Show that e1 = (2, 1, 0) and e2 = (1, 1, 0) form a basis of R3
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.
To show that e1 = (2, 1, 0) and e2 = (1, 1, 0) form a basis of R3, we need to show that they are linearly independent and span R3.
1. Linear Independence:
To show that e1 and e2 are linearly independent, we need to prove that the only solution to the equation a*e1 + b*e2 = 0 is a = b = 0.
Let's solve this equation:
a*2 + b*1 = 0 (for the first component)
a*1 + b*1 = 0 (for the second component)
a*0 + b*0 = 0 (for the third component)
Simplifying the equations, we get:
2a + b = 0 (equation 1)
a + b = 0 (equation 2)
0 = 0 (equation 3)
From equation 3, we can see that it does not provide any new information, so we will focus on equations 1 and 2.
We can use equation 2 to express a in terms of b: a = -b.
Substituting this value of a into equation 1, we get:
2*(-b) + b = 0
-2b + b = 0
-b = 0
b = 0
Therefore, b = 0 implies a = 0.
Hence, the only solution to the equation a*e1 + b*e2 = 0 is a = b = 0. This confirms that e1 and e2 are linearly independent.
2. Spanning R3:
To show that e1 and e2 span R3, we need to show that any vector in R3 can be expressed as a linear combination of e1 and e2.
Let's take an arbitrary vector in R3, say (x, y, z). We need to find scalars a and b such that a*e1 + b*e2 = (x, y, z).
Solving for a and b, we get:
2a + b = x (equation 1)
a + b = y (equation 2)
0 = z (equation 3)
From equation 3, we can see that z = 0.
Substituting z = 0 into equations 1 and 2, we get:
2a + b = x (equation 4)
a + b = y (equation 5)
We can express b in terms of a using equation 5: b = y - a.
Substituting this value of b into equation 4, we get:
2a + (y - a) = x
2a + y - a = x
a + y = x
Therefore, we can set a = x - y. Substituting this back into equation 5, we get:
b = y - (x - y) = 2y - x.
So, any vector (x, y, 0), where x and y are real numbers, can be expressed as a linear combination of e1 and e2: a*e1 + b*e2 = (x, y, 0).
This shows that e1 and e2 span R3.
Since e1 and e2 are linearly independent and span R3, they form a basis of R3.