Problem 7 - Differential Equations and Linear Algebra

Question

determine whether the given set of vectors is linearly independent or linearly dependent in \(\mathbb{R}^{n} .\) In the case of linear dependence, find a dependency relationship. $$\\{(1,-1,2),(2,1,0)\\}$$.

Step-by-Step Solution

Answer

The given set of vectors is linearly independent since the system of linear equations has only the trivial solution \((a, b) = (0, 0)\). There is no dependency relationship to find.

Step 1: Set up a Linear Combination

Write down a linear combination of the vectors. We want to see if there is a scalar combination of the vectors that adds up to the zero vector. Set two unknown scalars, \(a\) and \(b\), to multiply each vector, and set the resulting linear combination equal to the zero vector as follows: $$a(1, -1, 2) + b(2, 1, 0) = (0, 0, 0)$$

Step 2: Write Down the System of Linear Equations

Rewritten the given linear combination as a system of linear equations with variables \(a\) and \(b\): $$ \begin{cases} a + 2b = 0 \\ -a + b = 0 \\ 2a + 0b = 0 \end{cases} $$

Step 3: Solve the System of Linear Equations

To solve the system of linear equations, use any preferred method such as substitution or elimination. Notice there is an easy way to simplify the system: From the third equation, we can see that \(2a = 0\), and therefore \(a = 0\). Substituting \(a=0\) in the first equation, we get \(2b = 0\), which means \(b = 0\) as well. Therefore, our solution for the system of linear equations is \((a, b) = (0, 0)\).

Step 4: Determine Linear Independence

Since the system of linear equations has the trivial solution \((0, 0)\), it means that the given set of vectors is linearly independent. There is no nontrivial linear combination of the vectors that results in the zero vector. So, the given set of vectors is: $$\text{Linearly Independent}$$ Since the set of vectors is linearly independent, there is no dependency relationship to find.

Key Concepts

Vectors

Vectors are objects that have both magnitude and direction. They can be visualized as arrows pointing from one location to another in space. We often represent vectors in a coordinate system with components, such as

• In 2-dimensional space: \(x, y\)
• In 3-dimensional space: \(x, y, z\)

The set of vectors provided in this exercise are from \(\mathbb{R}^3\), specifically \( (1, -1, 2) \) and \( (2, 1, 0) \).
These components represent their positions and directions in a 3D space.
Understanding vectors through these components helps identify relationships like linear independence, where no vector in the set is a combination of the others. For example, in our step-by-step solution, we tested for linear combinations of vectors to see if such a dependency exists.

Linear Combination

To determine whether vectors are linearly dependent or independent, we often use linear combinations. A linear combination involves multiplying each vector by a scalar and summing them up:

• This operation helps identify if vectors can form another vector (often the zero vector), showing dependence.
• If the only combination that results in zero is the trivial one (all scalars are zero), the vectors are linearly independent.

In our task, the expression \(a(1, -1, 2) + b(2, 1, 0) = (0, 0, 0)\) helps explore combinations of these vectors to see if they yield the zero vector.
The solution shows how linear combination is set up with scalars \(a\) and \(b\), highlighting the dependency test for this vector set.

System of Linear Equations

The heart of solving linear dependence or independence lies in the system of linear equations. When trying to find if a set of vectors is independent, we construct a system of equations that represents the linear combination set to zero:

• Each component of the vector equation leads to an individual equation.
• The goal is to solve these equations and determine the values of the scalars involved.

In this scenario, the system \[\begin{cases}a + 2b = 0 -a + b = 0 2a + 0b = 0\end{cases}\] helps us determine whether there exists a non-trivial solution (non-zero values for \(a\) and \(b\)) indicating dependence.
After solving, we find that only the trivial solution exists (\(a = 0, b = 0\)), which confirms the vectors' independence.