Problem 2 - Determinants

Let

$$A=\left(\begin{tabular}{ccc} $a_{11}$ & $a_{12}$ & $a_{13}$ \\ ... ... & $a_{22}$ & $a_{23}$ \\ $a_{31}$ & $a_{32}$ & $a_{33}$ \end{tabular}\right)$$

be an orthogonal matrix of order 3 satisfying $\vert A\vert=1$, where $\vert A\vert$ is the determinant of the matrix $A$.

1.

Prove the following equalities.

$$a_{11}=\left\vert\begin{tabular}{cc} $a_{22}$ & $a_{23}$ \\ $a_{... ...lar}{cc} $a_{21}$ & $a_{22}$ \\ $a_{31}$ & $a_{32}$ \end{tabular}\right\vert.$$

2.

Prove that $A$ has an eigenvalue $\lambda = 1$.

1.

$A$ is an orthogonal matrix. That means in particular that the system of its column vectors is an orthonormal base of the Euclidian space $\mathbb{R}^3$. Let us call that base $\mathcal{B}'$. $A$ is the matrix that goes from the canonical base $\mathcal{B}$ to the base $\mathcal{B}'$: $A=P_{\mathcal{B}\mathcal{B}'}$. We know that $\det(A)=1>0$. It means moreover that the base is direct. Thus, we have the following property:

$$\left(\begin{tabular}{c} $a_{11}$ \\ $a_{21}$ \\ $a_{31}$ \end{... ...left(\begin{tabular}{c} $a_{13}$ \\ $a_{23}$ \\ $a_{33}$ \end{tabular}\right)$$

$$\left(\begin{tabular}{c} $a_{12}$ \\ $a_{22}$ \\ $a_{32}$ \end{... ...left(\begin{tabular}{c} $a_{11}$ \\ $a_{21}$ \\ $a_{31}$ \end{tabular}\right)$$

$$\left(\begin{tabular}{c} $a_{13}$ \\ $a_{23}$ \\ $a_{33}$ \end{... ...left(\begin{tabular}{c} $a_{12}$ \\ $a_{22}$ \\ $a_{32}$ \end{tabular}\right)$$

If we calculate explicitly the vectorial product for each of this property, then we could express the first coordinate of each right-hand vector in function of the coordinates of the corresponding left-hand vectors. That leads to the following relations:

$$a_{13} = a_{21}a_{32} - a_{31}a_{22}$$

$$a_{11} = a_{22}a_{33} - a_{32}a_{23}$$

$$a_{12} = a_{23}a_{31} - a_{21}a_{33}$$

which are the relations we are asking.

2.

We shall determine the characteristic polynomial of $A$:

$$\chi_A(x) = \left\vert \begin{tabular}{ccc} $a_{11} - \lambda$ & ... ...$a_{23}$ \\ $a_{31}$ & $a_{32}$ & $a_{33} - \lambda$ \end{tabular}\right\vert$$

$$=(a_{11} - \lambda)(a_{22}a_{33} + \lambda^2 - \lambda(a_{22}+a_{... ...a a_{12} - a_{32}a_{13}) + a_{31}(a_{12}a_{23} + a_{13}a_{22} - \lambda a_{13})$$

$$=-\lambda^3 + \lambda^2\hbox{tr}(A) + \lambda(-a_{11}a_{22}+a_{11}a_{33}-a_{22}a_{33} + a_{32}a_{23} + a_{12}a_{21}-a_{31}a_{13}) + \det(A)$$

In particular:

$$\chi_A(1)=\hbox{tr}(A) -a_{11}a_{22}+a_{11}a_{33}-a_{22}a_{33} + a_{32}a_{23} + a_{12}a_{21}-a_{31}a_{13}$$

From the calculus of the vectorial products in the question above, we can also deduce the following formulas:
$a_{22} = a_{13}a_{31} - a_{33}a_{11}$
$a_{33} = a_{11}a_{22} - a_{21}a_{12}$
We already know that:
$a_{11} = a_{22}a_{33} - a_{32}a_{23}$
We can therefore eliminate the following terms (in bold below):

$$\chi_A(1)=a_{11} + a_{22} \boldsymbol{+ a_{33}} \boldsymbol{-a_{1... ...{11}a_{33}-a_{22}a_{33} + a_{32}a_{23} \boldsymbol{+ a_{12}a_{21}}-a_{31}a_{13}$$

$$\chi_A(1)=\boldsymbol{a_{11}} + a_{22} +a_{11}a_{33} \boldsymbol{-a_{22}a_{33}} \boldsymbol{+ a_{32}a_{23}} -a_{31}a_{13}$$

$$\chi_A(1)=\boldsymbol{a_{22} +a_{11}a_{33} -a_{31}a_{13}}$$

$$\chi_A(1)=0$$

Thus, $\lambda = 1$ is an eigenvalue of $A$.