Question

Let V be the subspace of P3 spanned by the the following set
{1-x^2+x^3 , 2+x-x^2+x^3 , 1+2x+x^2-x^3}

(a) Show that f(x) = x + x^2 - x^3 ∈ V .
(b) Show that g(x) = 1 + x - x^2 +x^3 is not an element of V .
(c) Find a basis for V which contains f(x).
(d) Find a basis for P3 which contains g(x).

Answer 1

Answer on Question #68577 - Math - Linear Algebra

Let $V$ be the subspace of $P_3$ spanned by the following set

$$\{1 - x^2 + x^3, 2 + x - x^2 + x^3, 1 + 2x + x^2 - x^3\}$$

a) Show that $f(x) = x + x^2 - x^3 \in V$.

b) Show that $g(x) = 1 + x - x^2 + x^3$ is not an element of $V$.

c) Find a basis for $V$ which contains $f(x)$.

d) Find a basis for $P_3$ which contains $g(x)$.

Solution:

a)

$$\begin{array}{l} [[v_1] \quad [v_2] \quad [v_3] \quad [f]]] = \begin{bmatrix} 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ -1 & -1 & 1 & 1 \\ 1 & 1 & -1 & -1 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 0 & 1 & 2 & 1 \\ 0 & -1 & -2 & -1 \end{bmatrix} \rightarrow \\ \rightarrow \begin{bmatrix} 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \end{array}$$

Column $[f]$ is not a pivot column, so $f(x) \in V$

b)

$$\begin{array}{l} [[v_1] \quad [v_2] \quad [v_3] \quad [g]]] = \begin{bmatrix} 1 & 2 & 1 & 1 \\ 0 & 1 & 2 & 1 \\ -1 & -1 & 1 & -1 \\ 1 & 1 & -1 & 1 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 0 & 1 & 2 & 0 \\ 0 & -1 & -2 & 0 \end{bmatrix} \rightarrow \\ \rightarrow \begin{bmatrix} 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \end{array}$$

Column $[g]$ is a pivot column, so $g(x) \notin V$

c) Columns $[f]$ and $[v_1]$ are pivot columns of the matrix $[[f] \quad [v_1] \quad [v_2] \quad [v_3]]$. So $f(x)$ and $v_1(x)$ form a basis for $V$.

d) Any three vectors of the standard basis together with $g(x)$ form a basis for $P_3$.

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