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

Question

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

$Q^TQ = I$ and
$QQ^T = I$,

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
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 $Q^T Q = I$? — , Jul 25 '20 at 22:32

Rodrigo de Azevedo · Accepted Answer · 2017-02-20T15:19:49.220

10

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 $\mathrm Q^{-1} = \mathrm Q^{\top}$.

edited Feb 20 '17 at 15:19

answered Jun 19 '16 at 17:09

Rodrigo de Azevedo

1

Anton Grudkin · Answer 2 · 2016-06-19T17:17:32.347

4

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, $\det (QQ^T) = 1 \Rightarrow (\det Q)^2 = 1\Rightarrow \det Q \neq 0)$ .

edited Jun 19 '16 at 17:17

answered Jun 19 '16 at 16:57

Anton Grudkin

2,955

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

2 Answers2