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Chapter 4: Problem 13

In Exercises 13 and 14 , assume that \(A\) is row equivalent to \(B .\) Find bases for Nul \(A\) and \(\operatorname{Col} A .\) $$ A=\left[\begin{array}{rrrr}{-2} & {4} & {-2} & {-4} \\ {2} & {-6} & {-3} & {1} \\ {-3} & {8} & {2} & {-3}\end{array}\right], B=\left[\begin{array}{llll}{1} & {0} & {6} & {5} \\ {0} & {2} & {5} & {3} \\\ {0} & {0} & {0} & {0}\end{array}\right] $$

Short Answer

Expert verified
Col A basis: \([-2, 2, -3]^T, [4, -6, 8]^T\); Nul A basis: \([-6, -\frac{5}{2}, 1, 0]^T, [-5, -\frac{3}{2}, 0, 1]^T\).

Step by step solution

01

Understand the Relationship between A and B

Matrix A is row equivalent to matrix B, meaning they have the same row space. This implies they have the same column and null space dimensions, but B is in row echelon form, which is helpful in finding these spaces.
02

Identify the Pivot Columns in Matrix B

Matrix B is already in row echelon form. The pivot columns in matrix B are the first and second columns because they contain the leading entries 1 and 2, respectively.
03

Determine Column Space Basis from Matrix A

The column space of matrix A is spanned by the columns corresponding to the pivot columns of matrix B. Thus, the basis for the column space, Col A, is given by the first and second columns of matrix A: \[\text{Col A} = \begin{bmatrix} -2 \ 2 \ -3 \ \end{bmatrix}, \begin{bmatrix} 4 \ -6 \ 8 \end{bmatrix}.\]
04

Solve for the Null Space of Matrix B

The null space of matrix A, Nul A, is the set of solutions to the system \(B\mathbf{x} = \mathbf{0}\). Performing back substitution on B gives the following equations:1. \(x_1 + 6x_3 + 5x_4 = 0\) 2. \(2x_2 + 5x_3 + 3x_4 = 0\)3. \(x_3 = \text{free variable}, \; x_4 = \text{free variable}\).By solving these equations, we express the pivot variables in terms of free variables, \(x_3\) and \(x_4\):\[x_1 = -6x_3 - 5x_4\]\[x_2 = -\frac{5}{2}x_3 - \frac{3}{2}x_4\]This gives two basis vectors for Nul A:\[\begin{bmatrix} -6 \ -\frac{5}{2} \ 1 \ 0 \end{bmatrix}, \; \begin{bmatrix} -5 \ -\frac{3}{2} \ 0 \ 1 \end{bmatrix}.\]
05

Verify the Result

Ensure that the calculated basis vectors are linearly independent and satisfy the matrix equation \(B\mathbf{x} = \mathbf{0}\), confirming that they correctly span the null space of matrix A.

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Key Concepts

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

Column Space
The **column space** of a matrix is the set of all possible linear combinations of its column vectors. Basically, it is a subspace formed by the columns of the matrix.
In the given problem, the column space of matrix A (\( \operatorname{Col} A \)) is determined by identifying pivot columns in matrix B, which is row equivalent to A and already in row echelon form.
  • Since the first and second columns of matrix B are pivot columns, they correspond to the columns of A that form the basis for the column space.
  • Thus, the basis for \( \operatorname{Col} A \) is given by the first and second columns of matrix A: \(\begin{bmatrix} -2 \ 2 \ -3 \end{bmatrix}\) and \(\begin{bmatrix} 4 \ -6 \ 8 \end{bmatrix}\).
Finding the column space involves understanding the role of pivot columns as they dictate which columns retain the significant linear combinations needed to span the column space.
Null Space
The **null space** of a matrix consists of all vectors that, when multiplied by the matrix, result in the zero vector. This space reflects the solutions to the homogeneous equation \(A\mathbf{x} = \mathbf{0}\).
To find \(\operatorname{Nul} A\) in the problem, we used matrix B, as it is row equivalent to A and in a simpler form for calculations.
  • Performing back substitution on matrix B leads to equations describing relationships among the variables.
  • As found in the solution, there are two free variables, which means the null space will have two basis vectors corresponding to these free variables.
  • The null space basis can thus be expressed as: \(\begin{bmatrix} -6 \ -\frac{5}{2} \ 1 \ 0 \end{bmatrix}\) and \(\begin{bmatrix} -5 \ -\frac{3}{2} \ 0 \ 1 \end{bmatrix}\).
By understanding the free and pivot variables, one can systematically construct the basis for the null space, showing all vectors that map to zero given the original matrix.
Pivot Columns
**Pivot columns** are crucial in understanding the structure of a matrix. They correspond to columns containing leading entries (usually 1s) in its row-echelon form, and they effectively represent original converging points for linear transformations.
In the exercise, B is already in row echelon form, making it easier to identify the pivot columns.
  • Here, the first and second columns of B contain the leading non-zero entries (1 and 2).
  • These indicate which columns in A will form the basis for the column space.
  • These leading entries show how each subsequent entry depends on the free variables; they are the backbone for solving vector space problems.
Thus, recognizing pivot columns is essential for determining both column and null spaces as they enable us to identify which parts of the data structure are essential for spanning the respective spaces.
Row Echelon Form
The **row echelon form** of a matrix is a key step in simplifying matrix problems. It is achieved by transforming the matrix into a form where each successive row has more leading zeros, making calculations such as finding pivot columns straightforward.
Given that B is already in row echelon form in the exercise, processes become streamlined.
  • The steps involve using row operations to transform the matrix so that all the non-zero rows appear above any rows of all zero elements.
  • Additionally, each leading entry (pivot) in a row is to the right of the leading entry in the previous row.
  • This format helps identify dependencies between columns and assists in easy computation of both the column and null spaces.
Understanding the row echelon form is fundamental for solving systems of linear equations efficiently, allowing for clearer insights into the kernel and image of linear transformations.

