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.
4 Answers
If $$Q^TQ = I$$ $$R^TR = I,$$ then $$(QR)^T(QR) = (R^TQ^T)(QR) = R^T(Q^TQ)R = R^TR = I.$$ Of course, this can be extended to $n$ many matrices inductively.
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Very succinct answer, thank you. But you used commutative property in the proof, and in general matrices don't obey multiplicative commutativity. Could you please elaborate on that? – Kashmiri Nov 16 '20 at 05:25

@YasirSadiq hm? Nowhere did I use commutativity. I did use associativity and the fact that $(AB)^T = B^TA^T$  I added an extra expression to clarify. Both associativity and the transposeofproductisreverseproductoftransposes can be verified by writing out the elements of the matrices. – user217285 Nov 17 '20 at 05:01

@YasirSadiq You should learn what the commutative property is before littering *all* the answers with the same erroneous belief. – Ted Shifrin Nov 17 '20 at 05:07

@user217285Yes it's clear now ,Thank you so much. – Kashmiri Nov 17 '20 at 05:58

@Ted Shifrin I apologize shall I delete all my comments? – Kashmiri Nov 17 '20 at 05:58
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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3Intuitively, I think of it as a "hyperdimensional rotation". Is this roughly an apt expression? – MackTuesday Sep 01 '15 at 17:35

4@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. – Iwillnotexist Idonotexist Sep 02 '15 at 01:08
Let the orthogonal matrices be known as $M$ and $N$. By the definition of orthogonal matrices, $M \cdot N$ must be orthogonal, as
$$(M \cdot N)^T \cdot (M\cdot N) = N^T \cdot M^T \cdot M \cdot N = N^T \cdot N = I $$ $$(M \cdot N) \cdot (M\cdot N)^T = M \cdot N \cdot N^T \cdot M^T = M \cdot M^T = I $$

1Nice answer. But you used commutative property in the proof, and in general matrices don't obey multiplicative commutativity. Could you please elaborate on that – Kashmiri Nov 16 '20 at 05:51

@YasirSadiq This doesn't actually use commutativity. One property of taking the transpose of a product of matrices is that the order of those matrix factors is reversed, in addition to them individually being transposed. E.g. (A * B)^T = B^T * A^T. This answer just takes advantage of that property. – Starkle Nov 18 '20 at 00:59
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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1+1 but You used commutative property in the proof, and in general matrices don't obey multiplicative commutativity. Could you please elaborate on that – Kashmiri Nov 16 '20 at 05:51

@YasirSadiq (copied from my other comment) This doesn't actually use commutativity. One property of taking the transpose of a product of matrices is that the order of those matrix factors is reversed, in addition to them individually being transposed. E.g. (A * B)^T = B^T * A^T. This answer just takes advantage of that property. – Starkle Nov 18 '20 at 01:00