Problem 4 - Differential Equations and Linear Algebra

Question

Determine the null space of \(A\) and verify the Rank-Nullity Theorem. $$A=\left[\begin{array}{rrr}1 & 1 & -1 \\\3 & 4 & 4 \\\1 & 1 & 0\end{array}\right]$$

Step-by-Step Solution

Answer

The null space of matrix A is \(\mathrm{N}(A) = \{0\}\), as there are no free variables in the RREF and the only solution for the homogeneous system Ax = 0 is the trivial solution x = 0. The Rank-Nullity Theorem holds for matrix A since the rank (3) plus the dimension of the null space (0) is equal to the number of columns (3): \(3 + 0 = 3\).

Step 1: 1. Convert matrix A into Reduced Row Echelon Form using Gaussian elimination

Perform elementary row operations to change the matrix into its RREF. \[A =\left[\begin{array}{rrr}1 & 1 & -1 \\3 & 4 & 4 \\1 & 1 & 0 \end{array}\right] \rightarrow \left[\begin{array}{rrr}1 & 1 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right] \]

Step 2: 2. Identify the free variables

In the RREF of A, there are no free variables, as all the columns contain leading ones. This indicates that the system Ax = 0 has a unique solution.

Step 3: 3. Write the general solution of the system in terms of the free variables

Since there are no free variables, the only solution for the homogeneous system Ax = 0 is the trivial solution x = 0. Thus, the null space of A is just the set containing the zero vector: \[\mathrm{N}(A) = \{0\}\]

Step 4: 4. Verify the Rank-Nullity Theorem

The Rank-Nullity Theorem states that the dimension of the null space plus the rank of the matrix is equal to the number of columns in the matrix. In our case: Rank of A (number of leading ones in RREF): 3 Dimension of the null space (number of free variables): 0 Number of columns in matrix A: 3 The Rank-Nullity Theorem holds for matrix A since the rank plus the dimension of the null space is equal to the number of columns: \(3 + 0 = 3\).

Key Concepts

Null Space

The null space is a key concept in linear algebra, often denoted as \( \mathrm{N}(A) \) for a matrix \( A \). It consists of all the vectors that, when multiplied by \( A \), result in the zero vector. Essentially, it's the solution set to the equation \( Ax = 0 \). These vectors are sometimes called the "kernel" of the matrix.

• For the given matrix \( A \), discovering its null space involves solving \( Ax = 0 \).
• After using Gaussian elimination, if we find that there are no free variables—like in this case—the null space only comprises the zero vector.
• This means the matrix \( A \) has full column rank, contributing to a null space of dimension zero.

Understanding the null space helps to determine linear independence and analyze the solution types for linear systems.

Reduced Row Echelon Form

Reduced Row Echelon Form (RREF) is a form of a matrix where it has been simplified through row operations. The goal is to identify leading ones that help in determining the rank of the matrix and the solutions to a system of linear equations. Here’s what characterizes an RREF matrix:

• Every leading entry in each row is 1, and it’s the only non-zero entry in its column.
• Each leading 1 is to the right of any leading ones from previous rows.
• Rows consisting entirely of zeroes, if any, are at the bottom.

Applying this form to matrix \( A \) shows there are 3 leading 1s, indicating full rank. This simplifies analysis and solving of systems of equations, showing a unique solution in this instance.

Gaussian Elimination

Gaussian Elimination is a process used to transform any matrix into an upper triangular form, and eventually into Reduced Row Echelon Form (RREF). This method involves a series of row operations:

• Swap: Exchange two rows.
• Scale: Multiply a row by a non-zero constant.
• Add: Add a multiple of one row to another row to eliminate variables and simplify the system.

This systematic approach helps to solve linear equations and provides a clear path to find the rank of the matrix. In the exercise, Gaussian Elimination led the matrix \( A \) to reveal its rank directly, confirming that it has no free variables, resulting in a unique solution to the system \( Ax = 0 \). This technique not only aids in simplification but also in visualizing the structure and dependencies within a matrix.