Question

In each case find bases for the row and column spaces of \(A\) and determine the rank of \(A\). a. \(\left[\begin{array}{rrrr}2 & -4 & 6 & 8 \\ 2 & -1 & 3 & 2 \\ 4 & -5 & 9 & 10 \\ 0 & -1 & 1 & 2\end{array}\right]\) b. \(\left[\begin{array}{rrr}2 & -1 & 1 \\ -2 & 1 & 1 \\ 4 & -2 & 3 \\ -6 & 3 & 0\end{array}\right]\) c. \(\left[\begin{array}{rrrrr}1 & -1 & 5 & -2 & 2 \\ 2 & -2 & -2 & 5 & 1 \\\ 0 & 0 & -12 & 9 & -3 \\ -1 & 1 & 7 & -7 & 1\end{array}\right]\) d. \(\left[\begin{array}{rrrr}1 & 2 & -1 & 3 \\ -3 & -6 & 3 & -2\end{array}\right]\)

Short answer

Ranks: a=2, b=2, c=2, d=2. Bases: Row & Column space bases from original matrices.

Step-by-step solution

Determine Row Echelon Form for Matrix A (a)

Given the matrix \(A = \begin{bmatrix} 2 & -4 & 6 & 8 \ 2 & -1 & 3 & 2 \ 4 & -5 & 9 & 10 \ 0 & -1 & 1 & 2 \end{bmatrix}\), we apply row operations to transform it into row echelon form (REF):First, we subtract Row 1 from Row 2:\[ \begin{bmatrix} 2 & -4 & 6 & 8 \ 0 & 3 & -3 & -6 \ 4 & -5 & 9 & 10 \ 0 & -1 & 1 & 2 \end{bmatrix} \]Next, subtract 2 times Row 1 from Row 3:\[ \begin{bmatrix} 2 & -4 & 6 & 8 \ 0 & 3 & -3 & -6 \ 0 & 3 & -3 & -6 \ 0 & -1 & 1 & 2 \end{bmatrix} \]Subtract Row 2 from Row 3:\[ \begin{bmatrix} 2 & -4 & 6 & 8 \ 0 & 3 & -3 & -6 \ 0 & 0 & 0 & 0 \ 0 & -1 & 1 & 2 \end{bmatrix} \]Add Row 2 to Row 4:\[ \begin{bmatrix} 2 & -4 & 6 & 8 \ 0 & 3 & -3 & -6 \ 0 & 0 & 0 & 0 \ 0 & 0 & -2 & -4 \end{bmatrix} \]

Identify Pivot Columns and Determine Rank of A (a)

In the row echelon form, the pivot positions are in the first and second columns. Therefore, the rank of \(A\) is 2.

Find Basis for Row Space of A (a)

The non-zero rows in row echelon form \(\begin{bmatrix}2 & -4 & 6 & 8\0 & 3 & -3 & -6\end{bmatrix}\) form a basis for the row space of \(A\). Thus, the basis is \(\{[2, -4, 6, 8], [0, 3, -3, -6]\}\).

Find Basis for Column Space of A (a)

The pivot columns of the original matrix \(A\) are columns 1 and 2. Extract these columns: \([2, 2, 4, 0]^T\) and \([-4, -1, -5, -1]^T\).These columns form a basis for the column space of \(A\).Hence, the basis is \(\{[2, 2, 4, 0], [-4, -1, -5, -1]\}\).

Repeat Process for Matrix A (b)

Use the same procedure for matrix \(b\), \(c\), and \(d\). Apply row operations to transform into row echelon form, identify pivot columns and non-zero rows, and determine bases for row and column spaces.

Solve for Matrix A (b)

Start with matrix \(B = \begin{bmatrix} 2 & -1 & 1 \ -2 & 1 & 1 \ 4 & -2 & 3 \ -6 & 3 & 0 \end{bmatrix} \).Perform row operations to obtain the REF:Row 2 + Row 1: \[ \begin{bmatrix} 2 & -1 & 1 \ 0 & 0 & 2 \ 4 & -2 & 3 \ -6 & 3 & 0 \end{bmatrix} \]Row 3 - 2 times Row 1: \[ \begin{bmatrix} 2 & -1 & 1 \ 0 & 0 & 2 \ 0 & 0 & 1 \ -6 & 3 & 0 \end{bmatrix} \]Row 4 + 3 times Row 1: \[ \begin{bmatrix} 2 & -1 & 1 \ 0 & 0 & 2 \ 0 & 0 & 1 \ 0 & 0 & 3 \end{bmatrix} \].

Rank and Bases for Row and Column Spaces of A (b)

In this REF, pivot columns are 1 and 3. The rank is 2.Row space basis: \([2, -1, 1], [0, 0, 2]\).Column space basis: Original columns 1 and 2, i.e., \([2, -2, 4, -6]\) and \([1, 1, 3, 0]\).

