P2.7B

• • Given a basis v1, v2, v3, and v4 in V show that v1+v2, v2+v3, v3+v4, v4 is also a basis of V.

  • Answer:

    Suppose \[ a_1(v_1+v_2)+a_2(v_2+v_3)+a_3(v_3+v_4)+a_4 v_4=0. \] Collecting coefficients of the original basis: \[ (a_1)v_1+(a_1+a_2)v_2+(a_2+a_3)v_3+(a_3+a_4)v_4=0. \] Since \(\{v_1,v_2,v_3,v_4\}\) is a basis, each coefficient must vanish: \[ a_1=0,\quad a_2=0,\quad a_3=0,\quad a_4=0. \] Hence the new vectors are linearly independent.

    Because \(V\) is 4-dimensional, 4 linearly independent vectors form a basis. Therefore \(\{v_1+v_2,\; v_2+v_3,\; v_3+v_4,\; v_4\}\) is indeed a basis.

    Alternative view:

    Writing the new vectors in terms of the old basis gives the change-of-basis matrix \[ A= \begin{bmatrix} 1&0&0&0\\ 1&1&0&0\\ 0&1&1&0\\ 0&0&1&1 \end{bmatrix}. \] This is lower triangular with determinant \(1\ne 0\). Since \(A\) is invertible, the new set is also a basis.