Problem 7 In each case find a basis of the... [FREE SOLUTION]

Chapter 5: Problem 7

In each case find a basis of the null space of \(A\). Then compute rank \(A\) and verify (1) of Theorem 5.4.2. a. \(A=\left[\begin{array}{rrr}3 & 1 & 1 \\ 2 & 0 & 1 \\ 4 & 2 & 1 \\ 1 & -1 & 1\end{array}\right]\) b. \(A=\left[\begin{array}{rrrrr}3 & 5 & 5 & 2 & 0 \\ 1 & 0 & 2 & 2 & 1 \\ 1 & 1 & 1 & -2 & -2 \\ -2 & 0 & -4 & -4 & -2\end{array}\right]\)

Short Answer

Expert verified

Null space basis for (a): \( \begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix} \). Rank is 2.\nNull space basis for (b): \( \begin{bmatrix} -2 \\ -1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 2 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 2 \\ 0 \\ 0 \\ 1 \end{bmatrix} \). Rank is 2.

Step by step solution

Analyze matrix A for part (a)

Given matrix:\[A=\left[\begin{array}{rrr}3 & 1 & 1 \ 2 & 0 & 1 \ 4 & 2 & 1 \ 1 & -1 & 1\end{array}\right]\]First, find the null space by solving the equation \(A\mathbf{x} = \mathbf{0}\). We perform row reduction (Gaussian elimination) to obtain a row-echelon form of the matrix.

Perform Gaussian elimination on A for part (a)

Start by using row operations to simplify matrix to row-echelon form. The reduced form is \[\left[\begin{array}{rrr}1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0\end{array}\right]\]

Find basis for null space of A for part (a)

The free variable is \(x_3\). Expressing variables in terms of \(x_3\):\[x_1 = -x_3, \quad x_2 = -x_3, \quad x_3 = x_3\].The general solution to \(A\mathbf{x} = \mathbf{0}\) is:\[\mathbf{x} = x_3 \begin{bmatrix} -1 \ -1 \ 1 \end{bmatrix}\]Thus, a basis for the null space is \(\begin{bmatrix} -1 \ -1 \ 1 \end{bmatrix}\).

Calculate rank of A for part (a)

Rank of matrix A is the number of pivot columns, which can be seen from row-echelon form:Rank \(A = 2\).

Verify Theorem 5.4.2 for part (a)

The theorem states: \( \text{rank }A + \text{nullity } A = \text{number of columns of } A\).Here, number of columns = 3, rank \( = 2\), nullity \( = 1\). \[2 + 1 = 3\].Thus, the check verifies correctly.

Analyze matrix A for part (b)

Given matrix:\[A=\left[\begin{array}{rrrrr}3 & 5 & 5 & 2 & 0 \ 1 & 0 & 2 & 2 & 1 \ 1 & 1 & 1 & -2 & -2 \ -2 & 0 & -4 & -4 & -2\end{array}\right]\]Same process as before; solve \(A\mathbf{x} = \mathbf{0}\) using Gaussian elimination.

Perform Gaussian elimination on A for part (b)

Row-reduce matrix A to its row-echelon form. The reduced form is \[\left[\begin{array}{rrrrr}1 & 0 & 2 & 2 & 1 \ 0 & 1 & 1 & -2 & -2 \ 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0\end{array}\right]\]

Find basis for null space of A for part (b)

Free variables are \(x_3, x_4\), and \(x_5\). Express variables in terms of \(x_3, x_4,\) and \(x_5\):\[x_1 = -2x_3 - 2x_4 - x_5, \quad x_2 = -x_3 + 2x_4 + 2x_5\].The general solution is:\[\mathbf{x} = x_3 \begin{bmatrix} -2 \ -1 \ 1 \ 0 \ 0 \end{bmatrix} + x_4 \begin{bmatrix} -2 \ 2 \ 0 \ 1 \ 0 \end{bmatrix} + x_5 \begin{bmatrix} -1 \ 2 \ 0 \ 0 \ 1 \end{bmatrix}\]Thus, a basis for the null space is \(\begin{bmatrix} -2 \ -1 \ 1 \ 0 \ 0 \end{bmatrix}, \begin{bmatrix} -2 \ 2 \ 0 \ 1 \ 0 \end{bmatrix}, \begin{bmatrix} -1 \ 2 \ 0 \ 0 \ 1 \end{bmatrix}\).

Calculate rank of A for part (b)

The number of pivot columns is the rank:Rank \(A = 2\).

Verify Theorem 5.4.2 for part (b)

Check if \( \text{rank }A + \text{nullity } A = \text{number of columns of } A\).Number of columns = 5, rank \( = 2\), nullity \( = 3\).\[2 + 3 = 5\].The check holds true.

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Key Concepts

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

Basis

The term "basis" in linear algebra refers to a set of vectors in a vector space that are linearly independent and span the entire space. When focusing on the null space of a matrix, a basis encompasses vectors that solve the equation \(A\mathbf{x} = \mathbf{0}\). To put it simply, the null space of a matrix consists of all vector solutions to this equation.

Each vector in the basis is independent from the others, meaning no vector can be written as a combination of others.
The vectors span the null space: any vector in the null space is a linear combination of the basis vectors.

Finding the basis of the null space helps us understand the dimension of the null space, known as the nullity of the matrix. For example, in part (a) of our exercise, the basis for the null space is \(\begin{bmatrix} -1 \ -1 \ 1 \end{bmatrix}\). This single vector implies the nullity of the matrix is 1, since it captures all solutions for \(A\mathbf{x} = \mathbf{0}\). In part (b), three vectors form the basis, indicating a nullity of 3.

Gaussian elimination

Gaussian elimination is a step-by-step process used to simplify a matrix into a more manageable form, specifically the row-echelon form or its further refined version, the reduced row-echelon form. This technique is crucial for solving systems of linear equations. It helps us identify the solutions for variables by transforming the original matrix into a form where back substitution can easily be applied.

Start by selecting a pivot in the matrix, which is a non-zero element ideally at the top of a column.
Use row operations such as swapping, scaling, and adding multiples of rows to create zeros below the pivot, making the matrix triangular.

For instance, in step 2 of our exercise for matrix \(A\) in part (a), we used Gaussian elimination to convert \[A = \left[\begin{array}{rrr}3 & 1 & 1 \ 2 & 0 & 1 \ 4 & 2 & 1 \ 1 & -1 & 1\end{array}\right]\] into its row-echelon form \(\left[\begin{array}{rrr}1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0\end{array}\right]\). This facilitates the identification of free variables and consequently the basis for the null space.

Rank-nullity theorem

The rank-nullity theorem is a fundamental concept that connects various aspects of a matrix, including its columns, rank, and nullity. This theorem states:\[\text{rank}(A) + \text{nullity}(A) = \text{number of columns of } A\]In simpler terms:

Rank refers to the number of leading (or pivot) columns in a matrix's row-echelon form.
Nullity is the dimension of the null space: the number of free variables.
The sum of rank and nullity equals the total number of columns in the matrix.

For example, in part (a) of our exercise, the matrix has three columns. When row-reduced, we see two leading columns, giving a rank of 2, and one free variable, indicating a nullity of 1. As per the theorem, the sum is \(2 + 1 = 3\), matching the number of columns. This theorem provides an essential check for verifying the correctness of our matrix analysis.

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