Solve for Matrix A (c)

Matrix \(C = \begin{bmatrix} 1 & -1 & 5 & -2 & 2 \ 2 & -2 & -2 & 5 & 1 \ 0 & 0 & -12 & 9 & -3 \ -1 & 1 & 7 & -7 & 1 \end{bmatrix} \).Apply row operations to obtain REF:Subtract Row 1 from Row 2: \[ \begin{bmatrix} 1 & -1 & 5 & -2 & 2 \ 0 & 0 & -7 & 7 & -1 \ 0 & 0 & -12 & 9 & -3 \ -1 & 1 & 7 & -7 & 1 \end{bmatrix} \]Add Row 1 to Row 4:\[ \begin{bmatrix} 1 & -1 & 5 & -2 & 2 \ 0 & 0 & -7 & 7 & -1 \ 0 & 0 & -12 & 9 & -3 \ 0 & 0 & 12 & -9 & 3 \end{bmatrix} \].

Rank and Bases for Row and Column Spaces of A (c)

The REF shows pivot columns at positions 1 and 3, so the rank is 2.Row space basis: \([1, -1, 5, -2, 2], [0, 0, -7, 7, -1]\).Column space basis: Original columns 1 and 3: \([1, 2, 0, -1]\) and \([5, -2, -12, 7]\).

Solve for Matrix A (d)

Matrix \(D = \begin{bmatrix} 1 & 2 & -1 & 3 \ -3 & -6 & 3 & -2 \end{bmatrix} \).Apply a row operation, add 3 times Row 1 to Row 2:\[ \begin{bmatrix} 1 & 2 & -1 & 3 \ 0 & 0 & 0 & 7 \end{bmatrix} \].

Rank and Bases for Row and Column Spaces of A (d)

The REF shows pivot columns at positions 1 and 4, so the rank is 2.Row space basis: \([1, 2, -1, 3], [0, 0, 0, 7]\).Column space basis: Original columns 1 and 4: \([1, -3]\) and \([3, -2]\).

Key concepts

Row Space

The row space of a matrix refers to the set of all possible linear combinations of its rows. When you transform a matrix into its row echelon form (REF), you identify the rows that can serve as a basis for this space. These are the non-zero rows in the REF. The row space can be seen as describing the dimensions spanned by these rows in a vector space. If the matrix row echelon form looks like a staircase, the non-zero rows are the steps.

• A basis for the row space is made up of these non-zero rows.
• The number of such non-zero rows is equal to the rank of the matrix.

For example, when we convert matrix "a" to its REF, the resulting non-zero rows are \[\begin{bmatrix}2 & -4 & 6 & 8\0 & 3 & -3 & -6\ \end{bmatrix}\]These rows form the basis for the row space. This process allows us to see how the rows are dependent on or independent from each other.

Column Space

Column space is the set of all linear combinations of the columns of the matrix. To determine the column space, identify the pivot columns from the row echelon form of the matrix. However, the actual columns forming the basis are taken from the original matrix.To find the column space, do the following:

• Transform the matrix to row echelon form to find the pivot columns.
• Return to the original matrix and extract these pivot columns.
• The columns extracted form a basis for the column space.

For instance, in matrix "a", the row echelon form indicates that columns 1 and 2 are pivot columns. By selecting the first and second columns from the original matrix, which are \[\begin{bmatrix}2\ 2\ 4\ 0\end{bmatrix} \text{and}\begin{bmatrix}-4\-1\-5\-1\end{bmatrix}\],these vectors span the column space. The concept of column space helps in applications like solving linear equations and understanding the range of the associated linear transformation.

Row Echelon Form

Row echelon form (REF) is a simplified version of a matrix obtained through a series of transformations called row operations. These operations include switching two rows, multiplying a row by a non-zero scalar, and adding or subtracting multiples of rows from each other. The result is a matrix with a "staircase" pattern, where each leading (non-zero) entry is to the right of the leading entry in the row above it.

• An entry of zero below a leading entry is required.
• Rows of all zeros, if they exist, are at the bottom.

Obtaining the REF provides insight into the properties of a matrix such as rank and helps find the basis for the row and column spaces. Using our example "a", the original matrix is transformed to the REF:\[\begin{bmatrix}2 & -4 & 6 & 8\ 0 & 3 & -3 & -6\ 0 & 0 & 0 & 0\ 0 & 0 & -2 & -4 \end{bmatrix}\]With this form, it becomes easier to identify linearly independent vectors and understand the structure of linear equations. Understanding how to transform a matrix to REF is critical for solving various linear algebra problems.