Question

1.Let B = (a1,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.

2.Suppose al= (1, 0, 1), a2= (0, 1, -2) and
a3 = (-1, -1, 0) are vectors in R3and
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

We need to write the vector $"v"$ linear combination of the basis vectors of B

From the given data

$a_1 = (1,0,1), \space a_2 = (1, 1, 1) \space and \space a_3 =(1, 0, 0)$

Now we can write the vector $v$ as linear combination of the vectors $a_1, a_2 \space and \space a_3$

$$v = xa_1 + ya_2 + za_3$$

$$(a, b, c)= x(1,0,1) + y (1,1,1) + z (1, 0, 0)$$

$$(a, b, c)= (x,0,x) + (y, y, y) + (z, 0, 0)$$

$$(a, b, c)= (x+ y + z, y ,x+ y)$$

compare both sides ,

$$x + y + z = a\\ y=b \\ x + y = c$$

Plug $y = b$ in the equation $x + y = c$ then $x = c - y = c - b$

Now

$$x + y + z =a$$

$$c - b + b + z = a$$

$$z = a - c$$

Now we can write the linear combination as

$(a, b, c)= (c-b)( 1,0,1) + b (1,1,1) + (a-c) (1, 0, 0)$

2).

$a_1 =(1, 0, 1), a_2 = (0, 1, 2) \space and \space a_3 = (-1, -1, 0)$

Given, f is a linear function

and $f (a_1) = 1, f(a_2) = -1 , f(a_3) = 3$

Let the function f(x) = ax+by + c z

$f (a_1) = 1$ It means,

$f (1, 0, 1) =1$

$a(1) + b(0) +c(1) =1 \\ a + c =1 ................................(A)$

$f(a_2) = -1$

$f(0, 1, 2) = -1\\ a(0) + b(1) +c(-2) =-1\\ b -2 c = -1 .................................(B)$

$f(a_3) = 3 \\ f(-1, -1, 0) = 3\\ a(-1) +b(-1) +c(0) = 3\\ -a -b = 3 .............................(C)$

From the equations A, B and C

$c = 1- a \space$

Plug this c in the equation B, then

$b -2 + 2a =- 1 \\ b + 2a = 1 .....................(D)$

Equation (C) is $- a - b = 3$

From equation C and D

$$b + 2a - a - b = 1 + 3 \\ a =4$$

$$c = 1- a = 1- 4 = - 3$$

$$b - 2c = -1 \\ b - 2(-3) = -1 \\ b + 6 = -1\\ b = -7$$

So, the answer is

$$(a, b, c) = ( 4, -7, - 3 )$$

$f(a) = f(a, b, c) = f(4, -7, -3) = 4x - 7y -3z$