Linear Independence Of Vectors Is The Set {(4, 0, 2, 1), (2, 1, 3, 4), (2, 3, 4, 7), (2, 3, 1, 4)} Linearly Independent?
Determining whether a set of vectors is linearly independent is a fundamental concept in linear algebra. A set of vectors is linearly independent if no vector in the set can be written as a linear combination of the others. In simpler terms, it means that none of the vectors are redundant; they each contribute unique information to the space they span. When dealing with a set of vectors like {(4, 0, 2, 1), (2, 1, 3, 4), (2, 3, 4, 7), (2, 3, 1, 4)}, we need a systematic way to check for this independence. This often involves setting up a matrix, performing row operations, and analyzing the resulting form to see if there are any free variables, which would indicate linear dependence.
Understanding Linear Independence
Before diving into the specifics of our vector set, let's solidify our understanding of linear independence. A set of vectors {v1, v2, ..., vn} is said to be linearly independent if the only solution to the equation c1v1 + c2v2 + ... + cnvn = 0 (where c1, c2, ..., cn are scalars) is the trivial solution c1 = c2 = ... = cn = 0. If there exists any non-trivial solution (i.e., at least one ci ≠0), then the vectors are linearly dependent. Linear dependence implies that at least one vector in the set can be expressed as a combination of the others, making it redundant in terms of the space spanned by the set.
To illustrate this, consider two vectors in a 2D plane. If these vectors are linearly independent, they point in different directions and span the entire 2D plane. If they are linearly dependent, one is a scalar multiple of the other, meaning they lie on the same line and only span a 1D subspace. This geometric intuition extends to higher dimensions, although it becomes harder to visualize.
Methods to Determine Linear Independence
There are several methods to determine whether a set of vectors is linearly independent. One common approach involves forming a matrix with the vectors as columns and then performing Gaussian elimination (row reduction) to bring the matrix to its row-echelon form or reduced row-echelon form. The rank of the matrix, which is the number of non-zero rows in its row-echelon form, is crucial. If the rank of the matrix equals the number of vectors, then the vectors are linearly independent. If the rank is less than the number of vectors, they are linearly dependent. This method efficiently reveals whether any of the vectors can be written as a linear combination of the others. The process of row reduction systematically eliminates redundant information, making it clear whether each vector contributes uniquely to the span.
Another method involves calculating the determinant of the matrix formed by the vectors (if the matrix is square). If the determinant is non-zero, the vectors are linearly independent. If the determinant is zero, they are linearly dependent. The determinant provides a single number that encapsulates the linear independence of the vectors. However, calculating determinants can be computationally intensive for large matrices, making row reduction a more practical approach in many cases. The choice of method often depends on the specific problem and the tools available.
Applying Row Reduction to the Given Vectors
Let's apply the row reduction method to the given set of vectors: {(4, 0, 2, 1), (2, 1, 3, 4), (2, 3, 4, 7), (2, 3, 1, 4)}. We'll form a matrix with these vectors as columns:
| 4 2 2 2 |
| 0 1 3 3 |
| 2 3 4 1 |
| 1 4 7 4 |
Our goal is to perform elementary row operations to transform this matrix into its row-echelon form. These operations include swapping rows, multiplying a row by a non-zero scalar, and adding a multiple of one row to another. These operations do not change the linear dependence relationships between the columns. We start by swapping row 1 and row 4 to get a leading 1 in the first row:
| 1 4 7 4 |
| 0 1 3 3 |
| 2 3 4 1 |
| 4 2 2 2 |
Next, we eliminate the 2 and 4 in the first column by performing the operations R3 -> R3 - 2R1 and R4 -> R4 - 4R1:
| 1 4 7 4 |
| 0 1 3 3 |
| 0 -5 -10 -7 |
| 0 -14 -26 -14 |
Now, we eliminate the -5 and -14 in the third column by performing the operations R3 -> R3 + 5R2 and R4 -> R4 + 14R2:
| 1 4 7 4 |
| 0 1 3 3 |
| 0 0 5 8 |
| 0 0 16 28 |
Finally, we eliminate the 16 in the fourth column by performing the operation R4 -> R4 - (16/5)R3:
| 1 4 7 4 |
| 0 1 3 3 |
| 0 0 5 8 |
| 0 0 0 -6/5 |
The matrix is now in row-echelon form. We have four non-zero rows, which means the rank of the matrix is 4. Since we started with four vectors, and the rank of the matrix is 4, the vectors are linearly independent.
Conclusion: The Vectors Are Linearly Independent
Through the process of row reduction, we have demonstrated that the set of vectors {(4, 0, 2, 1), (2, 1, 3, 4), (2, 3, 4, 7), (2, 3, 1, 4)} is linearly independent. This means that none of these vectors can be written as a linear combination of the others. Each vector contributes unique information and expands the space spanned by the set. The rank of the matrix formed by these vectors is equal to the number of vectors, confirming their linear independence. Understanding linear independence is crucial in various applications of linear algebra, such as solving systems of equations, finding bases for vector spaces, and analyzing transformations.
Further Exploration of Linear Independence
Beyond the row reduction method, it's worth exploring other perspectives on linear independence. Geometrically, in four-dimensional space, these four linearly independent vectors form a basis. This means that any vector in 4D space can be expressed as a linear combination of these four vectors. The concept of a basis is fundamental in understanding the structure of vector spaces. A basis provides a minimal set of vectors needed to span the entire space, and linear independence is a key property of any basis.
Another related concept is the null space of a matrix. The null space of a matrix A is the set of all vectors x such that Ax = 0. If the columns of A are linearly independent, the null space contains only the zero vector. This is because the only solution to the equation c1v1 + c2v2 + ... + cnvn = 0 is the trivial solution when the vectors are linearly independent. The null space provides additional insights into the properties of the matrix and the linear transformations it represents.
In practical applications, linear independence is essential in fields like computer graphics, data analysis, and engineering. For instance, in computer graphics, linearly independent vectors are used to define coordinate systems and transformations. In data analysis, linearly independent features in a dataset provide unique information and avoid redundancy in the model. In engineering, understanding linear independence is crucial in analyzing the stability of systems and designing control mechanisms.
Moreover, the Gram-Schmidt process is a method for orthonormalizing a set of linearly independent vectors. Given a set of linearly independent vectors, the Gram-Schmidt process produces a set of orthonormal vectors that span the same subspace. This process is used in various applications, including eigenvalue computations and solving least-squares problems. Orthonormal vectors are particularly useful because they simplify many calculations and provide a stable basis for numerical computations.
In summary, the determination of linear independence is a cornerstone of linear algebra with far-reaching implications. By applying techniques like row reduction, understanding the geometric interpretation, and exploring related concepts such as the null space and the Gram-Schmidt process, we gain a deeper appreciation for the power and versatility of linear algebra in solving real-world problems.
While we've established the linear independence of our vector set, let's delve deeper into the concept of linear dependence to further solidify our understanding. Linear dependence arises when one or more vectors in a set can be expressed as a linear combination of the other vectors. This implies a certain redundancy within the set, as the dependent vector(s) do not contribute unique directional information to the span.
Consider a scenario where we have a set of vectors in 3D space, and one vector lies in the plane formed by the other two. This vector is linearly dependent because it can be written as a sum of scaled versions of the other two vectors. Geometrically, it doesn't
