I understand that if A is orthogonal, then $A^{-1} = A^T$. However, i cannot seem to understand the proof.
Can someone put in some numbers in the proof. This site tries to explain it but i cannot seem to get $A^TA$ becoming an identity matrix.
Thanks.
On
Let $A$ be an orthogonal matrix. Let us denote $A$ using columns: $$A = \begin{pmatrix} e_1 & e_2 & \ldots & e_n \end{pmatrix}$$ where the columns $e_1, e_2, \ldots, e_n$ are orhtonormal. This means that $e_i \cdot e_j = \delta_{ij}$, where $\cdot$ denotes the inproduct and $$\delta_{ij} = \begin{cases} 1 &\text{if } i = j\\ 0 &\text{otherwise} \end{cases}$$ Note that $A^{t}$, the transposed matrix is equal to $$\begin{pmatrix} e_1^{t} \\ e_2^{t}\\ \ldots \\ e_n^{t} \end{pmatrix}$$ so $e_i^t$ is a rowvector. If we now compute the following product $$A^tA = \begin{pmatrix} e_1^{t} \\ e_2^{t}\\ \ldots \\ e_n^{t} \end{pmatrix}$$ we find $$\begin{pmatrix} e_1^t e_1 & e_1^te_2 & \ldots &e_1^te_n\\ e_2^te_1 & e_2^te_2 & \ldots &e_2^te_n\\ \vdots & \vdots & \ldots & \vdots\\ e_n^te_1 & e_n^te_2 & \ldots &e_n^te_n \end{pmatrix} = \begin{pmatrix} e_1 \cdot e_1 & e_1 \cdot e_2 & \ldots &e_1 \cdot e_n\\ e_2\cdot e_1 & e_2\cdot e_2 & \ldots &e_2\cdot e_n\\ \vdots & \vdots & \ldots & \vdots\\ e_n\cdot e_1 & e_n\cdot e_2 & \ldots & e_n\cdot e_n \end{pmatrix}$$
which is the identity matrix. Therefore we have that (by uniqueness of the inverse matrix) that $A^t = A^{-1}$.
Suppose $A$ an orthogonal matrix. Then we can write $A$ as $A = (a^1,a^2,...,a^n)$. With $a^i$ the column vectors of $A$.
Then we know that $a^1,a^2,...,a^n$ are pairwise orthogonal. E.g. $(a^i,a^j) = 0$ with $i≠j$ and $(a^i,a^j)= 1$ with $i=j$.
Now if you do the matrix multiplication $A^TA$ or $AA^T$ only the positions $a^{i,i}$ become ones, all others zeros.