How to prove that a matrix is orthogonal given that How to prove that a matrix is orthogonal given that 
abs(det(Q)) = 1 and its columns have unit norm?
my work 

I tried with Gram matrix  but not correct , please help 
 A: Notation: $B_{n\times n}=(B_1,\ldots,B_n)$, where $B_i$ is the column $i$ of $B$.
Since $1=\det(Q)$ then $1=\det(Q^tQ)$. Therefore $Q^tQ$ is a positive definite symmetric matrix. 
It is well known that $Q^tQ=U^tDU$, where $U$ is a upper triangular matrix with diagonal $1,\ldots,1$ and $D$ is a diagonal matrix with diagonal $d_1,\ldots,d_n$ (see Cholesky decomposition).
Notice that $Q_i^tQ_i=d_iU_i^tU_i$, for every $i$. Since $Q_i^tQ_i=1($by hypothesis$)$ and $U_i^tU_i\geq 1($since there is a coordinate of $U_i$ equal to $1)$ then $0<d_i\leq 1$. 
Therefore $1=\det(Q^tQ)=\det(U^t)\det(D)\det(U)=\det(D)=\prod_{i=1}^n d_i$. Since $0<d_i\leq 1$ and $\prod_{i=1}^n d_i=1$ then $d_i=1$ for every $i$. Hence, $U_i^tU_i=\frac{1}{d_i}=1$ for every $i$. Since $U_i$ has one coordinate equal to 1 then the other coordinates of $U_i$ must be zero. Therefore $U=Id$ and $Q^tQ=U^tDU=Id^3=Id$.
A: This is for a matrix with real entries.
The diagonal elements of $Q^T Q$ are the norms of the columns of $Q$.  So 
$S = Q^T Q$ is a positive semidefinite matrix of determinant $1$ whose diagonal elements are all $1$.  In particular, its trace is $n$ (if the matrix is $n \times n$), so the arithmetic mean of its eigenvalues is $1$.
The determinant is the product of the eigenvalues, so the geometric mean of the eigenvalues is also $1$.   The arithmetic and geometric means of a list of nonnegative numbers are equal if and only if all the numbers are equal: see AM-GM inequality.  Thus all the eigenvalues are $1$, and the only normal matrix whose eigenvalues are all $1$ is $I$. Therefore $Q^T Q = I$, i.e. $Q$ is orthogonal.
