Problem 10 - Differential Equations and Linear Algebra

Question

Determine a basis for the subspace of \(\mathbb{R}^{n}\) spanned by the given set of vectors by (a) using the concept of the row space of a matrix, and (b) using the concept of the column space of a matrix. $$\\{(1,-1,2),(5,-4,1),(7,-5,-4)\\}$$

Step-by-Step Solution

Answer

A basis for the subspace of ℝⁿ spanned by the given set of vectors \(\{(1,-1,2),(5,-4,1),(7,-5,-4)\}\) is \(\{(1,-1,2),(5,-4,1)\}\) using both the concept of the row space and column space of a matrix.

Step 1: Form a matrix with given vectors as rows

Form a matrix A where the given vectors are the rows: $$ A = \begin{bmatrix} 1 & -1 & 2 \\ 5 & -4 & 1 \\ 7 & -5 & -4 \end{bmatrix} $$

Step 2: Perform Gaussian elimination to find the reduced row echelon form

Add/subtract multiples of one row to/from another row: $$ \begin{cases} R_2 = R_2 - 5R_1 \\ R_3 = R_3 - 7R_1 \end{cases} $$ After the above operations, we have: $$ A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & -9 \\ 0 & 2 & -18 \end{bmatrix} $$ Now subtract twice the second row from the third row: $$ R_3 = R_3 - 2R_2 $$ We get: $$ A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & -9 \\ 0 & 0 & 0 \end{bmatrix} $$ This is the reduced row echelon form of A.

Step 3: Form a basis from the nonzero rows

The nonzero rows of the matrix in reduced row echelon form are: $$\\{(1,-1,2),(0,1,-9)\\}$$ These rows form a basis for the subspace spanned by the given set of vectors. b) Using the concept of the column space of a matrix

Step 1: Form a matrix with given vectors as columns

Form a matrix B where the given vectors are the columns: $$ B = \begin{bmatrix} 1 & 5 & 7 \\ -1 & -4 & -5 \\ 2 & 1 & -4 \end{bmatrix} $$

Step 2: Perform Gaussian elimination to find the reduced row echelon form

We perform Gaussian elimination as in part (a): $$ \begin{cases} R_2 = R_2 + R_1 \\ R_3 = R_3 - 2R_1 \end{cases} $$ We have: $$ B = \begin{bmatrix} 1 & 5 & 7 \\ 0 & 1 & 2 \\ 0 & -9 & -18 \end{bmatrix} $$ Now add 9 times the second row to the third row: $$ R_3 = R_3 + 9R_2 $$ We get: $$ B = \begin{bmatrix} 1 & 5 & 7 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix} $$ This is the reduced row echelon form of B.

Step 3: Form a basis from the nonzero columns corresponding to the pivot columns

The pivot columns in the reduced row echelon form matrix are the first and second columns. Corresponding to these columns, the columns from the original matrix B are: $$\\{(1,-1,2), (5,-4,1)\\}$$ These columns form a basis for the subspace spanned by the given set of vectors. Both methods (a) and (b) give us the same basis for the subspace. Thus, a basis for the subspace of ℝⁿ spanned by the given set of vectors is: $$\\{(1,-1,2),(5,-4,1)\\}$$

Key Concepts

Row Space

In linear algebra, the row space of a matrix is a central concept when determining a basis for a subspace. The row space is composed of all possible linear combinations of its row vectors. Each row in a matrix can be seen as a vector in

• **Dimension**: The dimension of the row space is identical to the rank of the matrix, which is the number of leading non-zero rows in its reduced row echelon form.
• **Basis**: To find a basis for the row space, one typically performs Gaussian elimination to reach the reduced row echelon form (RREF). The non-zero rows in this form represent a set of linearly independent vectors that span the row space.

For example, for our matrix \( A \):\[A = \begin{bmatrix} 1 & -1 & 2 \ 5 & -4 & 1 \ 7 & -5 & -4 \end{bmatrix}\]We convert it into RREF to derive a basis:\[\{ (1, -1, 2), (0, 1, -9) \}\]

Column Space

The column space of a matrix is the set of all possible linear combinations of its column vectors. Unlike the row space, the column space is primarily concerned with the span of the matrix's column vectors themselves.

• **Dimension**: Similarly to row space, the dimension of the column space, also known as the column rank, is equal to the number of pivot columns in the reduced row echelon form.
• **Basis**: To determine a basis for the column space, the key is to look at the pivot columns (columns that contain leading entries) in the reduced row echelon form and focus on the corresponding columns in the original matrix.

Considering matrix \( B \):\[B = \begin{bmatrix} 1 & 5 & 7 \ -1 & -4 & -5 \ 2 & 1 & -4 \end{bmatrix}\]After performing Gaussian elimination, it's evident that columns corresponding to pivots are those of the original matrix \( B \):\[\{ (1, -1, 2), (5, -4, 1) \}\]These columns form a basis for the column space.

Gaussian Elimination

Gaussian elimination is a method used to solve systems of linear equations, but its utility extends to finding the reduced row echelon form (RREF) of a matrix, which is crucial for identifying bases of row and column spaces. Here’s how it works:

• **Row Operations**: The process involves three types of row operations - swapping two rows, multiplying a row by a non-zero scalar, and adding or subtracting multiples of rows from each other.
• **Goal**: The aim is to transform the matrix into a form where each leading entry of a row is the only non-zero number in its column, resembling a diagonal pattern as much as possible.

For example, through Gaussian elimination on matrix \( A \):\[A = \begin{bmatrix} 1 & -1 & 2 \ 5 & -4 & 1 \ 7 & -5 & -4 \end{bmatrix} \]The final RREF looks like:\[\begin{bmatrix} 1 & -1 & 2 \ 0 & 1 & -9 \ 0 & 0 & 0 \end{bmatrix} \] This form simplifies recognizing linearly independent rows and columns.

Reduced Row Echelon Form

Reduced Row Echelon Form (RREF) is a simplified version of a matrix achieved through Gaussian elimination.

• **Properties**: A matrix in RREF has several key attributes: - The leading entry of each non-zero row is 1. - Each leading 1 is the only non-zero entry in its column. - The leading 1 of each row is to the right of any leading 1s in previous rows.
• **Utility**: RREF is especially useful in easily identifying the rank of the matrix and providing a direct way to discern a basis for the row space among its non-zero rows.

For instance, when our matrix \( B \) was transformed:\[B = \begin{bmatrix} 1 & 5 & 7 \ 0 & 1 & 2 \ 0 & 0 & 0 \end{bmatrix} \]The RREF gives clear indicators of the matrix's structure and the independent vectors within.This clarity helps in selecting the correct basis vectors that define the associated subspaces.