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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 $c_{i}$.

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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 $$ A^TA = \pmatrix{c_1^T \\ \vdots \\ c_n^T}\pmatrix{c_1 & \cdots & c_n} = \pmatrix{c_1^Tc_1 & \cdots & c_1^Tc_n\\ \vdots & \ddots & \vdots \\ c_n^Tc_1 & \cdots & c_n^Tc_n}. $$

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  • $\begingroup$ Thanks for the clarification on matrix multiplication. So $A^TA$ is a diagonal matrix since dot product of two different column vectors is 0. Since the basis is orthonormal, norm of each column vector is 1 which means all the diagonal elements are 1. Thus $ A^{-1}$ =$A^T$. Is this right @Ben Grossmann ? $\endgroup$ Sep 21, 2021 at 19:10
  • $\begingroup$ That's right.${}$ $\endgroup$ Sep 21, 2021 at 19:11

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