

$$\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}$$

This represents a rotation and has therefore complex eigenvalues.

But if $A$ is orthogonal, then $A^tA=AA^t=I$, therefore applying the $\det$ to both sides and using the multiplication law for determinants, we obtain

$$(\det A)^2 = 1$$

Therefore $\det A=\pm 1$.

Answer 2

Let $\lambda$ be $A$ eigenvalue and $Ax=\lambda{x}$.

(1) ${x}^{t}Ax=\lambda{x^tx}>0.$ Because $x^tx>0$, then $\lambda>0$

(2) $|\lambda|^2x^tx=(Ax)^{t}Ax={x}^{t}A^{t}Ax=x^tx.$ So $|\lambda|=1$. Then $\lambda=e^{i\phi}$ for some $\phi\in\mathbb{R}$; i.e. all the eigenvalues lie on the unit circle.

Answer 3

Both statements are false as currently written. The following matrix serves as a counter-example for both. $$R = \begin{pmatrix}\cos 1 & -\sin 1 \\ \sin 1 & \cos 1\end{pmatrix}$$

The first statement needs to be modified so that the matrix has all real eigenvalues, otherwise it is false as noted by Algebraic Pavel in the comments. Let us express an arbitrary non-zero vector $\mathbf{b}$ in polar form $$\mathbf{b} = r\begin{pmatrix}\cos \phi \\ \sin\phi \end{pmatrix}$$ Then for the above matrix $R$, we get $$\mathbf{b}^\mathrm{T}R\mathbf{b}= r^2\left( \cos(\phi + 1)\cos\phi + \sin(\phi+1)\sin\phi\right) = r^2\cos 1 > 0 $$ The eigenvalues are both complex however. If we assume that all eigenvalues are real, then the arguments given by TooOldForMath and gaoxinge works fine.

The second statement should say that the determinant of an orthogonal matrix is $\pm 1$ and not the eigenvalues themselves. $R$ is an orthogonal matrix, but its eigenvalues are $e^{\pm i}$.

The eigenvalues of an orthogonal matrix needs to have modulus one. If the eigenvalues happen to be real, then they are forced to be $\pm 1$. Otherwise though, they are free to lie anywhere on the unit circle.