Question

Can any of the given sets of 3 -vectors below span the 3 -space? Why or why not? \(\left.\begin{array}{lllllllll}(a)[1 & 2 & 1] & {\left[\begin{array}{llll}2 & 3 & 1\end{array}\right]} & {\left[\begin{array}{lll}1 & 4 & 2\end{array}\right]}\end{array}\right]\) \(\begin{array}{lllllllll}(b) & {[8} & 1 & 3] & {\left[\begin{array}{lll}1 & 2 & 8\end{array}\right]} & {\left[\begin{array}{llll}-7 & 1 & 5\end{array}\right]}\end{array}\)

Short answer

Both sets (a) and (b) can span \(\mathbb{R}^3\) since they are linearly independent.

Step-by-step solution

Understand the Problem

The question asks whether the given sets of vectors can span the 3-dimensional space, or \(\mathbb{R}^3\). A set of vectors spans \(\mathbb{R}^3\) if they are linearly independent and the number of vectors is equal to the dimension of the space, which is 3 in this case.

Check the Number of Vectors

Each set contains exactly 3 vectors, which suggests that they might be able to span \(\mathbb{R}^3\) if they are linearly independent. We must now check their linear independence.

Construct the Matrix for Set (a)

For set (a), construct the matrix \( A = \begin{bmatrix} 1 & 2 & 1 \ 2 & 3 & 1 \ 1 & 4 & 2 \end{bmatrix} \). The next step is to determine if the columns of this matrix are linearly independent by finding its determinant.

Calculate the Determinant for Set (a)

Calculate the determinant of matrix \(A\): \[\text{det}(A) = \begin{vmatrix} 1 & 2 & 1 \ 2 & 3 & 1 \ 1 & 4 & 2 \end{vmatrix} = 1(3 \times 2 - 1 \times 4) - 2(2 \times 2 - 1 \times 1) + 1(2 \times 4 - 3 \times 1) = -1 -2 + 5 = 2.\]Because the determinant is not zero, vectors of set (a) are linearly independent and span \(\mathbb{R}^3\).

Construct the Matrix for Set (b)

For set (b), construct the matrix \( B = \begin{bmatrix} 8 & 1 & -7 \ 1 & 2 & 1 \ 3 & 8 & 5 \end{bmatrix} \). Calculate its determinant to check for linear independence of the vectors.

Calculate the Determinant for Set (b)

Calculate the determinant of matrix \(B\): \[\text{det}(B) = \begin{vmatrix} 8 & 1 & -7 \ 1 & 2 & 1 \ 3 & 8 & 5 \end{vmatrix} = 8(2 \times 5 - 1 \times 8) - 1(1 \times 5 - 1 \times 3) -7(1 \times 8 - 2 \times 3) = 16 - 2 + 7 = 21.\]Because the determinant is not zero, vectors of set (b) are linearly independent and span \(\mathbb{R}^3\).

Conclusion

Both sets (a) and (b) consist of linearly independent vectors. Therefore, each set is capable of spanning \(\mathbb{R}^3\).

Key concepts

3-Dimensional Space

In the world of Linear Algebra, the 3-dimensional space, also known as \(\mathbb{R}^3\), is an essential concept that refers to the set of all ordered triples of real numbers. These triples can be visualized as points in space, akin to a universe with three axes: x, y, and z. The 3-dimensional space is crucial for modeling physical phenomena in our world, providing a mathematical framework to describe objects in a space with width, height, and depth.

When dealing with vectors in \(\mathbb{R}^3\), we often ask whether a given set of three vectors can fill this space, known as 'spanning' the space. This essentially means that any vector in the 3-dimensional space can be represented as a combination of the vectors in the set, using scalar multiplication and addition. For a set of vectors to span \(\mathbb{R}^3\), they need to be both three in number and linearly independent. This criterion is derived from the definition of the dimension itself, where the dimension refers to the minimum number of vectors needed to span the space.

Linear Independence

Linear independence is a vital concept in understanding if a set of vectors can provide a 'basis' for a vector space like \(\mathbb{R}^3\). A set of vectors is said to be linearly independent if no vector in the set can be written as a linear combination of the others.

To check for linear independence, we often rely on calculating the determinant of a matrix formed by the vectors in question. This determinant is a numerical value that gives insight into the properties of the matrix. If the determinant is non-zero, it indicates that the vectors are linearly independent. In the context of the exercise given, both sets of vectors (a) and (b) in \(\mathbb{R}^3\) are verified to be linearly independent through determinant calculation, allowing them to span the space.

Determinant

The determinant is a critical tool in linear algebra, particularly when dealing with issues of spanning and linear independence. It's a scalar value that can be computed from the elements of a square matrix, and in the context of our exercise, it is used to ascertain whether a matrix's columns are linearly independent.

The steps involve constructing a matrix from the vectors and then performing a calculation, which for a 3x3 matrix involves a specific formula. For example, the determinant can be found using cofactor expansion. If the calculated determinant is not equal to zero, the vectors used to make the matrix are linearly independent. In our exercise, determinants were calculated to be non-zero for both sets of vectors, confirming that they can span \(\mathbb{R}^3\).

Vector Spaces

A vector space is a fundamental concept in linear algebra, providing a framework where vectors can be added together and multiplied by scalars, all while staying within the same space. These operations are guided by specific rules and properties, making vector spaces a critical aspect of understanding more complex mathematical constructs.

For a set of vectors to span a vector space such as \(\mathbb{R}^3\), the vectors must adhere to the space's dimension requirements and showcase linear independence. Vector spaces allow for the flexibility to express any vector in the space as a linear combination of a basis set of vectors, which in this case should be three vectors that span the entire 3-dimensional space. This basis forms the building blocks of the vector space, enabling comprehensive manipulation and transformation of vectors within that space.