How many Hecke operators span the Hecke algebra?

Linear Algebra Question (Linear independence, subspace vectors span, linear combination)?

• Let v_1 = (1, 2, 2, 1), v_2 = (0, 2, 0, 1), v_3 = (-2, 0, -4, 3). a) Show that these vectors are linearly independent. b) What is the subspace of E^4 that they span, that is, given v = (y_1, y_2, y_3, y_4) how can we tell when v is a linear combination of v_1, v_2, and v_3? Please explain all steps. Thank you!

• Answer:

  a) Suppose that there exist scalars A, B, C such that A(1, 2, 2, 1) + B(0, 2, 0, 1) + C(-2, 0, -4, 3) = (0, 0, 0, 0). Combining the vectors together: (A - 2C, 2A + 2B, 2A - 4C, A + B + 3C) = (0, 0, 0, 0). Hence, we need A - 2C = 0 2A + 2B = 0 2A - 4C = 0 A + B + 3C = 0. Since A = 2C from the first equation, substituting this into the third equation yields C = 0. Thus, A = 0. Then the second equation yields B = 0. Since A = B = C = 0 is the only solution, the vectors are linearly independent. -------------- B) This is related to part a, but a trifle more general. Suppose that there exist scalars A, B, C such that A(1, 2, 2, 1) + B(0, 2, 0, 1) + C(-2, 0, -4, 3) = (y₁, y₂, y₃, y₄). Combining the vectors together: (A - 2C, 2A + 2B, 2A - 4C, A + B + 3C) = (y₁, y₂, y₃, y₄). Hence, we need A - 2C = y₁ 2A + 2B = y₂ 2A - 4C = y₃ A + B + 3C = y₄. Now, solve this via Gaussian Elimination: [1 0 -2 | y₁] [2 2 0 | y₂] [2 0 -4 | y₃] [1 1 3 | y₄] 2R1 - R3 --> R3 and 2R4 - R2 --> R4, and finally (1/2) R2 --> R2: [1 0 -2 | y₁] [1 1 0 | y₂/2] [0 0 0 | 2y₁-y₃] [0 0 6 | 2y₄-y₂] R2 - R1 --> R2 and R3<-->R4 followed by (1/6)R3 --> R3:: [1 0 -2 | y₁] [0 1 2 | y₂/2 - y₁] [0 0 1 | (2y₄-y₂)/6] [0 0 0 | 2y₁-y₃]. In order for this to have a solution, we need 2y₁ - y₃ = 0 <==> y₃ = 2y₁. So, the desired subspace is {(y₁, y₂, y₃, y₄) in E^4 : y₃ = 2y₁}. I hope this helps!