Proof of orthogonal matrix property: $A^{-1} = A^t$ I have prooved this orthogonal property. Please correct it or show your version of the proof if I am wrong:
$A^{-1} = A^t$
$A^{-1} \times A = A^t \times A$
$I = I$
I appreciate your answer.
 A: Further to Kaster's VERY good proof above, we can continue as follows
$$
Q^T \cdot Q  = I
$$

But we know
$$
 Q^{-1} \cdot Q = I
$$

So

$$
Q^T \cdot Q  = I
$$
$$
(Q^T \cdot Q) \cdot Q^{-1}  = (I) \cdot Q^{-1}
$$
$$
Q^T \cdot (Q\cdot Q^{-1})  = Q^{-1}
$$
$$
Q^T \cdot (I)  = Q^{-1}
$$
$$
Q^T = Q^{-1}
$$


Which was ultimately what you wanted to show.
A: There are two main definitions of orthogonality. Accepting one you can prove another. Since you need to prove $Q^T = Q^{-1}$, you should define orthogonality as follows:

An orthogonal matrix is a square matrix with real entries whose columns and rows are orthogonal unit vectors.

So, let's say you have a matrix $Q = [q_1, q_2, \ldots, q_n]$, where $q_i$ is a unit column vector and $q_i^T q_j = \delta_{ij}$ due to the orthogonality. Now, find its transpose
$$
Q^T = \left [ \begin{array}{c}
q_1^T \\ q_2^T \\ \vdots \\ q_n^T
\end{array} \right ]
$$
where $q_i^T$ is a row vector. So
$$
Q^T \cdot Q = \left [ \begin{array}{c}
q_1^T \\ q_2^T \\ \vdots \\ q_n^T
\end{array} \right ] \cdot [q_1, q_2, \ldots, q_n] = \left [ \begin{array}{ccccc}
q_1^T q_1 & q_1^T q_2 & \cdots  & q_1^T q_n \\
q_2^T q_1 & q_2^T q_2 & \cdots & q_2^T q_n \\
\vdots & \vdots & \ddots & \vdots \\
q_n^T q_1 & q_n^T q_2 & \cdots & q_n^T q_n
\end{array}\right ] = \left [ \begin{array}{ccccc}
1 & 0 & \cdots  & 0 \\
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1
\end{array}\right ] = I
$$
which means $Q^T = Q^{-1}$.
