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Let \(T: V \rightarrow W\) be a linear transformation between finite-dimensional vector spaces \(V\) and \(W\). Let \(B\) and \(C\) be bases for \(V\) and \(W,\) respectively, and let \(A=[T]_{c+B}\) Show that nullity \((T)=\) nullity \((A)\)

Short Answer

Expert verified
Nullity \((T)\) equals nullity \((A)\) because \(T\)'s kernel and \(A\)'s null space are equivalent.

Step by step solution

01

Understand the Nullity of a Linear Transformation

The nullity of a linear transformation \(T: V \rightarrow W\) is defined as the dimension of its kernel, \(\text{nullity}(T) = \dim(\ker(T))\). The kernel of \(T\) is the set of all vectors \(v \in V\) such that \(T(v) = 0\).
02

Matrix Representation of Linear Transformation

The matrix \(A = [T]_{C+B}\) is the representation of \(T\) with respect to the bases \(B\) for \(V\) and \(C\) for \(W\). The columns of \(A\) represent the images of the basis vectors of \(V\) under \(T\), expressed in terms of the basis \(C\).
03

Kernel of the Matrix Representation

The nullity of the matrix \(A\), \(\text{nullity}(A)\), is defined as the dimension of the null space (kernel) of \(A\), which is the set of all vectors \(x\) such that \(Ax=0\). This corresponds to the same set of vectors as \(\ker(T)\).
04

Equivalence of Kernels

Since the linear transformation \(T\) and its matrix representation \(A\) over bases \(B\) and \(C\) are equivalent encapsulations of the same linear mapping, the kernel of \(T\) in \(V\) is mapped to the null space of its matrix representation \(A\) under the basis \(B\). This implies \(\dim(\ker(T)) = \dim(\ker(A))\).
05

Conclusion

Thus, the nullity of the linear transformation \(T\) and the nullity of its matrix representation \(A\) are equal. Hence, \(\text{nullity}(T) = \text{nullity}(A)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Finite-Dimensional Vector Spaces
A vector space is a collection of vectors where you can perform operations such as addition and scalar multiplication. When we say a vector space is finite-dimensional, we are referring to the presence of a finite basis. A basis is a set of vectors that are linearly independent and span the entire space. For example, in a two-dimensional space like the plane, a basis might consist of two vectors that point in different directions that are not co-linear. Since the basis is finite, this means every vector in our space can be expressed as a linear combination of the basis vectors.

In our exercise, finite-dimensional vector spaces are denoted as \( V \) and \( W \) where \( T: V \rightarrow W \) is a linear transformation. Both \( V \) and \( W \) have bases \( B \) and \( C \) respectively, and each of these is composed of a finite number of vectors. This concept of finite dimensionality is crucial because it allows us to express vector transformations in terms of matrices, which is more manageable for computations compared to operations with infinite-dimensional vector spaces.
Matrix Representation
When we talk about the matrix representation of a linear transformation, we are looking at a convenient way to describe a transformation using a matrix. Consider the transformation \( T: V \rightarrow W \) again, where \( V \) and \( W \) are finite-dimensional vector spaces with bases \( B \) and \( C \). The matrix \( A = [T]_{C+B} \) is formed by expressing the images of the basis vectors of \( V \) under \( T \), in terms of the basis \( C \) of \( W \).

Essentially, this means that each column of matrix \( A \) represents where a basis vector of \( V \) ends up when mapped by \( T \). This mapping is captured in terms of the elements of the basis \( C \) in \( W \). So, the matrix is a compact snapshot of the transformation \( T \), making it easier to perform and verify operations such as adding transformations, finding kernels, or determining invertibility. It translates the abstract concept of a transformation into concrete numerical data.
Kernel of a Transformation
The kernel of a transformation \( T \) is a vital concept in understanding how transformations work in vector spaces. The kernel \( \ker(T) \) is the set of all vectors \( v \) in \( V \) such that \( T(v) = 0 \) in \( W \). If some vectors in the space \( V \) are mapped to the zero vector in \( W \), this indicates the presence of redundancy or collapse in the transformation. Identifying the kernel helps to determine how many dimensions of the original space are 'lost' when moving through the transformation \( T \).

