Let
v1 = (1, 0, −1)
v2 = (0, 2, 2)
v3 = (−3, 4, 7)
and let W = Span{v1, v2, v3}.
1. Show that v3 is a linear combination c1v1 + c2v2 of v1 and v2 by finding
the constants c1 and c2.
2. Show that W = Span{v1, v2}.
3. Show that v1 and v2 are linearly independent.
4. What is the dimension of W?
a-3c=0
2b+4c=0
-a+2b+7c=0
3v1 - 2v2 + v3 = 0
v3 = 2v2 - 3v1
...
To answer these questions, we need to understand the concepts of linear combinations, spanning sets, linear independence, and dimension. Here's how you can find the answers to each of the questions:
1. To show that v3 is a linear combination of v1 and v2, we need to find constants c1 and c2 such that v3 = c1v1 + c2v2.
We can set up the following system of equations:
c1(1, 0, -1) + c2(0, 2, 2) = (-3, 4, 7)
This system can be written as:
c1 = -3
2c2 = 4
-c1 + 2c2 = 7
From the second equation, we can see that c2 = 2. Substituting this value into the first equation, we find c1 = -3. Therefore, v3 = -3v1 + 2v2, and we have shown that v3 is a linear combination of v1 and v2.
2. To show that W = Span{v1, v2}, we need to prove that any vector in W can be written as a linear combination of v1 and v2 and vice versa.
First, let's prove that W is a subset of Span{v1, v2}. Any vector in W can be written as c1v1 + c2v2 + c3v3. We've already shown that v3 = -3v1 + 2v2, so we can substitute this into the expression for W:
c1v1 + c2v2 + c3(-3v1 + 2v2) = (c1 - 3c3)v1 + (c2 + 2c3)v2
From this equation, we can see that any vector in W can indeed be written as a linear combination of v1 and v2. Therefore, W is a subset of Span{v1, v2}.
Next, let's prove that Span{v1, v2} is a subset of W. Since v1 and v2 are part of the spanning set for W, any linear combination of v1 and v2 will exist in W. Therefore, Span{v1, v2} is a subset of W.
Since W is a subset of Span{v1, v2} and Span{v1, v2} is a subset of W, we can conclude that W = Span{v1, v2}.
3. To show that v1 and v2 are linearly independent, we need to prove that the only solution to the equation c1v1 + c2v2 = 0 is c1 = c2 = 0.
Let's set up the following system of equations:
c1(1, 0, -1) + c2(0, 2, 2) = (0, 0, 0)
This system can be written as:
c1 = 0
2c2 = 0
-c1 + 2c2 = 0
From the second equation, we can see that c2 = 0. Substituting this value into the first and third equations, we find c1 = 0. Therefore, the only solution to the equation is c1 = c2 = 0, and v1 and v2 are linearly independent.
4. The dimension of W is the number of linearly independent vectors in the spanning set. In this case, we have shown that v1 and v2 are linearly independent, and v3 is a linear combination of v1 and v2. Therefore, the dimension of W is 2.