Let A =  a b c d , B =  e f g h ∈ M2,2(R) and define hA, Bi = ae + bf + 2cg + 2dh. (a) Show hA, Bi is an inner product of M2,2(R).

To show that hA, Bi is an inner product on M2,2(R), we need to verify the four properties of an inner product:

1. Conjugate Symmetry: hA, Bi = hB, Ai
2. Linearity in the first argument: hλA + μB, Ci = λhA, Ci + μhB, Ci for all λ, μ in R and all A, B, C in M2,2(R)
3. Positive-definiteness: hA, Ai ≥ 0 with equality if and only if A = 0
4. Non-degeneracy: hA, Ai = 0 implies A = 0

Let's verify each property:

1. Conjugate Symmetry:

hA, Bi = ae + bf + 2cg + 2dh = hB, Ai

2. Linearity in the first argument:

Let λ, μ be real numbers and C be another 2x2 matrix with elements i, j, k, l. Then,

hλA + μB, Ci = hλ[a b; c d] + μ[e f; g h], [i j; k l] = λ(ai + bj + 2ck + 2dl) + μ(ei + fj + 2gk + 2hl) = λhA, Ci + μhB, Ci

3. Positive-definiteness:

hA, Ai = a^2 + b^2 + 2c^2 + 2d^2 ≥ 0, with equality if and only if a = b = c = d = 0, i.e., A = 0

4. Non-degeneracy:

If hA, Ai = 0, then a^2 + b^2 + 2c^2 + 2d^2 = 0, which implies a = b = c = d = 0, i.e., A = 0

Therefore, hA, Bi is an inner product on M2,2(R).