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.
 A: Let the orthogonal matrices be known as $M$ and $N$. By the definition of orthogonal matrices, $M \cdot N$ must be orthogonal, as
$$(M \cdot N)^T \cdot (M\cdot N) = N^T \cdot M^T \cdot M \cdot N = N^T \cdot N = I $$
$$(M \cdot N) \cdot (M\cdot N)^T = M \cdot N \cdot N^T \cdot M^T = M \cdot M^T = I $$
A: Let $A$ and $B$ be two orthogonal matrices. You have $$AA^T = A^TA = I$$ and $$BB^T = B^TB =I.$$
So, we have $$(AB)^T(AB) = B^TA^TAB = I.$$
A: 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.
A: 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 $x$.)