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Most popular questions from this chapter

Let \(V\) be a vector space, and let \(T : V \rightarrow V\) be a linear trans- formation. Given \(\mathbf{z}\) in \(V,\) suppose \(\mathbf{x}_{p}\) in \(V\) satisfies \(T\left(\mathbf{x}_{p}\right)=\mathbf{z}\) , and let \(\mathbf{u}\) be any vector in the kernel of \(T .\) Show that \(\mathbf{u}+\mathbf{x}_{p}\) satisfies the nonhomogeneous equation \(T(\mathbf{x})=\mathbf{z} .\)

In Exercises \(7-12\) , assume the signals listed are solutions of the given difference equation. Determine if the signals form a basis for the solution space of the equation. Justify your answers using appropriate theorems. $$ 2^{k}, 4^{k},(-5)^{k} ; y_{k+3}-y_{k+2}-22 y_{k+1}+40 y_{k}=0 $$

\([\mathbf{M}]\) Let \(H=\operatorname{Span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) and \(\mathcal{B}=\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\} .\) Show that \(\mathbf{x}\) is in \(H\) and find the \(\mathcal{B}\) -coordinate vector of \(\mathbf{x},\) for $$ \mathbf{v}_{1}=\left[\begin{array}{r}{11} \\ {-5} \\ {10} \\\ {7}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{c}{14} \\ {-8} \\\ {13} \\ {10}\end{array}\right], \mathbf{x}=\left[\begin{array}{r}{19} \\\ {-13} \\ {18} \\ {15}\end{array}\right] $$

Let \(\mathbf{v}_{1}=\left[\begin{array}{r}{7} \\ {4} \\ {-9} \\\ {-5}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r}{4} \\ {-7} \\\ {2} \\ {5}\end{array}\right], \mathbf{v}_{3}=\left[\begin{array}{r}{1} \\\ {-5} \\ {3} \\ {4}\end{array}\right] .\) It can be verified that \(\mathbf{v}_{1}-3 \mathbf{v}_{2}+5 \mathbf{v}_{3}=\mathbf{0} .\) Use this information to find a basis for \(H=\operatorname{Span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\)

\([\mathbf{M}]\) Let \(\mathbf{a}_{1}, \ldots, \mathbf{a}_{5}\) denote the columns of the matrix \(A,\) where $$ A=\left[\begin{array}{rrrrr}{5} & {1} & {2} & {2} & {0} \\ {3} & {3} & {2} & {-1} & {-12} \\ {8} & {4} & {4} & {-5} & {12} \\ {2} & {1} & {1} & {0} & {-2}\end{array}\right], \quad B=\left[\begin{array}{ccc}{\mathbf{a}_{1}} & {\mathbf{a}_{2}} & {\mathbf{a}_{4}}\end{array}\right] $$ a. Explain why \(a_{3}\) and \(a_{5}\) are in the column space of \(B\) b. Find a set of vectors that spans Nul \(A .\) c. Let \(T : \mathbb{R}^{5} \rightarrow \mathbb{R}^{4}\) be defined by \(T(\mathbf{x})=A \mathbf{x} .\) Explain why \(T\) is neither one-to-one nor onto.
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