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I've seen the statement "The matrix product of two orthogonal matrices is another orthogonal matrix. " on Wolfram's website but haven't seen any proof online as to why this is true. By orthogonal matrix, I mean an $n \times n$ matrix with orthonormal columns. I was working on a problem to show whether $Q^3$ is an orthogonal matrix (where $Q$ is orthogonal matrix), but I think understanding this general case would probably solve that.

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If $$Q^TQ = I$$ $$R^TR = I,$$ then $$(QR)^T(QR) = R^T(Q^TQ)R = R^TR = I.$$ Of course, this can be extended to $n$ many matrices inductively.

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As an alternative to the other fine answers, here's a more geometric viewpoint:

Orthogonal matrices correspond to linear transformations that preserve the length of vectors (isometries). And the composition of two isometries $F$ and $G$ is obviously also an isometry.

(Proof: For all vectors $x$, the vector $F(x)$ has the same length as $x$ since $F$ is an isometry, and $G(F(x))$ has the same length as $F(x)$ since $G$ is an isometry; hence $G(F(x))$ has the same length as $x$.)

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    $\begingroup$ Intuitively, I think of it as a "hyperdimensional rotation". Is this roughly an apt expression? $\endgroup$ – MackTuesday Sep 1 '15 at 17:35
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    $\begingroup$ @MackTuesday With one catch; It is not only rotation but also reflection that can preserve vector lengths. I prefer the intuitive notion presented by this answer: There is no way, using pure rotation and reflection, to change the length of a vector. If I take a stick and swing it around or look at it in a (flat) mirror, no matter what I do the stick remains the same size. $\endgroup$ – Iwillnotexist Idonotexist Sep 2 '15 at 1:08
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Let the orthogonal matrices be known as $M$ and $N$. By the definition of orthogonal matrices, $M \cdot N$ must be orthogonal, as

$$(M \cdot N)^T \cdot (M\cdot N) = N^T \cdot M^T \cdot M \cdot N = N^T \cdot N = I $$ $$(M \cdot N) \cdot (M\cdot N)^T = M \cdot N \cdot N^T \cdot M^T = M \cdot M^T = I $$

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Let $A$ and $B$ be two orthogonal matrices. You have $$AA^T = A^TA = I$$ and $$BB^T = B^TB =I.$$

So, we have $$(AB)^T(AB) = B^TA^TAB = I.$$

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