Problem 50 - Differential Equations and Linear Algebra

Question

Generalizing the previous exercise, prove that if \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{k}\right\\}\) is linearly independent and \(\mathbf{v}_{k+1}\) is not in \(\operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{k}\right\\},\) then \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{k+1}\right\\}\) is linearly independent.

Step-by-Step Solution

Answer

The proof follows from the definition of linear independence. Given that the set \(\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\}\) is linearly independent and \(\mathbf{v}_{k+1}\) is not in their span, consider the linear combination \(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k + c_{k+1}\mathbf{v}_{k+1} = \mathbf{0}\). Rearranging the equation gives \(\mathbf{v}_{k+1} = -\frac{c_1}{c_{k+1}}\mathbf{v}_1 - \frac{c_2}{c_{k+1}}\mathbf{v}_2 - \cdots - \frac{c_k}{c_{k+1}}\mathbf{v}_k\), which implies that the only solution is when all coefficients, i.e., \(c_1, c_2, \ldots, c_k, c_{k+1}\), are zero, making the set \(\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k, \mathbf{v}_{k+1}\}\) linearly independent.

Step 1: Define linearly independence

Linear independence of a set of vectors means that no vector in the set can be represented as a linear combination of the other vectors. A set of vectors \(\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\}\) is linearly independent if and only if the following equation has only the trivial solution, where all coefficients are zero: \(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{0}\). In this problem, it is given that the set \(\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\}\) is linearly independent.

Step 2: Consider the new vector

We are given that the new vector \(\mathbf{v}_{k+1}\) is not in the span of the set \(\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\}\), i.e., it cannot be written as a linear combination of the vectors \(\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\}\).

Step 3: Form the linear combination including \(\mathbf{v}_{k+1}\)

Now we have to check if the set \(\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k, \mathbf{v}_{k+1}\}\) is linearly independent. Consider the following linear combination: \(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k + c_{k+1}\mathbf{v}_{k+1} = \mathbf{0}\).

Step 4: Rearrange the linear combination for \(\mathbf{v}_{k+1}\)

We can rearrange the above equation to get the expression for \(\mathbf{v}_{k+1}\) as follows: \(\mathbf{v}_{k+1} = -\frac{c_1}{c_{k+1}}\mathbf{v}_1 - \frac{c_2}{c_{k+1}}\mathbf{v}_2 - \cdots - \frac{c_k}{c_{k+1}}\mathbf{v}_k\).

Step 5: Contradiction and conclusion

Since \(\mathbf{v}_{k+1}\) is not in the span of the set \(\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\}\), there is no non-zero solution for the coefficients in the above equation. The only solution is when all coefficients, i.e., \(c_1, c_2, \ldots, c_k, c_{k+1}\), are zero. This means that the set \(\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k, \mathbf{v}_{k+1}\}\) is linearly independent. This completes the proof.

Key Concepts

span of vectors

The concept of the 'span of vectors' is crucial in understanding linear algebra. At its core, the span of a set of vectors is essentially all possible vectors you can create through combinations of these original vectors. Imagine each original vector as a separate direction in space, like directions or axes. When you combine them in varying amounts, using different coefficients, you get a whole "plane" or "space" of vectors.The span of vectors can be mathematically represented as all possible linear combinations of the set. For example, if you have vectors \(\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\), their span is the set of all combinations \(a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \cdots + a_k\mathbf{v}_k\), where \(a_1, a_2, \ldots, a_k\) are any real numbers.

• Importance: The span tells us all the directions and distances we can cover using our set of vectors.
• Application: It's often used to determine if a particular vector can be expressed as a combination of others, which helps in checking linear dependency.

If a new vector, like \( \mathbf{v}_{k+1} \), cannot be made using the linear combination of your original vectors, it is not in their span and this opens up a new direction or dimension for your vector space.

linear combination

A 'linear combination' is essentially a way to add vectors together so that they form a new vector. Picture it as adding several paths to form a new route. Each path reflects a particular vector and its 'weight' or importance is indicated by a coefficient. These coefficients can be seen as how much of each path you take. Here's how it works generally:For vectors \(\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\), a linear combination would be of the form \(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k\), where \(c_1, c_2, \ldots, c_k\) are real numbers.

• Flexibility: Linear combinations allow for great flexibility in creating new vectors from existing sets.
• Understanding Dependence: By seeing how freely we can manipulate these combinations we understand whether one vector is dependent on others.

When solving problems using linear combinations, the core goal is often to see if a given vector can be expressed as such a combination of others. If so, it's in the span and if not, it opens up new vector possibilities. This process aids greatly in proving whether a set of vectors is linearly independent.

trivial solution

The term 'trivial solution', in the context of linear independence, refers to the simplest possible solution to an equation where all unknowns are zeros. In linear algebra, checking for a trivial solution is vital when determining if a set of vectors is linearly independent.For instance, suppose you have the equation \(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{0}\). If the only solution is the trivial solution, where \(c_1 = c_2 = \ldots = c_k = 0\), then your set of vectors is linearly independent.

• Significance: Finding only the trivial solution proves no vector in the set can be composed from others, ensuring independence.
• Indicator of Independence: When no other solutions exist, the setup confirms independence inherently.

Detecting a trivial solution, thus, not only assures you of the independent nature of vectors but also affirms that there are no redundant vectors among them. In practice, linear independence checks often start by hunting for the trivial solution first.