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I am completely confused about this question. Any help would be greatly appreciated.

A square matrix A is called orthogonal if (A^T)A = I_n.

(a) Show that

A = [cos(theta), -sin(theta)]

[sin(theta), cos(theta)]

is orthogonal.

b) Assume that A, B are orthogonal matrices of the same size. Show that AB is also orthogonal.

(c) Let vector v_1, vector v_2,...., vector v_n be the columns of an orthogonal matrix A. Show that the vector v_i's are mutually perpendicular and unit vectors.

a) do the matrix multiplication and use the trig identity

cos^2(theta) + sin^2(theta) = 1.

b) AB orthogonal means (AB)^T(AB) = I. Using properties of the transpose

(AB)^T(AB) = B^TA^T(AB) = B^T(A^TA)B = B^TB = I.

c) let the matrix A be written (aij) where 1 ≤ i,j ≤ n are the indices of the matrix entries. Also, let v_i = (i1, i2 ... in)^T be the ith column vector of A. Matrix multiplication can be written

A^TA = (aij)^T(aij) = ∑∑vi･vj, sum over i and j

i.e., equals the dot-product of the ith row and jth column. Since the entries of A^TA are 1 along the diagonal (i=j) and zero otherwise (i≠j)

∑∑vi･vj = δij = Kronecker delta function ⇒ vi are mutually perpendicular and unit length.

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