# 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?

-
If a matrix A is orthogonal, it means $AA^T=A^TA=I$. (At least it is the usual terminology.) Perhaps you should clarify whether you use this definition of orthogonal matrix and what are your assumptions about S. Is S orthogonal too? –  Martin Sleziak Nov 13 '11 at 11:10
But if $A$ is orthogonal, how is $AA^T=A^TA=I$? I thought this is only possible if $A$ is orthonormal? –  xenon Nov 13 '11 at 11:19
Have a look at wikipedia link. Wiki says that: In linear algebra, an orthogonal matrix (less commonly called orthonormal matrix), is a square matrix with real entries whose columns and rows are orthogonal unit vectors (i.e., orthonormal vectors). –  Martin Sleziak Nov 13 '11 at 11:20
If you only assume that columns are orthogonal, then this set of matrices is not closed under product: $\begin{pmatrix} 2 & 0 \\ 0 & 1 \\ \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -1 \\ \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 1 & -1 \\ \end{pmatrix}$ –  Martin Sleziak Nov 13 '11 at 11:25
I don't know. Although google returns a few hits for "column orthogonal matrix and "row orthogonal matrix", I don't know whether it is standard terminology. –  Martin Sleziak Nov 13 '11 at 11:43

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.

-