Answer to Question #100468 in Linear Algebra for Santos

Question

1. Determine whether the set of all vectors of the form (a, b ,c) such that b=a+c+1 is a subspace of R3 .

2. Determine whether the vector p is in the span {S}. Given S = { p_1,p_2,p_3} where
p_1=2+x+〖4x〗^2, p_2=1-x+〖3x〗^2, p_3=3+2x+〖5x〗^2; p=7+8x+9x^2.

3. Using two methods, verify that the set S is linearly independent.
v1 = (2, -2, 0), v2=(6,1,4), v3=(2,0,-4)

4. Express v as a linear combination of the vectors v_(1,), v_2,v_3.

v=(2,-1,3), v1=(1,0,0), v2=(2,2,0), v3=(3,3,3)

Answer 1

1. For $V_1 = (a, a+c+1, c), a \in \R, c \in \R$ to be a subspace of $\R^3$ , it must be a vector space itself. Let us check the axioms: $\bold v_1 + \bold v_2 = (a_1, a_1 + c_1 + 1, c_1) + (a_2, a_2 + c_2 + 1, c_2)$ $\bold v_1 + \bold v_2 = (a_1 + a_2, (a_1+a_2) + (c_1+c_2) + 2, c_1 + c_2) \notin V_1$, hence $V_1$ is not a subspace.
2. For $\bold p \in span(\bold p_1, \bold p_2, \bold p_3)$, there must exist unique $\lambda_1, \lambda_2, \lambda_3$, not all equal to zero, so that $\bold p = \lambda_1 \bold p_1 + \lambda_2 \bold p_2 + \lambda_3 \bold p_3$. Writing the last equation in vector form, obtain: $\left(\begin{array}{ccc|c} 2 & 1 & 3 & 7\\ 1 & -1 & 2 & 8\\ 16 & 9 & 25 & 9 \end{array}\right)$. Performing Gauss-Jordan elimination on last matrix, obtain $\left(\begin{array}{ccc|c} 1 & 0 & 0 & 60\\ 0 & 1 & 0 & -14\\ 0 & 0 & 1 & -33 \end{array}\right)$, hence $\lambda_1 = 60, \lambda_2 = -14, \lambda_3 = -33$ and vector $\bold p$ is in span of $\bold p_1, \bold p_2, \bold p_3$.
3. First way: $\bold v_1, \bold v_2, \bold v_3$ are linearly independent, if $\lambda_1 \bold v_1 + \lambda_2 \bold v_2 + \lambda_3 \bold v_3 = 0$ only when all $\lambda_i = 0$. Rewriting the last equation in vector form, obtain $\left(\begin{array}{ccc} 2 & 6 & 2\\ -2 & 1 & 0\\ 0 & 4 & -4 \end{array}\right)\left(\begin{array}{c} \lambda_1\\ \lambda_2\\ \lambda_3 \end{array}\right) = \bold 0$. Reducing the matrix, obtain $\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right)\left(\begin{array}{c} \lambda_1\\ \lambda_2\\ \lambda_3 \end{array}\right) = \bold 0$, from where all $\lambda_i = 0$, hence $\bold v_1, \bold v_2, \bold v_3$ are linearly independent. Second way: Vectors are linearly independent, if the determinant of matrix, spanned by vectors is zero. $det \left(\begin{array}{ccc} 2 & 6 & 2\\ -2 & 1 & 0\\ 0 & 4 & -4 \end{array}\right) = -8 - 16 - (-48) = -72 \neq 0$, hence vectors are linearly independent.
4. As in Task 2, solve corresponding linear system of equations: $\left(\begin{array}{ccc|c} 1 & 2 & 3 & 2\\ 0 & 2 & 3 & -1\\ 0 & 0 & 3 &3 \end{array}\right)$, from where $\lambda_1 = 3, \lambda_2 = -2,$ $\lambda_3 = 1$, hence $\bold v = 3 \bold v_1 - 2 \bold v_2 + \bold v_3$.