# 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?

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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.

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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.

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