I want to prove this statement: if A is a matrix with real entries given by
\begin{equation*}
A_{n,n} =
\begin{bmatrix}
c_{1} & c_{2} & \cdots c_{n}
\end{bmatrix}
\end{equation*}
where $c_{i}$ is a column vector and {$c_{1}$,$c_{2}$,...,$c_{n}$} form an orthonormal basis of $\mathbb{R}^n$ then $A^{-1}$=$A^T$.
I thought of starting with $AA^T$=$A^TA$=$Id_n$ but
$\left(\begin{matrix}c_{1}....c_{n}\\\end{matrix}\right)\left(\begin{matrix}c_{1}\\.\\.\\.\\c_{n}\\\end{matrix}\right)$ = $\sum$ $c_{i}^2$
This seems erroneous because $AA^T$ is of order n$\times$n and the entries are different from the one big summation obtained here. Kindly provide me some directions to prove the above statement possibly using only $c_{i}$.
First of all, your matrix multiplication is incorrect. We have $$ AA^T = \pmatrix{c_1 & \cdots & c_n} \pmatrix{c_1^T\\ \vdots \\ c_n^T} = \sum_{i=1}^n c_ic_i^T. $$ This matrix is of the correct size since each matrix $c_ic_i^T$ has size $n \times n$.
Second, you have tried to show that $AA^T = I$, which I would say is not the most straightforward approach. Instead, consider the product $$ A^TA = \pmatrix{c_1^T \\ \vdots \\ c_n^T}\pmatrix{c_1 & \cdots & c_n} = \pmatrix{c_1^Tc_1 & \cdots & c_1^Tc_n\\ \vdots & \ddots & \vdots \\ c_n^Tc_1 & \cdots & c_n^Tc_n}. $$