Suppose Q^T equals Q^-1 (transpose equals inverse, so Q^TQ = I).

1. Show that the columns q₁, ..., q_n are unit vectors:

   \[ ||q_{i}||^{2} = 1 \]

2. Show that every two columns of Q are perpendicular:

   \[ q_{1}^{T}q_{2} = 0 \]

3. Find a 2 by 2 example with first entry q₁₁ = cos θ

1. Observe that:

   \[ ||q_{i}||^{2} \]

   is equal to the dot product of a column by itself, and every element in the resultant matrix of Q^TQ is the dot product of two columns in Q. Since Q^TQ = I, it means when the two columns are the same column in Q, the dot product is equal to 1, corresponding to the diagonal of I.


2. Based on similar observation, every other element in I that doesn't fall on the diagonal corresponds to the dot product of two different columns in Q, and is equal to zero.


3. One obvious example that would work:

   \[ Q = \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} \]

   as:

   \[ Q^T Q = \begin{vmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{vmatrix} \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} = I \]

# posted by rot13(Unafba Pune) @ 8:35 PM