Showing $AS$ is orthogonal if $A$ and $S$ are orthogonal Suppose matrix $A$ is an $n \times n$ orthogonal matrix and $S=\{\vec{u}_1, \vec{u}_2, \ldots, \vec{u}_n\}$ where $\vec{u}_1, \vec{u}_2, \ldots, \vec{u}_n$ are orthogonal to each other.
Now, since both $A$ and $S$ are orthogonal, by definition, I know that the matrix $$T=AS=\begin{bmatrix}
A\vec{u}_1 & A\vec{u}_2 & \cdots & A\vec{u}_n
\end{bmatrix}$$ is also orthogonal. But why is $T$ orthogonal too?
I tried to do this to prove that $T$ is orthogonal:
$$\begin{align*} 
T&=AS\\
T^TT&=(AS)^TAS\\
T^TT&=S^TA^TAS
\end{align*}$$
At this point, I am stuck because $A^TA$ don't give me the identity matrix unless $A$ is orthonormal, which it isn't ($A$ is only orthogonal). This goes the same for $S$.
How should I move on from here to show that $T$ is indeed orthogonal?
 A: Although discussion in the comments showed that this was more clarifying the terminology than a real question, I'm adding a few points from the comments here, so that the question is not left unanswered.
If a matrix A is orthogonal, it means $AA^T=A^TA=I$. (At least it is the usual terminology.) Hence if $S$ and $T$ are orthogonal, you get $T^TT=S^TA^TAS=S^TS=I$.
If you only assume that columns are orthogonal, then this set of
matrices is not closed under product:
$\begin{pmatrix}
 2 & 0 \\
 0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
 1 & 1 \\
 1 & -1 \\
\end{pmatrix}
=
\begin{pmatrix}
 2 & 2 \\
 1 & -1 \\
\end{pmatrix}
$.
Although google returns several hits for "column orthogonal matrix"
and "row orthogonal
matrix",
I don't know whether it is standard terminology.
A: One way to see that this ought to be a group, is that the set of orthogonal matrices is "equal" to the set of linear isometries of the Euclidean space (assuming you are indeed working with matrices whose entries are real numbers). Composing two isometries gives an isometry, and if they both are linear, the composition is.
P.S: The term "equal" is between quotation marks, because we need to impose an equivalence relation on the set of matrices to get a genuine equality.
