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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.

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    $\begingroup$ Just don't assume what you need to prove or its equivalent statements. $\endgroup$ – Kaster Oct 23 '13 at 18:55
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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 $Q^T = Q^{-1}$.

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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.

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