Question

In each case, find a basis for \(V\) that includes the vector \(\mathbf{v}\). a. \(V=\mathbb{R}^{3}, \mathbf{v}=(1,-1,1)\) b. \(V=\mathbb{R}^{3}, \mathbf{v}=(0,1,1)\) c. \(V=\mathbf{M}_{22}, \mathbf{v}=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]\) d. \(V=\mathbf{P}_{2}, \mathbf{v}=x^{2}-x+1\)

Short answer

The bases are: (a) \((1,-1,1), (1,0,0), (0,1,0)\), (b) \((0,1,1), (1,0,0), (0,0,1)\), (c) \([\begin{smallmatrix}1&1\\1&1\end{smallmatrix}], [\begin{smallmatrix}1&0\\0&0\end{smallmatrix}], [\begin{smallmatrix}0&1\\0&0\end{smallmatrix}], [\begin{smallmatrix}0&0\\1&0\end{smallmatrix}]\), (d) \(x^2-x+1, x, 1\).

Step-by-step solution

Understand the requirement for a basis

A basis for a vector space \(V\) is a set of vectors that are linearly independent and span \(V\). A set of vectors \(\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\}\) is a basis for \(V\) if every vector in \(V\) can be expressed as a linear combination of these vectors. We'll find a set that includes the given vector and forms a basis for each case.

Solve part (a)

For \(V = \mathbb{R}^3\) and \(\mathbf{v} = (1, -1, 1)\), we need two more vectors to form a basis for \(\mathbb{R}^3\). One possible basis is \(\{(1, -1, 1), (1, 0, 0), (0, 1, 0)\}\). This set of vectors is linearly independent and spans \(\mathbb{R}^3\).

Solve part (b)

For \(V = \mathbb{R}^3\) and \(\mathbf{v} = (0, 1, 1)\), we also need two additional vectors to form a basis. A suitable basis could be \(\{(0, 1, 1), (1, 0, 0), (0, 0, 1)\}\). These vectors are linearly independent and cover the entire space.

Solve part (c)

For \(V = \mathbf{M}_{22}\), the space of \(2 \times 2\) matrices, and \(\mathbf{v} = \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix}\), a basis can include \(\mathbf{v}\) as well as three other independent matrices. A possible basis is \(\left\{\begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix}\right\}\).

Solve part (d)

For \(V = \mathbf{P}_2\), the space of polynomials of degree at most 2, and \(\mathbf{v} = x^2 - x + 1\), a basis including \(\mathbf{v}\) would also need to include two polynomials of the form \(1\) (constant term) and \(x\). Therefore, a basis can be \(\{x^2 - x + 1, x, 1\}\).

Key concepts

Linear Independence

Understanding linear independence is crucial when discussing vector spaces and bases. Linear independence refers to a set of vectors within a vector space where no vector in the set can be represented as a linear combination of the others. Essentially, each vector contributes something uniquely "independent" to the overall set, without relying on the others.

To determine if vectors are linearly independent, you check whether a certain linear combination equals the zero vector only when all coefficients are zero. For instance, if you have vectors \( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \), they are linearly independent if the equation \( c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \ldots + c_n\mathbf{v}_n = \mathbf{0} \) holds true only when \( c_1 = c_2 = \ldots = c_n = 0 \).

Linear independence ensures that each vector adds something essential to the space, without being a mere replica or a combination of other vectors. When forming a basis, achieving linear independence among vectors is a key requirement.

Spanning Set

A spanning set for a vector space is a group of vectors that can be used to express any vector within that space. This means that if you have a vector space \( V \), a set of vectors \( \{\mathbf{v}_1, \mathbf{v}_2, \ldots , \mathbf{v}_n\} \) spans \( V \) if every vector \( \mathbf{w} \) in \( V \) can be written as a linear combination of these vectors.

Let's look at what this means in practical terms. In the context of a basis, the concept of a spanning set ensures that the combination of the basis vectors is enough to cover the entire space. For example, in \( \mathbb{R}^3 \), the standard basis \( \{(1, 0, 0), (0, 1, 0), (0, 0, 1)\} \) spans the space because any vector \( (x, y, z) \) in \( \mathbb{R}^3 \) can be represented as \( x(1, 0, 0) + y(0, 1, 0) + z(0, 0, 1) \).

When finding a basis, not only should the set of vectors be linearly independent, but they must also span the vector space. This is why a sufficient number of vectors are needed to cover all dimensions of the space.

Vector Spaces

Vector spaces, sometimes called linear spaces, are fundamental to understanding linear algebra. A vector space consists of a set of vectors along with two operations: vector addition and scalar multiplication. These operations must satisfy specific rules or axioms, such as being closed under addition and scalar multiplication and having an additive identity - usually represented by the zero vector.

Some common examples of vector spaces include \( \mathbb{R}^n \), the space of \( n \)-tuples of real numbers, and \( \mathbf{M}_{mn} \), the space of all \( m \times n \) matrices. More abstract examples include function spaces, like \( \mathbf{P}_2 \), which contains polynomials of degree up to 2.

A basis for a vector space, like \( \mathbb{R}^3 \) or \( \mathbf{M}_{22} \), must include a set of vectors that span the space and are linearly independent, meaning each vector contributes something unique. Understanding vector spaces and their properties is crucial when working with linear combinations, matrix operations, and transformations.