# Using Column vectors in proving a statement about real matrices

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 $$A^TA = \pmatrix{c_1^T \\ \vdots \\ c_n^T}\pmatrix{c_1 & \cdots & c_n} = \pmatrix{c_1^Tc_1 & \cdots & c_1^Tc_n\\ \vdots & \ddots & \vdots \\ c_n^Tc_1 & \cdots & c_n^Tc_n}.$$
• 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 ? Sep 21, 2021 at 19:10
• That's right.${}$ Sep 21, 2021 at 19:11