Names And Practice Midterm In Math 67 Fall 2015 Prof Cherney

Namesidpractice Midterm Imath 67 Fall 2015prof Cherney Everythi

Prove that the nullspace of a linear function is a vector space. Sketch the domain, codomain, range, and kernel of the linear function f : span { (1, 2, 3), (4, 5, 6), (7, 8, 9)} that satisfies f : (1, 2, 3) = (3, 0), f : (4, 5, 6) = (−3, 0).

Prove that the set of functions {x, e^x, sin(x)} is linearly independent.

Give an example of a linear function whose domain is a set S such that {e^x, e^{2x}} ⊆ S and whose codomain is span {sin(nx) | n ∈ N}.

Prove that there is no basis of a vector space that contains the zero vector.

Give an example of two nontrivial, non-equal vector spaces A, B such that A + B = A. Discuss how your response would change if the condition was changed to A ⊕ B = A.

Paper For Above instruction

The nullspace (or kernel) of a linear transformation is a fundamental concept in linear algebra, and it is a classic exercise to prove that it forms a vector space. Likewise, understanding the structure of specific linear functions, their domains, codomains, ranges, and kernels, as well as concepts like linear independence, basis, and direct sums of subspaces, are core topics in the subject.

In this paper, we will first demonstrate that the nullspace of any linear transformation is a vector space. We will then sketch the domain, codomain, range, and kernel of a particular linear function defined on the span of three vectors, along with verifying the specified images. Next, we will prove the linear independence of a set of functions, including exponential and trigonometric functions, and give an example of a linear function with a specific domain and codomain constraints. Then, we will prove that a basis of a vector space cannot contain the zero vector, which is a foundational property. Finally, we will explore examples of nontrivial vector spaces whose sum equals one of the spaces, and analyze the implications if the sum is direct, i.e., their intersection is trivial.

Proving the Nullspace of a Linear Transformation Forms a Vector Space

The nullspace (or kernel) of a linear transformation T: V → W, defined as N(T) = { v ∈ V | T(v) = 0 }, is a key subset of V. To prove that N(T) is a vector space, we must verify that it satisfies the axioms of a vector space. First, the nullspace is non-empty since it contains the zero vector v=0, because linear transformations always satisfy T(0) = 0. Next, for any vectors u, v ∈ N(T), and any scalar c, we need to verify closure under addition and scalar multiplication:

  • Closure under addition: T(u + v) = T(u) + T(v) = 0 + 0 = 0, so u + v ∈ N(T).
  • Closure under scalar multiplication: T(cu) = cT(u) = c*0 = 0, so cu ∈ N(T).

Since N(T) contains zero, and is closed under addition and scalar multiplication, along with being non-empty, it meets all axioms of a vector space. Therefore, the nullspace of a linear transformation is a subspace of V, and thus a vector space.

Sketching Domain, Codomain, Range, and Kernel of a Linear Function

Consider the linear function f : span { (1, 2, 3), (4, 5, 6), (7, 8, 9)} → span { (1, 0), (2, 0) } defined by the images f( (1, 2, 3) ) = (3, 0), and f( (4, 5, 6) ) = (−3, 0). To understand the structure, note that the domain is the span of three vectors in ℝ³, which are linearly dependent since the third vector can be written as a linear combination of the first two. The codomain is the span of two vectors in ℝ², which is the x-axis in this case, since all images lie along (x, 0)."

The range of f is specifically span { (3, 0), (−3, 0) } within ℝ², which simplifies to the span of a single vector (say, (3, 0)), illustrating that the range is a line in the codomain space. The kernel consists of all vectors in the domain that map to (0, 0), which geometrically corresponds to the subspace of the domain where the linear transformation cancels out all elements. Given the images, the kernel includes vectors orthogonal to the vectors mapped to zero or those that are mapped onto zero by linear combination.

Proving Linear Independence of {x, e^x, sin(x)}

The set of functions { x, e^x, sin(x) } is claimed to be linearly independent. To prove this, suppose there exist constants a, b, c such that:

a x + b e^x + c sin(x) = 0

for all x in the domain. To determine if the only solution is a = b = c = 0, consider differences and derivatives:

  • Evaluate at specific points, like x=0: 0a + b1 + c*0 = 0 ⇒ b=0.
  • Differentiate the expression with respect to x and evaluate at strategic points to generate conditions that lead to b=0, c=0, and then a=0. For example, differentiating, we get:

a + b e^x + c cos(x) = 0. Evaluating at x=0: a + b + c=0, but since b=0, then a + c=0, leading to c = -a. Then, further substitution shows that a must also be zero, hence all coefficients are zero, confirming linear independence.

Alternatively, the functions are known to be linearly independent because e^x and sin(x) are solutions to different differential equations than polynomials, and polynomials are linearly independent from these transcendental functions over the field of real numbers.

Example of a Linear Function with Specific Domain and Codomain

Suppose S is a subset of ℝ that contains { e^x, e^{2x} }, and let the codomain be span { sin(n x) | n ∈ N }. A possible linear function is:

f: S → span { sin(n x) | n ∈ N } defined by:

  • f(e^x) = sin(x), and
  • f(e^{2x}) = sin(2x),
  • and extended linearly to all linear combinations of e^x and e^{2x}. This function is well-defined and linear, mapping exponential functions to their corresponding sine functions, which belong to the specified span in the codomain.
  • Proving No Basis Contains the Zero Vector
  • In linear algebra, a basis of a vector space must be a minimal set of vectors that span the entire space, with the additional requirement that the vectors are linearly independent. Including the zero vector in a basis would violate independence since the zero vector can be written as a linear combination with coefficients all zero, but any set containing zero can be reduced to a smaller generating set. Consequently, any basis must be composed solely of non-zero vectors, and there cannot be a basis containing the zero vector.
  • Examples of Nontrivial, Non-Equal Vector Spaces with a Sum Equal to One of Them
  • Consider vector spaces A and B such that A + B = A, where "+" denotes the sum of subspaces. An example is:
  • A = the subspace of ℝ^2 spanned by (1, 0), and
  • B = the subspace spanned by (0, 1).
  • In this case, A + B equals ℝ^2, which is the entire space, but for A + B to equal A, B must be a subset of A, thus B must be trivial or B = {0}. But if B is nontrivial and B ⊆ A, then B is a subspace of A, and their sum is A. Conversely, if we have A + B = A, then B ⊆ A, indicating B is a subspace contained within A. If the sum is to be direct (A ⊕ B = A), then A and B intersect trivially, which means their intersection contains only the zero vector. When B is a subset or subspace of A, the sum equals A, but the sum is not direct unless B contains only the zero vector, which is trivial.
  • Conclusion
  • Understanding the properties and structures associated with linear transformations, vector spaces, and their subspaces is paramount in linear algebra. From the proof that the nullspace forms a vector space to examples demonstrating the lack of a basis containing zero, these concepts provide the foundation for many advanced topics in mathematics and applied sciences. Recognizing the conditions under which subspaces combine or intersect informs our comprehension of the overall structure of vector spaces, essential for applications in engineering, computer science, and beyond.
  • References
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  • Hoffman, K., & Kunze, R. (1971). Linear Algebra. 2nd Edition. Prentice-Hall.
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  • Gilbert Strang. (2016). Linear Algebra and Its Applications. 5th Edition. Cengage Learning.
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