Let $A$ be an orthogonal matrix . Prove that:
a) The determinant of the matrix $A$ is $\det A = \pm 1$.
b) The absolute value of the eigenvalues (if there exist) of the matrix $A$ is $1$.
Solution
a) Since the matrix is orthogonal we have that $AA^t=\mathbb{I}$. However $\det A = \det A^t$. The determinant of the product is the product of the determinants. Hence:
$$1=\det \mathbb{I} = \det (A A^t) = \det A \det A^t = (\det A)^2$$
and the result follows.
b) Since $A$ is orthogonal , we have that $|Au|=|u|$ . But if $u \neq 0$ is an eigenvector with eigenvalue $\lambda$ then:
$$Au=\lambda u \implies |u|= |Au|= |\lambda||u|$$
and the result follows.
Remarks:
a) The determinant of the matrix $A$ is $\det A = \pm 1$.
b) The absolute value of the eigenvalues (if there exist) of the matrix $A$ is $1$.
Solution
a) Since the matrix is orthogonal we have that $AA^t=\mathbb{I}$. However $\det A = \det A^t$. The determinant of the product is the product of the determinants. Hence:
$$1=\det \mathbb{I} = \det (A A^t) = \det A \det A^t = (\det A)^2$$
and the result follows.
b) Since $A$ is orthogonal , we have that $|Au|=|u|$ . But if $u \neq 0$ is an eigenvector with eigenvalue $\lambda$ then:
$$Au=\lambda u \implies |u|= |Au|= |\lambda||u|$$
and the result follows.
Remarks:
- Actually it is also true that each complex eigenvalue must have modulus 1.
- It is not true that from the fact that $|\lambda|=1$ we get $\lambda=\pm 1$ for the eigenvalue might be complex.
Application :
ReplyDeleteConsider the real matrix \(\displaystyle{A\in\mathbb{M}_{2}(\mathbb{R})}\) which represents the linear map
of the angle rotation by \(\displaystyle{\dfrac{\pi}{6}}\) .
Then,
\(\displaystyle{
A=\begin{pmatrix}
\cos\,\dfrac{\pi}{6} &\sin\,\dfrac{\pi}{6} \\
-\sin\,\dfrac{\pi}{6}&\cos\,\dfrac{\pi}{6}
\end{pmatrix}}\)
The matrix \(\displaystyle{A}\) is an orthogonal matrix with \(\displaystyle{\rm{det}(A)=1}\) .
According to the above result, the matrix \(\displaystyle{A}\) has not real eigenvalues and
geometrically, this result is obvious.