Answer:
Thus for the vectors v_1, v_2, v_p there are scalars c_1, c_2, c_p not all zeros, such that c_1v_1 +c_2v_2+... +c_pv_p = 0. It means that the vectors v_1, v_2, v_p are linearly dependent in contradiction with the fact that the vectors form a basis for H. So the assumption that T(v_1), T(v_2),..., T(v_p) are linearly dependent is false, proving the required.
Step-by-step explanation:
Let B = {v_1 ,v_2,..., v_p} be a basis of H, that is dim H = p and for any v ∈ H there are scalars c_1 , c_2, c_p, such that v = c_1*v_1 + c_2*v_2 +....+ C_p*V_p It follows that
T(v) = T(c_1*v_1 + c_2v_2 + ••• + c_pV_p) = c_1T(v_1) +c_2T(v_2) + c_pT(v_p)
so T(H) is spanned by p vectors T(v_1),T(v_2), T(v_p). It is enough to prove that these vectors are linearly independent. It will imply that the vectors form a basis of T(H), and thus dim T(H) = p = dim H.
Assume in contrary that T(v_1 ), T(v_2), T(v_p) are linearly dependent, that is there are scalars c_1, c_2, c_p not all zeros, such that
c_1T(v_1) + c_2T(v_2) +.... + c_pT(v_p) = 0
T(c_1v_1) + T(c_2v_2) +.... + T(c_pv_p) = 0
T(c_1v_1+ c_2v_2 ... c_pv_p) = 0
But also T(0) = 0 and since T is one-to-one, it follows that c_1v_1 + c_2v_2 +.... + c_pv_p = O.
Thus for the vectors v_1, v_2, v_p there are scalars c_1, c_2, c_p not all zeros, such that c_1v_1 +c_2v_2+... +c_pv_p = 0. It means that the vectors v_1, v_2, v_p are linearly dependent in contradiction with the fact that the vectors form a basis for H. So the assumption that T(v_1), T(v_2),..., T(v_p) are linearly dependent is false, proving the required.
