We were studying vector transformations using matrices and rotation matrices in our physics class and the professor's proof for some rotation matrix $R$ being orthogonal was to take a column vector $A$ and use the fact that $|A| = |RA|\implies (RA)^T(RA)=A^TA\implies A^TR^TRA=A^TA\implies R^TR=I$. He didn't provide an explanation for the last step, stating that it was obvious so I was wondering if I could prove it some other way. Here's what I came up with.
Let us say that we have a linear transformation matrix $M=\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}$. Multiplying this by any column vector $A=\begin{bmatrix}x\\y\\z\end{bmatrix}$ represents how $A$ will be transformed under a linear transformation where $\hat i,\hat j$ and $\hat k$ get transformed to $\begin{bmatrix}a\\d\\g\end{bmatrix}$, $\begin{bmatrix}b\\e\\h\end{bmatrix}$ and $\begin{bmatrix}c\\f\\i\end{bmatrix}$, respectively. Let's call these $v_1$, $v_2$ and $v_3$.
Now, if the linear transformation depicts a proper rotation, then we can come up with the following conditions :
- $|v_1|=|v_2|=|v_3|=1$
- $v_1\cdot v_2 = v_2\cdot v_3 = v_3\cdot v_1 = 0$ as the transformed vectors should be mutually perpendicular
- $[v_1~~~v_2~~~v_3]>0$ to preserve the system's handed-ness
Now, if we evaluate $M^TM$, we get $$\begin{bmatrix}a^2+d^2+g^2&&ad+be+cf&&ag+bh+ci\\ad+be+cf&&b^2+e^2+h^2&&gd+eh+fi\\ag+bh+ci&&gd+eh+fi&&c^2+f^2+i^2\end{bmatrix}$$ and making use of the previous conditions, we find out that $M^TM=I$
On top of that, from condition (3), we know that $|M|>0$ and $M^TM=I\implies|M|=\pm1$. So $|M|=1$, which is a condition for a rotation to be proper.
Is the proof valid? Are the conditions valid? Did I miss something?
Thank you!