Question

Let B = far a2, a3) be an ordered basis of
R3 with al = (1, 0, -1), a2 = (1, 1, 1),
a3 = (1, 0, 0). Write the vector v = (a, b, c) as
a linear combination of the basis vectors
from B.
(b) Suppose al = (1, 0, 1), a2 = (0, 1, -2) and
a3 = (-1, -1, 0) are vectors in R3 and
f : R3 -> R is a linear functional such that
f(al) = 1, f(a2) = -1 and f(a3) = 3. If
a = (a, b, c) E R3, find f(a).

Answer 1

Answer on Question #58697 – Math – Linear Algebra

Question

(a) Let $B = \{a1, a2, a3\}$ be an ordered basis of $R3$ with $a1 = (1, 0, -1)$, $a2 = (1, 1, 1)$, $a3 = (1, 0, 0)$. Write the vector $v = (a, b, c)$ as a linear combination of the basis vectors from $B$.

Solution

We note that the vectors $\{a1, a2, a3\}$ are linearly independent. If $v = (a, b, c)$ is the linear combination of $\{a1, a2, a3\}$, then $v = \begin{bmatrix} a \\ b \\ c \end{bmatrix} = x_1 a1 + x_2 a2 + x_3 a3 = x_1 \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} + x_2 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + x_3 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$, where unknown constants are $x_1, x_2, x_3 \in R$.

Answer: $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = x_1 \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} + x_2 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + x_3 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$.

Question

(b) Suppose $a1 = (1, 0, 1)$, $a2 = (0, 1, -2)$ and $a3 = (-1, -1, 0)$ are vectors in R3 and f: $R3 \to R$ is a linear functional such that $f(a1) = 1$, $f(a2) = -1$ and $f(a3) = 3$. If $a = (a, b, c) \in R3$, find f(a).

Solution

We note that the linear functional has properties: $f(\pmb{u} + \pmb{v}) = f(\pmb{u}) + f(\pmb{v})$ and $f(c\pmb{u}) = c f(\pmb{u})$, where $c$ is a constant. It is known that the inner product has these properties. Let the unknown vector $\pmb{w}$ be $\pmb{w}(w_1, w_2, w_3)$ and the dependent linear functional is $f(\pmb{u}) = \pmb{u} \cdot \pmb{w}$. We have conditions:

$$f (\boldsymbol {a} \mathbf {1}) = 1 \Leftrightarrow \boldsymbol {a} \mathbf {1} \cdot \boldsymbol {w} = 1 \Leftrightarrow (1, 0, 1) \cdot \left(w _ {1}, w _ {2}, w _ {3}\right) = 1 \Leftrightarrow 1 w _ {1} + 0 + 1 w _ {3} = 1;$$ $$f (\boldsymbol {a} \mathbf {2}) = - 1 \Leftrightarrow \boldsymbol {a} \mathbf {2} \cdot \boldsymbol {w} = - 1 \Leftrightarrow (0, 1, - 2) \cdot \left(w _ {1}, w _ {2}, w _ {3}\right) = - 1 \Leftrightarrow 0 + 1 w _ {2} - 2 w _ {3} = - 1;$$ $$f (\boldsymbol {a} \mathbf {3}) = 3 \Leftrightarrow \boldsymbol {a} \mathbf {3} \cdot \boldsymbol {w} = 3 \Leftrightarrow (- 1, - 1, 0) \cdot \left(w _ {1}, w _ {2}, w _ {3}\right) = 3 \Leftrightarrow - 1 w _ {1} - 1 w _ {2} + 0 = 3.$$

We have the system of linear equations:

$$\left\{ \begin{array}{l} 1 w _ {1} + 0 + 1 w _ {3} = 1, \\ 0 + 1 w _ {2} - 2 w _ {3} = - 1, \\ - 1 w _ {1} - 1 w _ {2} + 0 = 3. \end{array} \right.$$

that gives the solution

$$\left\{ \begin{array}{l} w _ {1} = 4, \\ w _ {2} = - 7, \\ w _ {3} = - 3. \end{array} \right.$$

Then

$$\boldsymbol {w} \left(w _ {1}, w _ {2}, w _ {3}\right) = (4, - 7, - 3) \text{ and}$$ $$f (\boldsymbol {a}) = \boldsymbol {a} \cdot \boldsymbol {w} = (a, b, c) (4, - 7, - 3) = 4 a - 7 b - 3 c.$$

Answer: $f(\pmb{a}) = 4a - 7b - 3c$.

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