Problem 3 - Differential Equations and Linear Algebra

Question

Determine the null space of \(A\) and verify the Rank-Nullity Theorem. $$A=\left[\begin{array}{rr}2 & -1 \\\\-4 & 2\end{array}\right]$$

Step-by-Step Solution

Answer

The null space of matrix A is the set of all vectors of the form \(t\left[\begin{array}{c}\frac{1}{2} \\ 1 \end{array}\right]\), where t is any scalar. The rank of A is 1 and the nullity of A is 1. Thus, the Rank-Nullity Theorem holds true for this matrix as rank(A) + nullity(A) = number of columns in A (1 + 1 = 2).

Step 1: Determine the null space of matrix A

To find the null space of matrix A, we want to find all the solutions to the equation Ax=0. Write down the augmented matrix by putting the right side to 0 and then perform row operations to obtain the row echelon form: \[ \left[\begin{array}{cc|r}2 & -1 & 0 \\ -4 & 2 & 0 \end{array}\right] \rightarrow \left[\begin{array}{cc|r}1 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 \end{array}\right] \] The second row represents a trivial equation (0=0), so we only need to focus on the first row. The first-row equation is \(x_1 -\frac{1}{2}x_2 = 0\) or \(x_1 = \frac{1}{2}x_2\). Since \(x_1 = \frac{1}{2}x_2\), let \(x_2=t\) for some scalar t. Then \(x_1=\frac{1}{2}t\). We can write the solution as: \(x=t\left[\begin{array}{c}\frac{1}{2} \\ 1 \end{array}\right]\) Thus, the null space of matrix A is the set of all vectors of the form \(t\left[\begin{array}{c}\frac{1}{2} \\ 1 \end{array}\right]\), where t is any scalar.

Step 2: Calculate the rank of matrix A

To calculate the rank of matrix A, find the number of linearly independent rows in the row echelon form: \[\left[\begin{array}{cc}1 & -\frac{1}{2} \\ 0 & 0 \end{array}\right]\] There is only 1 linearly independent row, so rank(A) = 1.

Step 3: Verify the Rank-Nullity Theorem

The Rank-Nullity Theorem states that the rank(A) + nullity(A) = number of columns in A. In this case, the number of columns in A is 2. Calculate the nullity of A: nullity(A) = number of columns in A - rank(A) = 2 - 1 = 1 So, the nullity of A is 1. Now, verify that rank(A) + nullity(A) = number of columns in A: 1 + 1 = 2 Since the left side equals the right side, the Rank-Nullity Theorem holds true for this matrix A.

Key Concepts

Rank-Nullity Theorem

The Rank-Nullity Theorem is a crucial concept in linear algebra. It provides an insightful relationship between the rank of a matrix and its nullity. This theorem states that for any matrix \( A \) with dimensions \( m \times n \), the sum of its rank and nullity equals the number of its columns.

• Rank is the number of linearly independent rows or columns in the matrix.
• Nullity is the dimension of the null space of the matrix, which essentially measures the number of free variables.

In our example matrix \( A = \left[\begin{array}{rr}2 & -1 \ -4 & 2\end{array}\right] \), we calculated:

• Rank(A) is 1.
• The nullity is calculated as 1 (because nullity = number of columns - rank).

Thus, applying the theorem: \( 1 + 1 = 2 \), which matches the number of columns. This verification confirms the Rank-Nullity Theorem in our calculation.

Linear Independence

Linear independence is an important property of matrices and sets of vectors. It tells us if vectors are not simple combinations of each other.

• A set of vectors is linearly independent if no vector can be formed as a linear combination of the others.
• In matrices, this property relates to rows or columns, helping determine the rank.

When transforming Matrix \( A \) into its row echelon form, we identify the number of linearly independent rows. In this case, with the row echelon form \( \left[\begin{array}{cc}1 & -\frac{1}{2} \ 0 & 0 \end{array}\right] \), only the first row is non-zero, showing that there’s only one independent row, establishing that rank(A) = 1.

Row Echelon Form

Getting a matrix into row echelon form involves a series of row operations to simplify it. Row echelon form aids in solving linear equations and finding matrix properties, like rank and null space.

• In row echelon form, each leading entry of a row is to the right of the leading entry of the row above it.
• All zeros are at the bottom of the matrix.

For the matrix \( A \), with initial operations, we turned \( A \) into \( \left[\begin{array}{cc}1 & -\frac{1}{2} \ 0 & 0 \end{array}\right] \). Working with this form:

• We easily solved for the null space by setting the equation \( x_1 - \frac{1}{2}x_2 = 0 \).
• This showed the relation \( x_1 = \frac{1}{2}x_2 \), leading us to express the null space in parameter form.

Matrix Rank

The matrix rank represents the number of linearly independent rows or columns a matrix has. It’s instrumental in understanding the solutions to linear systems.

• Rank informs us about potential solution dimensions, such as if there are unique, infinite, or no solutions to a system.
• For an \( n \times n \) matrix, rank could be at most \( n \).

In the exercise, we derived that matrix \( A \) has 1 linearly independent row from its row echelon form. This tells us rank(A) = 1. It reflects the simplicity, as one row suffices to describe the pivotal elements of \( A \). Understanding rank is also vital in confirming properties like those in the Rank-Nullity theorem.