# Why is the inverse of an orthogonal matrix equal to its transpose? [duplicate]

I don't get why that's the case. Or is it a definition? The way the concept was presented to me was that an orthogonal matrix has orthonormal columns. And that's it.

• math.stackexchange.com/questions/537217/… Commented Sep 21, 2016 at 18:15
• A classical example (that does not "explain" the general case): rotation $R_a=\pmatrix{\cos(a)&-sin(a)\\ \sin(a)&\cos(a)}$ whose inverse is rotation by $-a$: $(R_a)^{-1}=R_{-a}=\pmatrix{\cos(-a)&-\sin(-a)\\ \sin(-a)&\cos(-a)}$=$\pmatrix{\cos(a)&\sin(a)\\ -\sin(a)&\cos(a)}=(R_a)^T.$ Commented Sep 21, 2016 at 18:36

## 2 Answers

Let $A$ be an $n\times n$ matrix with real entries. The matrix $A$ is orthogonal if the column and row vectors are orthonormal vectors. In other words, if $v_1,v_2,\cdots,v_n$ are column vectors of $A$, we have

$v_i^Tv_j=\begin{cases}1 \quad\text{if }i=j\\ 0\quad\text{if } i\neq j\end{cases}$

If $A$ is an orthogonal matrix, using the above information we can show that $A^TA=I$. Since the column vectors are orthonormal vectors, the column vectors are linearly independent and thus the matrix $A$ is invertible. Thus, $A^{-1}$ is well defined.

Since $A^TA=I$, we have $(A^TA)A^{-1}=IA^{-1}=A^{-1}$. Since matrix multiplication is associative, we have $(A^TA)A^{-1}=A^T(AA^{-1})$, which equals $A^T$. We therefore have $A^T=A^{-1}$.

• Thanks, this was the simple explanation I needed :) Commented Sep 25, 2016 at 15:16
• @SupreethNarasimhaswamy isn't it a circular reasoning, assuming $A^TA=I$ to prove that $A^T=A^{-1}$? Commented Nov 30, 2019 at 20:52
• @ZeeshanAli It's not an assumption -- it follows directly from the orthonormality of the columns. Commented Oct 28, 2020 at 21:37

Well, if $Q$ is any $n \times n$ matrix with real entries with column vectors $v_1, v_2, \ldots, v_n$, we have

$$(Q^T Q)_{ij} = \sum_{k=1}^n Q^T_{ik} Q_{kj} = \sum_{k=1}^n Q_{ki} Q_{kj} = \sum_{k=1}^n (v_i)_k (v_j)_k = \langle v_i, v_j \rangle$$

Now, assume that $Q$ is an orthogonal matrix, i.e its column vectors $v_i$ are orthonormal. What does this tell you about the entries of $Q^T Q$?

• To put this into plain English, the $ij$-th element of $Q^TQ$ is the dot product of the $i$th and $j$th columns of $Q$.
– amd
Commented Sep 22, 2016 at 5:21