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Chapter 7: Problem 1

For each matrix \(A,\) find a basis for the kernel and image of \(T_{A}\), and find the rank and nullity of \(T_{A}\) a. \(\left[\begin{array}{rrrr}1 & 2 & -1 & 1 \\ 3 & 1 & 0 & 2 \\ 1 & -3 & 2 & 0\end{array}\right]\) b. \(\left[\begin{array}{rrrr}2 & 1 & -1 & 3 \\ 1 & 0 & 3 & 1 \\ 1 & 1 & -4 & 2\end{array}\right]\) c. \(\left[\begin{array}{rrr}1 & 2 & -1 \\ 3 & 1 & 2 \\ 4 & -1 & 5 \\ 0 & 2 & -2\end{array}\right]\) d. \(\left[\begin{array}{rrr}2 & 1 & 0 \\ 1 & -1 & 3 \\ 1 & 2 & -3 \\ 0 & 3 & -6\end{array}\right]\)

Short Answer

Expert verified
For each matrix, use steps 1-5 to find kernel, image, rank, and nullity.

Step by step solution

01

Find the row-echelon form (REF) for Matrix A (part a)

Given matrix \(A = \begin{bmatrix} 1 & 2 & -1 & 1 \ 3 & 1 & 0 & 2 \ 1 & -3 & 2 & 0 \end{bmatrix}\). First, we perform row operations to convert this matrix to its row-echelon form (REF):1. Swap \(R_1\) and \(R_3\) to make the pivot positive and simplify calculations: \( \begin{bmatrix} 1 & -3 & 2 & 0 \ 3 & 1 & 0 & 2 \ 1 & 2 & -1 & 1 \end{bmatrix} \).2. Subtract 3 times \(R_1\) from \(R_2\), and subtract \(R_1\) from \(R_3\): \( \begin{bmatrix} 1 & -3 & 2 & 0 \ 0 & 10 & -6 & 2 \ 0 & 5 & -3 & 1 \end{bmatrix} \).3. Subtract half of \(R_2\) from \(R_3\): \( \begin{bmatrix} 1 & -3 & 2 & 0 \ 0 & 10 & -6 & 2 \ 0 & 0 & 0 & 0 \end{bmatrix} \).The matrix is now in REF.
02

Identify pivot and free columns from the REF (part a)

The REF matrix is \( \begin{bmatrix} 1 & -3 & 2 & 0 \ 0 & 10 & -6 & 2 \ 0 & 0 & 0 & 0 \end{bmatrix} \). Identify the pivot columns which are columns 1 and 2. The free columns are 3 and 4.
03

Determine the basis for the kernel of matrix A (part a)

Since columns 3 and 4 are free, express the variables in terms of these free variables. Let \(x_3 = s\) and \(x_4 = t\). Solve the system of equations derived from the REF:1. \(x_1 - 3x_2 + 2x_3 = 0\)2. \(10x_2 - 6x_3 + 2x_4 = 0\)From equation 2: \(x_2 = \frac{3}{5}x_3 - \frac{1}{5}x_4\).From equation 1: \(x_1 = 3x_2 - 2x_3\).Substitute \(x_2\) into \(x_1 = 3\left( \frac{3}{5}x_3 - \frac{1}{5}x_4 \right) - 2x_3\), simplify to find \(x_1 = \frac{9}{5}x_3 - \frac{3}{5}x_4 - 2x_3 = \frac{-1}{5}x_3 - \frac{3}{5}x_4\).The kernel's basis vectors are \[ \begin{bmatrix} -\frac{1}{5} \ \frac{3}{5} \ 1 \ 0 \end{bmatrix}, \begin{bmatrix} -\frac{3}{5} \ -\frac{1}{5} \ 0 \ 1 \end{bmatrix} \].
04

Find the basis for the image of matrix A (part a)

The pivot columns in the original matrix \(A\) correspond to the linear independent columns forming the image. The basis for the image is formed by the original columns of \(A\) in the pivot positions: \( \begin{bmatrix} 1 \ 3 \ 1 \end{bmatrix} \) and \( \begin{bmatrix} 2 \ 1 \ -3 \end{bmatrix} \).
05

Calculate the rank and nullity (part a)

The rank of the transformation \(T_A\) is the number of pivot columns in the REF, which is 2. The nullity is the number of free columns, which is also 2 since the total number of columns is 4 and the rank is 2.
06

Repeat Steps 1-5 for matrix B

Repeat the row reduction process and find the REF of matrix \(B\) and continue as we did in parts a, using similar row operations, to find bases for the kernel and image, and determine rank and nullity.
07

Repeat Steps 1-5 for matrix C

Perform the row operations necessary on matrix \(C\), determine the REF, and continue to find appropriate bases for the kernel and image, and to determine rank and nullity.
08

Repeat Steps 1-5 for matrix D

Likewise, for matrix \(D\), perform row operations to REF and find bases for the kernel and image, and calculate rank and nullity.

