Question

Show that if the columns of a matrix \(Q\) form an orthonormal basis for \(C(Q)\), then \(Q^TQ=1\)

Answer 1

that one should be an I, the identity matrix

Answer 2

it's really tough to do these proofs as my textbook doesn't really do them

Answer 3

If the columns form an orthonormal basis, then

Q = (e_1,e_2,e_3,....e_N)

where each of the e_i is a basis vector. Since the basis is orthonormal, all the dot products obey e^T_i dot e_j = 1 if i=j, 0 otherwise. All that's left to do then is for Q^T
and do the multiplications......

Answer 4

thanks, just throwing the brackets on your formulas! If the columns form an orthonormal basis, then \(Q = (e_1,e_2,e_3,....e_N)\) where each of the \(e_i\) is a basis vector. Since the basis is orthonormal, all the dot products obey \[e^T_i \cdot e_j = 1 \text{ if } i=j, 0 \text{ otherwise}\]. All that's left to do then is for \(Q^T\) and do the multiplications......