Problem 33 - Differential Equations and Linear Algebra

Question

Determine a spanning set for the null space of the given matrix \(A\) $$A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 3 & 4 & 5 \\ 5 & 6 & 7 \end{array}\right]$$

Step-by-Step Solution

Answer

The spanning set for the null space of the given matrix A is \(\{ \begin{bmatrix} -7\\ 2\\ 1 \end{bmatrix} \}\).

Step 1: Form the augmented matrix [A|0] and perform Gaussian elimination

First, we need to form the augmented matrix: $$ [A|0] = \left[\begin{array}{lll|l} 1 & 2 & 3 & 0 \\ 3 & 4 & 5 & 0 \\ 5 & 6 & 7 & 0 \end{array}\right] $$ Now, we perform Gaussian elimination to get [A|0] in reduced row echelon form (RREF): $$ [A|0]_{RREF} = \left[\begin{array}{lll|l} 1 & 2 & 3 & 0 \\ 0 & -2 & -4 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] $$

Step 2: Write down the equations corresponding to the RREF

With the RREF in hand, we write down the corresponding system of linear equations: $$ \begin{cases} x_{1}+2x_{2}+3x_{3}=0 \\ -2x_{2}-4x_{3}=0 \end{cases} $$

Step 3: Solve for the general solution of the system

We will express x1 and x2 in terms of x3, which is the free variable: From the second equation, we can solve for x2: $$ x_{2}=2x_{3} $$ Now, substitute x2 in the first equation and solve for x1: $$ x_{1}+2(2x_{3})+3x_{3}=0 $$ $$ x_{1}=-7x_{3} $$

Step 4: Write the general solution as a vector and find the spanning set for the null space

Now, express the general solution of the system in vector form: $$ x=\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix} = \begin{bmatrix} -7x_{3}\\ 2x_{3}\\ x_{3} \end{bmatrix} = x_{3} \begin{bmatrix} -7\\ 2\\ 1 \end{bmatrix} $$ So, the spanning set for the null space of A is: $$ \{ \begin{bmatrix} -7\\ 2\\ 1 \end{bmatrix} \} $$

Key Concepts

Gaussian elimination

Gaussian elimination is a method used to solve systems of linear equations. It transforms a matrix into a simpler form to facilitate finding solutions. Here's how it works:

• The process begins by forming an augmented matrix from the original matrix by adding a column of zeros—if solving for a null space like in this exercise.
• Next, we perform row operations to create zeros below each pivot position, which are the leading entries in a row.
• The main goal is to simplify the matrix, preferably into triangular form, so we can easily identify the relationships between different variables.

Remember, these operations do not change the solution set of the system of equations. Gaussian elimination is a powerful tool in understanding linear systems because it paves the way to further simplification into reduced row echelon form.

reduced row echelon form

Reduced row echelon form (RREF) is a simplified version of a matrix achieved through Gaussian elimination. A matrix is in RREF if it satisfies these conditions:

• All zero rows, if any, are at the bottom of the matrix.
• The leading entry of each non-zero row (called the pivot) is 1.
• Pivots are the only non-zero entry in their column.
• Each pivot is to the right of any pivots in the rows above it.

For the matrix in the example, RREF was used to identify clear relationships among the variables. Achieving RREF allows for straightforward insight into the dependency and behavior of variables, helping to extract the general solution easily.

general solution

The general solution of a system of linear equations is a description of all possible solutions. After achieving RREF, the equation system can be solved, showing variable dependencies. In our case:1. Identify free variables which can take any value (usually identified in the non-pivot columns).2. Express dependent variables in terms of the free variables. In this exercise:

• Start by solving the second equation in terms of the free variable, here, the third variable \(x_3\).
• Replace it in the first equation to find expressions for the remaining dependent variables, like \(x_1 = -7x_3\) and \(x_2 = 2x_3\).

Together, these solutions describe a line of solutions, a typical outcome for null spaces with a single free variable.

spanning set

A spanning set for a space is a collection of vectors that can be combined to form any vector in that space. Determining a spanning set for the null space of a matrix is particularly significant because it gives a complete picture of the solution set of a homogeneous system of linear equations.

• From the general solution, express it as a vector involving the free variable.
• In this case, the solution is \(x_3 \begin{bmatrix} -7 \ 2 \ 1 \end{bmatrix}\), showing that sets of linear combinations of this vector span the null space.
• A spanning set with fewer vectors is more efficient as it occupies minimal space.

The significance of finding a spanning set is that it serves to describe all possible solutions for the equations you're working with.