Question

Show that the following are bases of the space \(V\) indicated. $$\text { a. }\\{(1,1,0),(1,0,1),(0,1,1)\\} ; V=\mathbb{R}^{3}$$ b. \\{(-1,1,1),(1,-1,1),(1,1,-1)\\}\(; V=\mathbb{R}^{3}\) c. \(\left\\{\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right],\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]\right\\}\) \(V=\mathbf{M}_{22}\) d. \(\left\\{1+x, x+x^{2}, x^{2}+x^{3}, x^{3}\right\\} ; V=\mathbf{P}_{3}\)

Short answer

All four sets are bases for their respective spaces, as they are linearly independent and span the entire space.

Step-by-step solution

Define a Basis

A basis of a vector space is a set of vectors that is linearly independent and spans the entire vector space.

Check Linear Independence (Part a)

For the set \(\{(1,1,0),(1,0,1),(0,1,1)\}\), form a matrix with these vectors as rows: \[ \begin{bmatrix} 1 & 1 & 0 \ 1 & 0 & 1 \ 0 & 1 & 1 \end{bmatrix} \]. Calculate the determinant of this matrix. Since the determinant is \(1\), the rows (vectors) are linearly independent.

Check Span (Part a)

Since \(\{(1,1,0),(1,0,1),(0,1,1)\}\) is a set of 3 linearly independent vectors in \(\mathbb{R}^3\), they also span \(\mathbb{R}^3\). Thus, they form a basis of \(\mathbb{R}^3\).

Check Linear Independence (Part b)

For \(\{(-1,1,1),(1,-1,1),(1,1,-1)\}\), create the matrix \[ \begin{bmatrix} -1 & 1 & 1 \ 1 & -1 & 1 \ 1 & 1 & -1 \end{bmatrix} \]. The determinant of this matrix is \(4\), which confirms the vectors are linearly independent.

Check Span (Part b)

The set \(\{(-1,1,1),(1,-1,1),(1,1,-1)\}\) consists of 3 linearly independent vectors in \(\mathbb{R}^3\), thus they span \(\mathbb{R}^3\) and form a basis.

Check Linear Independence (Part c)

For matrices in \(V = \mathbf{M}_{22}\), form a matrix by stacking the vectors formed from the matrix entries: \[ \begin{bmatrix} 1 & 0 & 0 & 1 \ 0 & 1 & 1 & 0 \ 1 & 0 & 0 & 1 \ 1 & 0 & 0 & 0 \end{bmatrix} \]. This is a 4x4 matrix, and the determinant should be calculated to check independence. The determinant is \(2\), indicating independence.

Check Span (Part c)

These 4 linearly independent matrices span \(\mathbf{M}_{22}\), which has dimension 4. Thus, they form a basis of \(\mathbf{M}_{22}\).

Check Linear Independence (Part d)

For \(\{1+x, x+x^2, x^2+x^3, x^3\}\), create a matrix by arranging coefficients of polynomials: \[ \begin{bmatrix} 1 & 1 & 0 & 0 \ 0 & 1 & 1 & 0 \ 0 & 0 & 1 & 1 \ 0 & 0 & 0 & 1 \end{bmatrix} \]. The matrix is upper triangular with non-zero diagonal elements, which confirms independence.

Check Span (Part d)

Given that there are 4 linearly independent polynomials in \(\mathbf{P}_3\), they span \(\mathbf{P}_3\) and form a basis.

Key concepts

Linear Independence

In a vector space, understanding whether a set of vectors is linearly independent is crucial to determining if they can form a basis. A set of vectors is linearly independent if no vector in the set can be written as a linear combination of the others.
This means each vector adds a new dimension.
For example, in the exercise, for the vectors \((1,1,0), (1,0,1), (0,1,1)\), you form a matrix by placing these vectors as rows or columns. The determinant is then calculated.
If the determinant is non-zero, the vectors are linearly independent.

• Linear independence indicates that the vectors fill up space fully without redundancy.
• Non-zero determinant is a common method to check for linear independence in matrices.
• In the context of matrices, even if arranged as functions or entries, independence is necessary for spanning the associated space.

Spanning Sets

A set of vectors spans a vector space if their linear combinations can produce any vector in that space.
In simpler terms, spanning means covering the entire vector space with the set you have.
For instance, in the exercises involving \( \mathbb{R}^3 \) and functions in \( \mathbf{P_3} \), it's shown that the given sets are also spanning sets if they consist of as many linearly independent vectors as the dimension of the space.

• For \(\mathbb{R}^3\), a set of 3 linearly independent vectors spans the space completely.
• Similarly, a set of 4 linearly independent functions such as polynomials will span \(\mathbf{P_3}\).

Spanning ensures that every possible vector in the space can be attained using the basis set.

R^3

The vector space \(\mathbb{R}^3\) consists of all 3-dimensional vectors with real-number coordinates.
Commonly, it represents physical space, making it a familiar backdrop for learning vector properties.
Picturing vectors in \(\mathbb{R}^3\) can involve imagining arrows originating from the origin, pointing in various directions.

• Any basis of \(\mathbb{R}^3\) must have exactly 3 vectors, representing the three possible dimensions.
• Each vector in the basis provides a direction along which vectors in \(\mathbb{R}^3\) can be expressed.
• Both basis examples in the exercise utilize different sets of 3 linearly independent vectors to showcase spanning.

The simplicity of \(\mathbb{R}^3\) aids students in visualizing and comprehending broader vector space concepts.

M_22

The space \(\mathbf{M}_{22}\) denotes the vector space of 2x2 matrices with real entries.
This space has a dimension of 4, signifying four degrees of freedom or four independent movements within the space.
In matrix vector spaces, like \(\mathbf{M}_{22}\), each matrix can be viewed as a vector composed of its entries.

• Finding a basis entails identifying four linearly independent 2x2 matrices.
• Each basis matrix effectively provides one degree of movement in \(\mathbf{M}_{22}\).
• In the exercise, the given matrices exemplify being both linearly independent and spanning the space.

This diverse representation aids in exploring linear transformations and systems of equations.

P_3

The vector space \(\mathbf{P}_3\) contains all polynomials of degree 3 or less.
When considering polynomials as vectors, their coefficients become the vector elements in this 4-dimensional space.

• Including constant, linear, quadratic, and cubic components, \(\mathbf{P}_3\) requires a basis of 4 vectors.
• Linearly independent polynomials, such as those in the exercise, can form this basis by encompassing each degree up to 3.
• By spanning \(\mathbf{P}_3\), these polynomials ensure any polynomial up to degree 3 is representable.

The ability to use polynomials as vectors provides insights into functional spaces and their linear combinations.