2
$\begingroup$

The characteristic equation of matrix A is $$\lambda ^3 - I_1\lambda^2 + I_2\lambda-I_3 = 0 $$

For orthogonal matrix $$I_3 = det(A) = \pm1$$ $$I_1 = tr(A)$$

Taking examples of orthogonal matrices, it looks like $I_1 = I_2$. Is this true always for an orthogonal matrix? Is there some proof?

$\endgroup$
0

1 Answer 1

0
$\begingroup$

This isn't true. E.g. when $A$ is the negative of the identity matrix, its characteristic polynomial is $(x+1)^3=x^3+3x^2+3x+1$.

In general, the characteristic polynomial of a $3\times3$ nonsingular matrix $A$ is in the form of \begin{aligned} &x^3-\sum_i\lambda_i(A)x^2+\sum_{i\ne j}\lambda_i(A)\lambda_j(A)x-\prod_i\lambda_i(A)\\ =\,&x^3-\operatorname{tr}(A)x^2+\det(A)\operatorname{tr}(A^{-1})x-\det(A). \end{aligned} You are essentially asking whether $$ \operatorname{tr}(A)=\det(A)\operatorname{tr}(A^{-1}) $$ when $A$ is orthogonal. Since $A^{-1}=A^T$ in this case, the above equality is true if and only if $\operatorname{tr}(A)=0$ or $\det(A)=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.