Orthogonal matrix multiplication by orthonormal matrix???

Question

$a=6b_1-2b_2+3b_3$ vector given in orthonormal basis $\{b_i\} $where $i=1,2,3$

Orthogonal tensor : $Q=cosθb_1⊗b_1+sinθb_1⊗b_2-sinθb_2⊗b_1+cosθb_2⊗b_2+b_3⊗b_3$

Can you tell me how to matrix representation Qa?

Thank you...

Answer 1

Let us take some examples

$$b_1\otimes b_1=\pmatrix{1\\0\\0} \pmatrix{1&0&0}=\pmatrix{1&0&0\\0&0&0\\0&0&0}$$

$$b_1\otimes b_2=\pmatrix{1\\0\\0}\pmatrix{0&1&0}=\pmatrix{0&1&0\\0&0&0\\0&0&0}$$

$$b_2\otimes b_1=\pmatrix{0\\1\\0}\pmatrix{1&0&0}=\pmatrix{0&0&0\\1&0&0\\0&0&0}$$

etc...

Thus, gathering all these contributions:

$$Q=\pmatrix{\cos(\theta)&0&0\\ 0&0&0\\0&0&0}+\pmatrix{0&\sin(\theta)&0\\0&0&0\\0&0&0}+\cdots = \pmatrix{\ \ \cos(\theta)&\sin(\theta)&0\\ -\sin(\theta)&\cos(\theta)&0\\0&0&1}$$

Interpretation: it is a rotation matrix with angle $-\theta$ around the $z$-axis.

Thus, with $A=\pmatrix{6\\-2\\3}$,

$$QA=\pmatrix{\ \ \cos(\theta)&\sin(\theta)&0\\ -\sin(\theta)&\cos(\theta)&0\\0&0&1}\pmatrix{6\\-2\\3}=\pmatrix{6\cos(\theta)-2\sin(\theta)\\-6\sin(\theta)-2\cos(\theta)\\3}$$