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

Question

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.

Answer 1

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.

Answer 2

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

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

Answer 3

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

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

Answer 4

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 $x$.)