# Why is the matrix product of 2 orthogonal matrices also an orthogonal matrix?

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.

• 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 5:25
• @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 transpose-of-product-is-reverse-product-of-transposes can be verified by writing out the elements of the matrices. – user217285 Nov 17 '20 at 5:01
• @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 5:07
• @user217285Yes it's clear now ,Thank you so much. – Kashmiri Nov 17 '20 at 5:58
• @Ted Shifrin I apologize shall I delete all my comments? – Kashmiri Nov 17 '20 at 5:58

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 $x$.)

• Intuitively, I think of it as a "hyperdimensional rotation". Is this roughly an apt expression? – MackTuesday Sep 1 '15 at 17:35
• @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 2 '15 at 1:08

Let the orthogonal matrices be known as $M$ and $N$. By the definition of orthogonal matrices, $M \cdot N$ must be orthogonal, as

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

• 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 – Kashmiri Nov 16 '20 at 5:51
• @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 0:59

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

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

• +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 5:51
• @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 1:00