What is a Linear Combination?
In linear algebra, a linear combination is an expression created by multiplying a set of vectors by scalar constants and adding the results together. It is the basic operation used to move through or "span" a vector space.
If you have a set of vectors {v1, v2, ..., vn} and a corresponding set of scalars {c1, c2, ..., cn}, the linear combination is defined as:
w = c1v1 + c2v2 + ... + cnvn
Relationship to Linear Algebra
Linear combinations are the foundation upon which most of linear algebra is built:
Systems of Equations: Solving a system of linear equations is essentially the process of finding the specific scalars (the linear combination) required to reach a target vector.
Span: The "span" of a set of vectors is the collection of every possible linear combination that can be formed from them.
Basis Vectors: Any vector in a space can be represented as a unique linear combination of its basis vectors. For example, in 3D, the vector (x, y, z) is the linear combination: xi + yj + zk.
Worked Examples
Basic Vector Addition and Scaling
Given vectors u = (1, 2) and v = (-3, 5), find the linear combination w = 3u - 2v.
1. Multiply vector u by 3: 3(1, 2) = (3, 6).
2. Multiply vector v by 2: 2(-3, 5) = (-6, 10).
3. Subtract the results: (3, 6) - (-6, 10).
4. Combine the components: (3 - (-6), 6 - 10) = (9, -4).
Final Answer: w = (9, -4)
Finding Scalars for a Specific Vector
Express the vector w = (7, 11) as a linear combination of u = (1, 1) and v = (2, 3).
1. Set up the equation: c1(1, 1) + c2(2, 3) = (7, 11).
2. Create the system of equations:
c1 + 2c2 = 7
c1 + 3c2 = 11
3. Solve by subtraction: (c1 + 3c2) - (c1 + 2c2) = 11 - 7, which gives c2 = 4.
4. Substitute c2 = 4 back into the first equation: c1 + 2(4) = 7, so c1 + 8 = 7.
5. This gives c1 = -1.
Final Answer: w = -u + 4v
Verifying a Linear Combination in 3D
Determine if w = (1, 2, 3) can be written as a linear combination of u = (1, 0, 1) and v = (0, 1, 1).
1. Set up the vector equation: c1(1, 0, 1) + c2(0, 1, 1) = (1, 2, 3).
2. Break it into component equations:
x-component: c1 = 1
y-component: c2 = 2
z-component: c1 + c2 = 3
3. Check for consistency: Substitute c1 = 1 and c2 = 2 into the z-component equation.
4. 1 + 2 = 3. Since the equations are consistent, the combination is possible.
Final Answer: Yes, w = u + 2v
Solving for Scalars in 2D
Express the vector w = (8, 9) as a linear combination of u = (2, 5) and v = (-1, 3).
Step 1: Set up the vector equation We need to find two numbers (scalars), a and b, so that: a(2, 5) + b(-1, 3) = (8, 9)
Step 2: Create a system of linear equations By looking at the top and bottom rows separately, we get:
Top row: 2a - b = 8 (Equation 1)
Bottom row: 5a + 3b = 9 (Equation 2)
Step 3: Solve the system From Equation 1, we can see that: b = 2a - 8
Now, put this into Equation 2: 5a + 3(2a - 8) = 9 5a + 6a - 24 = 9 11a = 33 a = 3
Now find b: b = 2(3) - 8 b = -2
Final Solution: The linear combination is 3u - 2v = w.
Verifying a Linear Combination in 3D
Determine if w = (2, 5, 4) can be written as a linear combination of u = (1, 2, 1) and v = (0, 1, 2).
Step 1: Set up the equation a(1, 2, 1) + b(0, 1, 2) = (2, 5, 4)
Step 2: Extract the equations
1a + 0b = 2
2a + 1b = 5
1a + 2b = 4
Step 3: Solve and check for consistency From the first equation, we can see immediately that a = 2.
Now, put a = 2 into the second equation: 2(2) + b = 5 4 + b = 5 b = 1
Finally, we must check if these values work for the third equation: 1(2) + 2(1) = 2 + 2 = 4
Since 4 = 4, the values are consistent across all three rows.
Final Solution: Yes, w can be written as the linear combination 2u + v.
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