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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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5 Answers 5

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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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  • $\begingroup$ 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? $\endgroup$
    – Kashmiri
    Nov 16, 2020 at 5:25
  • $\begingroup$ @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 transpose-of-product-is-reverse-product-of-transposes can be verified by writing out the elements of the matrices. $\endgroup$
    – user217285
    Nov 17, 2020 at 5:01
  • $\begingroup$ @YasirSadiq You should learn what the commutative property is before littering all the answers with the same erroneous belief. $\endgroup$ Nov 17, 2020 at 5:07
  • $\begingroup$ @user217285Yes it's clear now ,Thank you so much. $\endgroup$
    – Kashmiri
    Nov 17, 2020 at 5:58
  • $\begingroup$ @Ted Shifrin I apologize shall I delete all my comments? $\endgroup$
    – Kashmiri
    Nov 17, 2020 at 5:58
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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? $\endgroup$ Sep 1, 2015 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. $\endgroup$ Sep 2, 2015 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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    $\begingroup$ Nice answer. But you used commutative property in the proof, and in general matrices don't obey multiplicative commutativity. Could you please elaborate on that $\endgroup$
    – Kashmiri
    Nov 16, 2020 at 5:51
  • $\begingroup$ @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. $\endgroup$
    – user573694
    Nov 18, 2020 at 0:59
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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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    $\begingroup$ +1 but You used commutative property in the proof, and in general matrices don't obey multiplicative commutativity. Could you please elaborate on that $\endgroup$
    – Kashmiri
    Nov 16, 2020 at 5:51
  • $\begingroup$ @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. $\endgroup$
    – user573694
    Nov 18, 2020 at 1:00
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A hint, and the approach is not tha straightforward, but may give a different perspective:

Check that the map from to group of invertible square matrices to itself is an automorphism

$$g \mapsto (g^{t})^{-1}$$

(an involution in fact).

Now, for every endomorphism of a group, the set of fixed points forms a subgroup. In our case, the subgroup is $O(n, \mathbb{R})$ ( real case), or $U(n)$ ( in the complex case). Note also that we have an automorphism of a Lie group, and the corresponding automorphism of Lie algebras is $X \mapsto -X^t$.

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