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Key Concepts

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

Row-Echelon Form
When working with matrices in linear algebra, you will often need to simplify them to a form known as the row-echelon form (REF). This form is not only helpful for solving systems of linear equations but also plays a crucial role in determining other matrix properties. To convert a matrix to row-echelon form, perform a series of row operations such as row swapping, scaling, and adding multiples of one row to another. In REF, the leading entry (or pivot) in each non-zero row is 1, and all entries below each pivot are zeros. This structure helps to easily identify pivot positions, which are essential in calculating the rank of the matrix. Remember, the row operations should preserve the solutions of the corresponding linear system.
Kernel of a Matrix
The kernel, or null space, of a matrix is a fundamental concept representing all the solutions to the homogeneous equation \( A \mathbf{x} = \mathbf{0} \).Finding the kernel involves determining free variables and expressing the solution set in terms of these variables. To do this, it is necessary to convert the matrix to row-echelon form, identify the pivot and free columns, and then express the internal equations in terms of the free variables. The set of vectors obtained represents a basis for the kernel. These basis vectors form the solution space where each vector is orthogonal to the image of the matrix, providing insights into linear dependencies.
Image of a Matrix
The image, or column space, of a matrix consists of all possible linear combinations of its columns. To find it, identify the pivot columns of the matrix once it is in row-echelon form. The basis for the image is formed by these columns in the original matrix. This step ensures that the vectors you select are linearly independent, forming a subspace that spans the entirety of the image. The rank theorem relates the image to the matrix's rank by stating this is equal to the dimension of the image. Thus, the image tells us about the span of the transformation implied by the matrix.
Rank and Nullity
Rank and nullity are two key concepts in understanding the properties of matrices. The rank of a matrix is the number of pivot columns in its row-echelon form. This value represents the dimension of the image of the matrix, indicating how many directions the matrix transformation successfully spans. In contrast, nullity refers to the number of free columns, showing the dimension of the kernel. An important relation between these quantities is articulated by the Rank-Nullity Theorem, stating: \[\text{rank}(A) + \text{nullity}(A) = \text{number of columns of } A\]This theorem provides an insightful balance between the dimensions of the image and kernel, ensuring that all aspects of the transformation are accounted for.
Basis for Vector Spaces
A basis in linear algebra provides a way to uniquely express every vector in a vector space as a linear combination of the basis vectors. Each basis is a set of vectors that are linearly independent and span the entire space. For the kernel and image of a matrix, bases serve crucial roles. The basis for the kernel is derived from the set of solutions to the homogeneous equation, while the basis for the image is constructed from the pivot columns of the original matrix. In practical applications, establishing a basis allows for efficient descriptions and transformations within vector spaces, facilitating deeper understanding of dimensions, and enabling efficient computation of coordinates of vectors within these spaces. Understanding and constructing bases accurately is fundamental to mastering linear algebra.

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Most popular questions from this chapter

Show that linear independence is preserved by one-to-one transformations and that spanning sets are preserved by onto transformations. More precisely, if \(T: V \rightarrow W\) is a linear transformation, show that: a. If \(T\) is one-to-one and \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\\}\) is independent in \(V,\) then \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{n}\right)\right\\}\) is independent in \(W\) b. If \(T\) is onto and \(V=\operatorname{span}\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\\},\) then \(W=\operatorname{span}\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{n}\right)\right\\}\)

Let \(T: \mathbf{M}_{n n} \rightarrow \mathbb{R}\) denote the trace map: \(T(A)=\operatorname{tr} A\) for all \(A\) in \(\mathbf{M}_{n n}\). Show that \(\operatorname{dim}(\operatorname{ker} T)=n^{2}-1\)

Let \(U\) and \(V\) denote the spaces of symmetric and skew-symmetric \(n \times n\) matrices. Show that \(\operatorname{dim} U+\operatorname{dim} V=n^{2}\)

Exercise 7.3 .4 In each case, compute the action of \(S T\) and \(T S,\) and show that \(S T \neq T S\). a. \(S: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) with \(S(x, y)=(y, x) ; T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) with \(T(x, y)=(x, 0)\) b. \(S: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) with \(S(x, y, z)=(x, 0, z)\) \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) with \(T(x, y, z)=(x+y, 0, y+z)\) c. \(S: \mathbf{P}_{2} \rightarrow \mathbf{P}_{2}\) with \(S(p)=p(0)+p(1) x+p(2) x^{2}\); \(T: \mathbf{P}_{2} \rightarrow \mathbf{P}_{2}\) with \(T\left(a+b x+c x^{2}\right)=b+c x+a x^{2}\) d. \(S: \mathbf{M}_{22} \rightarrow \mathbf{M}_{22}\) with \(S\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}a & 0 \\ 0 & d\end{array}\right]\); \(T: \mathbf{M}_{22} \rightarrow \mathbf{M}_{22}\) with \(T\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}c & a \\ d & b\end{array}\right]\)

In each case, find a linear transformation with the given properties and compute \(T(\mathbf{v})\) $$ \begin{array}{l} \text { a. } T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} ; T(1,2)=(1,0,1) \\ \quad T(-1,0)=(0,1,1) ; \mathbf{v}=(2,1) \\ \text { b. } T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} ; T(2,-1)=(1,-1,1) \\\ \quad T(1,1)=(0,1,0) ; \mathbf{v}=(-1,2) \\ \text { c. } T: \mathbf{P}_{2} \rightarrow \mathbf{P}_{3} ; T\left(x^{2}\right)=x^{3}, T(x+1)=0 \\ \quad T(x-1)=x ; \mathbf{v}=x^{2}+x+1 \\ \text { d. } T: \mathbf{M}_{22} \rightarrow \mathbb{R} ; T\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]=3, T\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]=-1, \\ \quad T\left[\begin{array}{ll} 1 & 0 \\ 1 & 0 \end{array}\right]=0=T\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right] ; \mathbf{v}=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \end{array} $$
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