Now, when we discuss the nullity of a transformation, this is just the dimension of the kernel of \( T \). For matrix representation, the nullity of the associated matrix \( A \), as described, corresponds to the dimension of the null space of \( A \). This set encompasses all the solutions to the matrix equation \( Ax = 0 \). By showing \( \text{nullity}(T) = \text{nullity}(A) \), we demonstrate that the matrix representation faithfully preserves the kernel properties of the linear transformation, validating that matrix-related operations can parallel those in vector spaces.

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Most popular questions from this chapter

A pendulum consists of a mass, called a bob, that is affixed to the end of a string of length \(L\) (see Figure 6.26 ). When the bob is moved from its rest position and released, it swings back and forth. The time it takes the pendulum to swing from its farthest right position to its farthest left position and back to its next farthest right position is called the period of the pendulum. (Figure can't copy) Let \(\theta=\theta(t)\) be the angle of the pendulum from the vertical. It can be shown that if there is no resistance, then when \(\theta\) is small it satisfies the differential equation \\[ \theta^{\prime \prime}+\frac{g}{L} \theta=0 \\] where \(g\) is the constant of acceleration due to gravity, approximately \(9.7 \mathrm{m} / \mathrm{s}^{2}\). Suppose that \(L=1 \mathrm{m}\) and that the pendulum is at rest (i.e., \(\theta=0\) ) at time \(t=0\) second. The bob is then drawn to the right at an angle of \(\theta_{1}\) radians and released. (a) Find the period of the pendulum. (b) Does the period depend on the angle \(\theta_{1}\) at which the pendulum is released? This question was posed and answered by Galileo in \(1638 .\) [Galileo Galilei \((1564-1642)\) studied medicine as a student at the University of Pisa, but his real interest was always mathematics. In \(1592,\) Galileo was appointed professor of mathematics at the University of Padua in Venice, where he taught primarily geometry and astronomy. He was the first to use a telescope to look at the stars and planets, and in so doing, he produced experimental data in support of the Copernican view that the planets revolve around the sun and not the earth. For this, Galileo was summoned before the Inquisition, placed under house arrest, and forbidden to publish his results. While under house arrest, he was able to write up his research on falling objects and pendulums. His notes were smuggled out of Italy and published as Discourses on Two New Sciences in 1638.]

Find the coordinate vector of \(p(x)=2-x+3 x^{2}\) with respect to the basis \(\mathcal{B}=\left\\{1,1+x,-1+x^{2}\right\\}\) of \(\mathscr{P}_{2}\)

Let \(U\) and \(W\) be subspaces of a finite-dimensional vector space \(V\). Prove Grassmann's Identity: \\[ \operatorname{dim}(U+W)=\operatorname{dim} U+\operatorname{dim} W-\operatorname{dim}(U \cap W) \\] [Hint: The subspace \(U+W\) is defined in Exercise 48 of Section 6.1. Let \(\mathcal{B}=\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\right\\}\) be a basis for \(U \cap W .\) Extend \(\mathcal{B}\) to a basis \(\mathcal{C}\) of \(U\) and a basis \(\mathcal{D}\) of \(W\) Prove that \(\mathcal{C} \cup \mathcal{D}\) is a basis for \(U+W\)

(a) Show that \(\mathscr{C}[0,1] \cong \mathscr{C}[2,3] .[\text { Hint: Define } T:\) \(\mathscr{C}[0,1] \rightarrow \mathscr{C}[2,3]\) by letting \(T(f)\) be the function whose value at \(x\) is \((T(f))(x)=f(x-2)\) for \(x\) in \([2,3] .]\) (b) Show that \(\mathscr{C}[0,1] \cong \mathscr{C}[a, a+1]\) for all \(a\).

Test the sets of matrices for linear independence in \(M_{22} .\) For those that are linearly dependent, express one of the matrices as a linear combination of the others. $$\left\\{\left[\begin{array}{rr} 1 & 1 \\ 0 & -1 \end{array}\right],\left[\begin{array}{rr} 1 & -1 \\ 1 & 0 \end{array}\right],\left[\begin{array}{ll} 1 & 0 \\ 3 & 2 \end{array}\right]\right\\}$$

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