Question

Determine whether the given set of vectors is linearly independent for \(-\infty

Short answer

Answer: Yes, the given vectors are linearly independent for all values of t except where sin(t) = 0.

Step-by-step solution

Check for Linear Independence

Let \(\mathbf{x}^{(1)}(t)=(2 \sin t, \sin t)\) and \(\mathbf{x}^{(2)}(t)=(\sin t, 2\sin t)\). To check for linear independence, we need to find the coefficients of the linear combination \(\alpha_1 \mathbf{x}^{(1)}(t) + \alpha_2 \mathbf{x}^{(2)}(t) = 0\). So, we have the following system of equations: \begin{align*} \alpha_1 (2\sin t) + \alpha_2 (\sin t) &= 0, \\ \alpha_1 (\sin t) + \alpha_2 (2\sin t) &= 0. \end{align*}

Solve the System of Equations

To find \(\alpha_1\) and \(\alpha_2\), we can solve the system of linear equations. For \(\sin t \neq 0\), we can divide both sides of the first equation by \(\sin t\): $$2\alpha_1 + \alpha_2 = 0.$$ Next, we can substitute this expression for \(\alpha_2\) into the second equation: $$\alpha_1 + 2(2\alpha_1) = 0 \implies \alpha_1 + 4\alpha_1 = 0 \implies 5\alpha_1 = 0 \implies \alpha_1 = 0.$$ Now we can find \(\alpha_2\) using the first equation: $$2(0) + \alpha_2 = 0 \implies \alpha_2 = 0.$$

Conclusion

Since the only solution for \(\alpha_1\) and \(\alpha_2\) is the trivial solution (both equal to 0), the given set of vectors \(\mathbf{x}^{(1)}(t)\) and \(\mathbf{x}^{(2)}(t)\) is linearly independent for \(-\infty<t<\infty\) and \(\sin t \neq 0\).

Key concepts

Linear Combination

When we talk about a linear combination, we refer to the process of adding together multiple vectors, each multiplied by a corresponding scalar. In a more technical sense, given vectors \(\mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_n\) and scalars \(\alpha_1, \alpha_2, ..., \alpha_n\), the linear combination would be \(\alpha_1 \mathbf{v}_1 + \alpha_2 \mathbf{v}_2 + ... + \alpha_n \mathbf{v}_n\).

In the context of the exercise, we are dealing with the vectors \(\mathbf{x}^{(1)}(t)\) and \(\mathbf{x}^{(2)}(t)\), and we are looking to find out if there exists a non-trivial set of scalars \(\alpha_1\) and \(\alpha_2\) such that \(\alpha_1 \mathbf{x}^{(1)}(t) + \alpha_2 \mathbf{x}^{(2)}(t) = 0\), which is essentially a linear combination equal to the zero vector. The process of checking linear independence involves setting up a linear combination equal to zero and determining if the only solution for the scalars is the trivial one (all scalars are zero), indicating independence, or if there exist non-zero scalars that satisfy the equation, pointing to dependence.

System of Linear Equations

In our exercise, we encounter a system of linear equations, which is a collection of two or more linear equations involving the same set of variables. To solve such a system is to find the values of the variables that satisfy all equations simultaneously.

For instance, the equations \(2\alpha_1 + \alpha_2 = 0\) and \(\alpha_1 + 2\alpha_2 = 0\) form a system of linear equations that can be solved using various methods, such as substitution, elimination, or matrix operations. If the system has at least one solution, it is called consistent; otherwise, it is inconsistent. The solutions to these systems can be a single point (unique solution), a line or plane (an infinite number of solutions), or no points at all (no solution).

To clarify the concept, consider the provided solution where we used substitution. After manipulating the first equation, we substituted the expression for one variable into the other equation. This technique simplifies the system to a single variable, making it easier to find the unique solution if it exists.

Trivial Solution

The term trivial solution often pops up when discussing systems of linear equations and their solutions. Specifically, the trivial solution is one where all the variables are equal to zero. This solution is always an option for homogeneous equations, where the constant terms are all zero.

In the context of linear independence, the presence of the trivial solution plays a pivotal role. If the only way to express the linear combination of vectors as the zero vector \(0\) is by having all scalars equal to zero \(\alpha_1 = \alpha_2 = ... = \alpha_n = 0\), then the set of vectors is deemed linearly independent. On the other hand, if there are non-zero values for the scalars that still produce the zero vector, we say the vectors are linearly dependent.

Applying this to the example at hand, we solved the system and found that \(\alpha_1 = \alpha_2 = 0\) was the only solution. It qualifies as the trivial solution, confirming that our vectors \(\mathbf{x}^{(1)}(t)\) and \(\mathbf{x}^{(2)}(t)\) are linearly independent. Any other outcome would have indicated a dependency between the vectors.