Problem 10 - 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. $$\begin{aligned} &\\{(2,-1,0,1),(1,0,-1,2),(0,3,1,2)\\\ &(-1,1,2,1)\\} \end{aligned}$$.

Step-by-Step Solution

Answer

The given set of vectors is linearly dependent since the rank of the matrix is 2 while the number of columns is 4. The dependency relationship between the given vectors is: $$(-2k)((2,-1,0,1))+(\frac{2}{3}k)((1,0,-1,2))+k(0,3,1,2))=(0,0,0,0)$$

Step 1: Write Down the Matrix

Write down the given matrix of the given vectors in columns. $$A=\begin{bmatrix} 2 & -1 & 0 & 1\\ 1 & 0 & -1 & 2\\ 0 & 3 & 1 & 2\\ -1 & 1 & 2 & 1 \end{bmatrix}$$

Step 2: Perform Row Reduction

Perform Gaussian elimination (row reduction) on the coefficient matrix to bring it to its row echelon form. After row reduction: $$A=\begin{bmatrix} 1 & 0 & -1 & 2\\ 0 & 1 & \frac{1}{3} & \frac{2}{3}\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}$$ Below are the operations applied during row reduction: 1. Swap Row 1 and Row 2 2. Add Row 1 to Row 4 3. Multiply Row 2 by 3 and Add Row 3

Step 3: Check the Rank

Now, compare the rank of the matrix (number of nonzero rows) to the number of columns. The rank of this matrix is 2, while the number of columns is 4. So, the set of vectors is linearly dependent.

Step 4: Find the Dependency Relationship

In the row reduced matrix, we have: $$\begin{cases} x_1 - x_3 = -2x_4 \\ x_2 + \frac{1}{3}x_3 = \frac{2}{3}x_4 \end{cases}$$ Set \(x_4=k\) (for some real number k). $$\begin{cases} x_1 - x_3 = -2k \\ x_2 + \frac{1}{3}x_3 = \frac{2}{3}k \end{cases}$$ So, the dependency relationship between the given vectors is: $$(-2k)((2,-1,0,1))+(\frac{2}{3}k)((1,0,-1,2))+k(0,3,1,2))=(0,0,0,0)$$

Key Concepts

Linear Independence

In linear algebra, determining linear independence is crucial when analyzing a set of vectors. A set of vectors is considered to be linearly independent if no vector in the set can be written as a linear combination of the others. In simple terms, if you try to express one vector using the others and the only solution is having all coefficients equal to zero, then the vectors are independent.

When vectors are linearly independent, they span a space where each dimension has its unique direction. This means no redundancy; each vector contributes something new to the vector space.

Linear Dependence

On the other hand, if a set of vectors is linearly dependent, it means there is at least one vector that can be expressed as a combination of the others. This means you could remove this vector without changing the span of the vector set, as it doesn't contribute new information or direction.

In practice, solving for linear dependence involves finding coefficients, not all zero, such that the linear combination of vectors equals the zero vector. In the problem given, after performing row reduction, the rank of the matrix was less than the number of columns, indicating linear dependence. The dependency relationship provides insight into how these vectors are interconnected.

Row Reduction

Row reduction is a method used to simplify matrices and involves elementary row operations. The goal is often to bring the matrix into a row-echelon form, where each leading coefficient (the first non-zero number from left to right in a row) is 1, and is the only non-zero entry in its column.

Basic steps include:

• Swapping rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting the multiples of rows

Row reduction is a powerful tool because it can simplify complex systems of linear equations, aiding in solving and understanding their structure.

Gaussian Elimination

Gaussian elimination is a systematic method used for solving systems of linear equations. It utilizes the process of row reduction to transform the matrix into a simpler form, typically the row-echelon form or even reduced row-echelon form.

The process involves:

• Using row operations to zero out elements below the pivot positions (the leading 1s)
• Proceeding systematically from the top to the bottom of the matrix and left to right column-wise

After reaching row-echelon form, it becomes easy to identify the solution can be identified, showing whether the system is consistent, and all relations between vectors, such as those indicating linear dependency.