Does an orthogonal matrix $A$ belong to the sett O(n)={all nxn-matrices, such that $A*A^T=I$

Question

How can i prove the question in the title?

Is it valid to say that a matrix $U$ is orthogonal when it has orthonormal columns if and only if $U^TU=I$?

Answer 1

Suppose, for instance, that $n=3$. Then, if$$A=\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix},$$then$$A.A^T=\begin{bmatrix}a^2+b^2+c^2 & a d+b e+c f & a g+b h+c i \\ a d+b e+c f & d^2+e^2+f^2 & d g+e h+f i \\ a g+b h+c i & d g+e h+f i & g^2+h^2+i^2\end{bmatrix}.$$Asserting that $A.A^T=\operatorname{Id}$ mens then $6$ things:$$\left\{\begin{array}{l}a^2+b^2+c^2=1\\d^2+e^2+f^2=1\\g^2+h^2+i^2=1\\a d+b e+c f=0\\a g+b h+c i=0\\d g+e h+f i=0.\end{array}\right.$$But this means that the rows of $A$ have norm $1$ and that they are orthogonal. And it follows from $A.A^T=\operatorname{Id}$ that $A^T.A=\operatorname{Id}$, which means the same thing for the columns.