Let the column vectors of a $3 × 3$ matrix $A$ form an orthonormal basis. Explain why $A^T = A^{−1}$ . Let the column vectors of a $3 × 3$ matrix $A$ form an orthonormal basis. Explain why $A^T = A^{−1}$
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My Attempt:  $AA^T=I$ if and only if $A^TA=I$. So $A$ is orthogonal if and only if $A^T$ is orthogonal.
Is this a correct explanation/ can it be put in a more better way?
 A: Let $A=\begin{pmatrix}a_1 & a_2 & a_3 \end{pmatrix}$. Since the columns form an orthonormal basis, $a_i^T a_j=\delta_{ij}$, so $A^TA=I$. Since $\mathrm{rank}(A)=3$, $A^{-1}$ exists, and thus $A^TAA^{-1}=IA^{-1}$, and therefore $A^T=A^{-1}$.
A: If you write $A=\begin{bmatrix}c_1&\cdots&c_n\end{bmatrix}$ in terms of its columns, then 
$$
(A^TA)_{kj}=c_k^Tc_j.
$$
The fact that the columns are orthonormal make $c_k^Tc_j=\delta_{k,j}$, and so $A^TA=I$. Then $A$ is invertible, and multiplying the equality on the right by $A^{-1}$ we get $$A^T=A^{-1}.$$ 
A: Recall $X\cdot Y = (x_i\cdot y_j)$ is a matrix of dot products. But then if the rows of $A$ form an orthonormal system, then $AA^T = (a_i\cdot a_j) = (\delta_{ij})=I$ and similarly for the transpose of this which is $A^TA=I^T=I$. So $A^T$ satisfies the definition of the inverse matrix.
A: A more detailed version of $AA^T = I \iff A^T A = I$:
\begin{align}
A A^T &= I \iff \\
A^{-1} (A A^T) &= A^{-1} I \iff \\
(A^{-1} A) A^T &= A^{-1} \iff \\
A^T &= A^{-1} \iff \\
A^T A &= A^{-1} A = I
\end{align}
What is missing in your agrument is a reason, why $A A^T = I$ holds here.
