# For an orthogonal matrix $Q$, why does $QQ^T = I$?

In my linear algebra text (Strang), an orthogonal matrix is defined to be a square matrix whose columns are orthonormal. In other words, an orthogonal matrix is a matrix $Q = [q_1 \cdots q_n]$ where each $q_i$ is a unit column vector of length $n$ with $$q_i^Tq_j = \begin{cases} 0 & \text{ when }i \neq j \\ 1 & \text{ when } i = j \end{cases}.$$

My book says that an orthogonal matrix $Q$ has the properties

1. $Q^TQ = I$ and
2. $QQ^T = I$,

from which it follows that $Q^T = Q^{-1}$. I can see how (1) follows from orthonormal columns, but I don't see why (2) is necessarily true. Can anyone provide some insight?

• Each element of $Q^TQ$ is the scalar product of two columns of $Q$. These vectors are orthonormal, so... – amd Jun 19 '16 at 16:56
• @amd My question was about $QQ^T$ though, not $Q^TQ$. – tddevlin Jun 19 '16 at 17:04
• Oops. Missed that. The other follows from the properties of square matrices, as per the accepted answer below. – amd Jun 20 '16 at 20:42
• Does this answer your question? Why, if a matrix $Q$ is orthogonal, then $Q^T Q = I$? – NNOX Apps Jul 25 '20 at 22:32

## 2 Answers

From $\mathrm Q^{\top} \mathrm Q = \mathrm I_n$, we conclude that $\mathrm Q^{\top}$ is the left inverse of $\mathrm Q$. Since $\mathrm Q$ is square, the left inverse is also the right inverse and, thus, $\mathrm Q \mathrm Q^{\top} = \mathrm I_n$. Hence, the inverse of $\mathrm Q$ is $\mathrm Q^{-1} = \mathrm Q^{\top}$.

If we know that $Q^TQ=I$ we may note that $$QQ^T = (QQ^T)(QQ^{-1}) = Q(Q^TQ)Q^{-1} = QIQ^{-1} = I.$$ Here we used the fact that matrix $Q^{-1}$ exist which is follows from $Q^TQ = I$ (indeed, $\det (QQ^T) = 1 \Rightarrow (\det Q)^2 = 1\Rightarrow \det Q \neq 0)$ .