Show that e1 = (2, 1, 0) and e2 = (1, 1, 0) form a basis of R3

Answer 1

To show that e1 = (2, 1, 0) and e2 = (1, 1, 0) form a basis of ℝ^3, we need to show two things: linear independence and span.

1) Linear Independence:
To show that e1 and e2 are linearly independent, we need to show that the only solution to the equation c1 * e1 + c2 * e2 = (0, 0, 0) is c1 = 0 and c2 = 0.

Let's denote c1 and c2 as the scalars that multiply each vector:

c1 * e1 + c2 * e2 = (0, 0, 0)

Substituting the values for e1 and e2:

c1 * (2, 1, 0) + c2 * (1, 1, 0) = (0, 0, 0)

Expanding the equation:

(2c1 + c2, c1 + c2, 0) = (0, 0, 0)

This gives us the following system of equations:

2c1 + c2 = 0 (Equation 1)
c1 + c2 = 0 (Equation 2)

We can solve this system of equations to find the values of c1 and c2:

From Equation 2, we can express c1 in terms of c2:

c1 = -c2

Substituting this into Equation 1:

2(-c2) + c2 = 0

-2c2 + c2 = 0

-c2 = 0

c2 = 0

Substituting c2 = 0 back into Equation 2:

c1 + 0 = 0

c1 = 0

Thus, the only solution to c1 * e1 + c2 * e2 = (0, 0, 0) is c1 = c2 = 0.

Since the only solution is the trivial solution, e1 and e2 are linearly independent.

2) Span:
To show that e1 and e2 span ℝ^3, we need to show that any vector in ℝ^3 can be expressed as a linear combination of e1 and e2.

Let (a, b, c) be an arbitrary vector in ℝ^3. We want to find c1 and c2 such that c1 * e1 + c2 * e2 = (a, b, c).

(c1 * 2, c1 * 1 + c2 * 1, 0) = (a, b, c)

This gives us the following system of equations:

2c1 = a (Equation 3)
c1 + c2 = b (Equation 4)
0 = c (Equation 5)

From Equation 5, we can see that c must be 0.

Substituting this back into Equation 3:

2c1 = a

c1 = a/2

Substituting this back into Equation 4:

(a/2) + c2 = b

c2 = b - (a/2)

So, for any vector (a, b, c) in ℝ^3, we can find c1 and c2 such that c1 * e1 + c2 * e2 = (a, b, c).

Therefore, e1 = (2, 1, 0) and e2 = (1, 1, 0) form a basis of ℝ^3.