33
$\begingroup$

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 $1$

$\endgroup$
4
  • 1
    $\begingroup$ for two use the fact that you can diagonalize orthogonal matrices and the determinant of orthogonal matrices is 1 $\endgroup$
    – Bman72
    Jan 27, 2014 at 10:54
  • 11
    $\begingroup$ Two is false. The determinant is $\pm 1$, not the eigenvalues in general. Take a rotation matrix for example. $\endgroup$
    – EuYu
    Jan 27, 2014 at 10:57
  • 1
    $\begingroup$ 1) does not need to to be true if $A$ is not symmetric. In that case, you could construct a real matrix with complex eigenvalues satisfying (1) but obviously with non-real eigenvalues (though their real parts might be positive). $\endgroup$ Jan 27, 2014 at 10:59
  • $\begingroup$ @Ale The link you give has counter-examples in it. The eigenvalues merely have to satisfy $|\lambda| = 1$. $\endgroup$
    – EuYu
    Jan 27, 2014 at 11:01

4 Answers 4

33
$\begingroup$

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.

$\endgroup$
2
  • 11
    $\begingroup$ If $|\lambda|=1$, it does not follow that $\lambda=\pm 1$. Only $\lambda=e^{i\phi}$ for some real $\phi$. $\endgroup$
    – J.R.
    Jan 27, 2014 at 11:07
  • $\begingroup$ Yes, you are very right.@TooOldForMath You must consider the real eigenvalue. $\endgroup$
    – gaoxinge
    Jan 27, 2014 at 11:08
15
$\begingroup$

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.

  1. 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 $\theta$).

  1. 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

$\endgroup$
7
  • $\begingroup$ Can you explain why $det(A^{T}-I) = det(A-I)$ $\endgroup$ Jan 2, 2017 at 18:14
  • 2
    $\begingroup$ The link you have provided is not working. Can you please give the new link? $\endgroup$ Aug 6, 2017 at 10:44
  • $\begingroup$ @SachchidanandPrasad : I appreciate your remark. I updated the link, it was changed. $\endgroup$ Aug 6, 2017 at 13:58
  • $\begingroup$ Looks like the link is broken again, though archive.org has a saved copy of the previous link web.archive.org/web/20150206015748/www2.bc.edu/~reederma/… $\endgroup$ Nov 17, 2018 at 3:12
  • $\begingroup$ @HermanJaramillo, so can I say that the eigenvalues of orthogonal matrices are real iff it produces a pure reflection? $\endgroup$
    – DatBoi
    Jun 20, 2023 at 15:49
7
$\begingroup$

(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 $\det A=\pm 1$.

$\endgroup$
1
  • $\begingroup$ This solution has problems. $|\lambda|=1$ holds for the eigenvalues of every orthogonal matrix Q. $\endgroup$
    – Yao Zhao
    Nov 12, 2020 at 17:14
6
$\begingroup$

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.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .