Problem 25 - Differential Equations and Linear Algebra

Question

Decide (with justification) whether or not the given set \(S\) of vectors (a) spans \(V,\) and (b) is linearly independent. $$V=\mathbb{R}^{3}, S=\\{(6,-3,2),(1,1,1),(1,-8,-1)\\}$$

Step-by-Step Solution

Answer

The given set of vectors $S=\{(6, -3, 2), (1, 1, 1), (1, -8, -1)\}$ both spans $V=\mathbb{R}^3$ and is linearly independent, as the rank of the matrix formed by these vectors is equal to the dimension of the vector space and the number of vectors in set $S$.

Step 1: Write the vectors as columns of a matrix

We are given set S: \({(6, -3, 2), (1, 1, 1), (1, -8, -1)}\). Let's write these vectors as columns of a matrix A: $$ A = \begin{pmatrix} 6 & 1 & 1\\ -3 & 1 & -8 \\ 2 & 1 & -1 \\ \end{pmatrix} $$

Step 2: Find the reduced row echelon form (RREF) of the matrix

Now let's find the RREF of matrix A by performing row operations. First, let's use row replacement and row scaling operations to make all elements below the first pivot (6) = 0: $$ A_1 = \begin{pmatrix} 6 & 1 & 1\\ 0 & \frac{1}{2} & -\frac{29}{2} \\ 0 & \frac{2}{3} & -\frac{5}{3} \\ \end{pmatrix} $$ Now, let's use row replacement and row scaling operations to make all elements below the second pivot (\(\frac{1}{2}\)) = 0: $$ A_2 = \begin{pmatrix} 6 & 1 & 1\\ 0 & \frac{1}{2} & -\frac{29}{2} \\ 0 & 0 & 2 \\ \end{pmatrix} $$ Finally, let's use row scaling to make all pivot elements equal to 1: $$ RREF(A) = \begin{pmatrix} 1 & \frac{1}{6} & \frac{1}{6} \\ 0 & 1 & -29 \\ 0 & 0 & 1 \\ \end{pmatrix} $$

Step 3: Count the number of non-zero rows in the RREF matrix

We now have the reduced row echelon form of matrix A. We can count the number of non-zero rows in the RREF matrix, which will give us the rank of the matrix: Rank(A) = 3

Step 4: Determine whether the set S spans V

Since the rank of the matrix A is equal to the dimension of the vector space V (\(\mathbb{R}^3\)), set S spans V. So, the answer to the first part of the question is yes.

Step 5: Determine whether the vectors are linearly independent

To determine whether the vectors in set S are linearly independent, compare the rank of the matrix A (which is 3) to the number of vectors in set S (also 3). Since these numbers are equal, the given set S of vectors is linearly independent. So, set S of vectors both spans V and is linearly independent.

Key Concepts

Linear Independence

Understanding linear independence is crucial when studying vector spaces. A set of vectors, like the one given in set \( S \), is said to be linearly independent if no vector in the set can be written as a combination of the others. In other words, there aren't any 'extra' vectors that can be omitted without changing the span of the set.

To determine linear independence, we translate the problem into matrix terms. Each vector becomes a column in a matrix. If after transforming the matrix into its reduced row echelon form (RREF) all the columns have leading 1's (pivots) and there are no free variables, it implies that each vector contributes something new—none of them is redundant. In the exercise at hand, the RREF of matrix A has three pivots, which matches the number of vectors; hence, the set \( S \) is linearly independent. This means that none of the vectors in \( S \) can be created by a combination of the others, showcasing the power of the RREF in solving linear independence concerns.

Reduced Row Echelon Form

The reduced row echelon form (RREF) of a matrix is a special form that is crucial for solving various linear algebra problems, including determining linear independence and spanning sets. To convert a matrix into RREF, a series of row operations are performed, which include swapping rows, multiplying a row by a non-zero scalar, and adding a multiple of one row to another row.

For the given matrix \( A \) representing vectors in set \( S \) as its columns, the RREF is achieved by such row operations, which aim to create leading 1's (pivots) and zeros everywhere else in each column containing a pivot. This process uncovers the essential structure of the matrix and, by extension, the relationships among the vectors. In our exercise, the transformations led to an RREF with three leading 1's and no free variables, clearly indicating that the matrix has full rank and the vectors span \( \mathbb{R}^3 \) without any redundancy.

Matrix Rank

The rank of a matrix is a fundamental concept, representing the maximum number of linearly independent rows or columns within the matrix. It can also be seen as a measure of the matrix's 'information content', telling us the dimension of the vector space spanned by its rows or columns. In the context of our exercise, we count the number of pivots in the RREF of matrix A to determine its rank.

In this case, the rank is 3, which is equal to both the number of vectors in set \( S \) and the dimension of the vector space \( \mathbb{R}^3 \). The rank being equal to the number of vectors and the dimension of \( \mathbb{R}^3 \) confirms that the vectors span the entire vector space, which is exactly what one would expect for a basis of \( \mathbb{R}^3 \). The concept of matrix rank gives us a powerful tool for understanding the structure of solutions to linear systems and the functionality of transformations represented by matrices.