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Chapter 4: Problem 16

Determine the dimensions of Nul \(A\) and \(\mathrm{Col} A\) for the matrices shown in Exercises \(13-18\) . $$ A=\left[\begin{array}{rr}{3} & {4} \\ {-6} & {10}\end{array}\right] $$

Short Answer

Expert verified
Dimension of Nul A is 0; dimension of Col A is 2.

Step by step solution

01

Determine the Dimension of Nul A

The null space (Nul A) consists of all solutions to the homogeneous equation \(A\mathbf{x} = \mathbf{0}\). To find the dimension of Nul A, we need to determine the number of free variables in the reduced row echelon form (RREF) of matrix \(A\). Start by transforming \(A\) into RREF using Gaussian elimination:\[\begin{bmatrix} 3 & 4 \ -6 & 10 \end{bmatrix} \to \begin{bmatrix} 1 & \frac{4}{3} \ 0 & 18 \end{bmatrix}\to \begin{bmatrix} 1 & 0 \ 0 & 1.33 \end{bmatrix}\to \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]Since there are no free variables (both variables are leading), the dimension of Nul A is 0.
02

Determine the Dimension of Col A

The column space (Col A) is spanned by the linearly independent columns of \(A\). The number of linearly independent columns is equal to the rank of \(A\), which is also the number of pivot columns in the RREF of \(A\). From the RREF \(\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\), we see there are two pivot columns. Therefore, the dimension of Col A is 2.

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Key Concepts

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

Null Space
In linear algebra, the null space (often denoted as Nul A) of a matrix consists of all the vectors \(\mathbf{x}\) that satisfy the homogeneous equation \(A\mathbf{x} = \mathbf{0}\). Simply put, it is the set of all solutions to this equation. Calculating the null space involves a few key steps.
  • First, take the matrix A and apply operations to transform it into its reduced row echelon form (RREF).
  • Then, identify the free variables. The number of free variables corresponds to the number of dimensions in the null space.
For example, given matrix \(A=\begin{bmatrix} 3 & 4 \ -6 & 10 \end{bmatrix}\) transformed into RREF becomes \(\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\). As both columns have leading ones, there are no free variables. Thus, the null space is trivial and its dimension is 0. This means the null space only contains the zero vector.
Column Space
The column space (symbolized as Col A) of a matrix A includes all possible linear combinations of its column vectors. In essence, it's the set of vectors that can be formed by combining the columns of the matrix.To determine the dimension of the column space:
  • Find the reduced row echelon form (RREF) of the matrix.
  • Count the number of pivot columns (columns with leading 1s) in the RREF.
For matrix \(A=\begin{bmatrix} 3 & 4 \ -6 & 10 \end{bmatrix}\), the RREF is \(\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\). Both columns are pivot columns, which makes them linearly independent. Therefore, the dimension of the column space is 2, meaning the column vectors span a two-dimensional plane.
Gaussian Elimination
Gaussian elimination is a pivotal method in linear algebra used to transform matrices into a simpler form known as row echelon form (REF) or further reduced row echelon form (RREF). This transformation is essential for solving linear systems of equations, finding determinants, and calculating inverses.Here's a brief rundown of the process:
  • Begin with a matrix and use a series of row operations to convert the matrix into an upper triangular form.
  • The operations include swapping rows, multiplying a row by a nonzero scalar, and adding or subtracting a scalar multiple of one row from another.
  • Once in row-echelon form, further operations yield the reduced row-echelon form, simplifying analysis and interpretation.
In our example, transforming \(A=\begin{bmatrix} 3 & 4 \ -6 & 10 \end{bmatrix}\) using Gaussian elimination led us to \(\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\), clarifying both the null space and column space directly.
Reduced Row Echelon Form
The reduced row echelon form (RREF) of a matrix is a specific kind of matrix that simplifies understanding and solving linear equations. Attaining the RREF from any matrix involves a sequence of row operations.Characteristics of RREF include:
  • Leading entries in each nonzero row are 1.
  • Each leading 1 is the only nonzero entry in its column.
  • The leading 1 in each row appears to the right of the leading 1 in the row above.
For instance, matrix \(A=\begin{bmatrix} 3 & 4 \ -6 & 10 \end{bmatrix}\) was transformed to \(\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\) in RREF, providing insight that all variables are leading ones (i.e., no free variables exist). This simplifies determining both the null and column spaces, emphasizing both the structure and constraints of the algebraic system the matrix represents.

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Most popular questions from this chapter

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. $$ 2^{k}, 4^{k},(-5)^{k} ; y_{k+3}-y_{k+2}-22 y_{k+1}+40 y_{k}=0 $$

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Suppose \(A\) is \(m \times n\) and \(\mathbf{b}\) is in \(\mathbb{R}^{m} .\) What has to be true about the two numbers rank \([A \text { b }]\) and rank \(A\) in order for the equation \(A \mathbf{x}=\mathbf{b}\) to be consistent?
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