Let $A\in M_3(\mathbb{R})$ be an orthogonal matrix with $det(A)=1$. Then to show $$(Tr(A)-1)^2+\sum_{i<j}(a_{ij}-a_{ji})^2=4$$ where $Tr$ denotes trace.
Suppose the matrix $A=\left(\begin{matrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} &a_{23}\\ a_{31} & a_{32} & a_{33} \end{matrix}\right)$ in $M_3(\mathbb{R})$ be orthogonal. Then $Tr(AA^T)=\sum_{i,j}a_{ij}^2=3$. If we expand the left hand side of the equation in question, we have $$LHS=4+2(a_{11}a_{22}+a_{11}a_{33}+a_{22}a_{33})-2(a_{12}a_{21}+a_{13}a_{31}+a_{23}a_{32})-2(a_{11}+a_{22}+a_{33})$$ $$=4+2(a_{11}a_{22}-a_{12}a_{21})+2(a_{11}a_{33}-a_{13}a_{31})+2(a_{22}a_{33}-a_{23}a_{32})-2(a_{11}+a_{22}+a_{33})$$ $$=4+minor(a_{33})+minor(a_{22})+minor(a_{11})-2(a_{11}+a_{22}+a_{33})$$
Is there any theorem/result that can be used to say that the remaining part of the last expression excluding 4 is zero? Or is there any way we can manipulate $a_{ij}$ so that it becomes zero? I tried hard but i see no way out. Any help is highly appreciated.
I think the LHS should be $$LHS=4+2(minor(a_{33})+minor(a_{22})+minor(a_{11}))−2(a_{11}+a_{22}+a_{33})$$ and you can simplify this expression with the help of the characteristic plynomial of $A$. You can write $$Det(tI-a)=t^3-c_2t^2+c_1t-c_0$$ where $I$ is the identity matrix and $c_i$ is the sum of all principal minor of $A$ of size $3-i$. In particular, $c_0=Det(A)=1$, $c_1=minor(a_{33})+minor(a_{22})+minor(a_{11})$ using your notations and $c_2=Tr(A)$. Finally, $A$ is a matrix of $M_3(\mathbb{R})$ so it has at least one real eigenvalue, $A$ is orthogonal so this eigenvalue is either $1$ or $-1$ and $det(A)=1$ so $A$ has at least an eigenvalue equal to $1$. This means that $Det(I-A)=0$ hence $1-c_2+c_1-c_0=0$ and then $$minor(a_{33})+minor(a_{22})+minor(a_{11})=a_{11}+a_{22}+a_{33}$$ and fially $LHS=4$.