# 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.

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? Nov 16, 2020 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. Nov 17, 2020 at 5:01
• @YasirSadiq You should learn what the commutative property is before littering all the answers with the same erroneous belief. Nov 17, 2020 at 5:07
• @user217285Yes it's clear now ,Thank you so much. Nov 17, 2020 at 5:58
• @Ted Shifrin I apologize shall I delete all my comments? Nov 17, 2020 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? Sep 1, 2015 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. Sep 2, 2015 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 Nov 16, 2020 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. Nov 18, 2020 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 Nov 16, 2020 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. Nov 18, 2020 at 1:00