Problem 42 - Differential Equations and Linear Algebra

Question

Find a basis and the dimension for the row space, column space, and null space of the given matrix \(A\) $$A=\left[\begin{array}{rrr} -4 & 0 & 3 \\ 0 & 10 & 13 \\ 6 & 5 & 2 \\ -2 & 5 & 10 \end{array}\right]$$

Step-by-Step Solution

Answer

For the given matrix A, the basis for the column space is \(\left\{ \begin{bmatrix}-4 \\ 0 \\ 6 \\ -2\end{bmatrix}, \begin{bmatrix}0 \\ 10 \\ 5 \\ 5\end{bmatrix}\right\}\) with dimension 2; the basis for the row space is \(\left\{ \begin{bmatrix}1 & 0 & -\frac{3}{4}\end{bmatrix}, \begin{bmatrix}0 & 1 & \frac{13}{10}\end{bmatrix}\right\}\) with dimension 2; and the basis for the null space is \(\left\{\begin{bmatrix}\frac{3}{4} \\ -\frac{13}{10} \\ 1\end{bmatrix}\right\}\) with dimension 1.

Step 1: Given matrix A is: $$A=\left[\begin{array}{rrr} -4 & 0 & 3 \\\ 0 & 10 & 13 \\\ 6 & 5 & 2 \\\ -2 & 5 & 10 \end{array}\right]$$ #Step 2: Perform Gaussian elimination and find RREF#

After performing Gaussian elimination, we find the row-reduced echelon form (RREF) of A: $$RREF(A)=\left[\begin{array}{rrr} 1 & 0 & -\frac{3}{4} \\\ 0 & 1 & \frac{13}{10} \\\ 0 & 0 & 0 \\\ 0 & 0 & 0 \end{array}\right]$$ #Step 3: Identify pivot columns and basis for column space#

Step 2: The first and second columns of RREF(A) are pivot columns. Therefore, we can form a basis for the column space of A using the linearly independent corresponding columns in matrix A: $$Basis_{Col}(A) = \left\{ \begin{bmatrix}-4 \\ 0 \\ 6 \\ -2\end{bmatrix}, \begin{bmatrix}0 \\ 10 \\ 5 \\ 5\end{bmatrix}\right\}$$ The dimension of the column space Dim(Col(A)) is 2 (the number of vectors in the basis). #Step 4: Identify basis for row space and its dimension#

Non-zero rows in RREF(A) form the basis for the row space of A: $$Basis_{Row}(A) = \left\{ \begin{bmatrix}1 & 0 & -\frac{3}{4}\end{bmatrix}, \begin{bmatrix}0 & 1 & \frac{13}{10}\end{bmatrix}\right\}$$ The dimension of the row space Dim(Row(A)) is 2 (the number of vectors in the basis). #Step 5: Find the general solution and basis for null space#

Step 3: The general solution of the system in RREF form is: $$\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} = x_3\begin{bmatrix}\frac{3}{4} \\ -\frac{13}{10} \\ 1\end{bmatrix} = k\begin{bmatrix}\frac{3}{4} \\ -\frac{13}{10} \\ 1\end{bmatrix}$$ where \(k\) is a scalar. Therefore, the basis for the null space is: $$Basis_{Null}(A) = \left\{\begin{bmatrix}\frac{3}{4} \\ -\frac{13}{10} \\ 1\end{bmatrix}\right\}$$ The dimension of the null space is 1 (the number of vectors in the basis and the number of free variables). #Summary: Basis and dimensions for given matrix A#

For the given matrix A, - Basis for column space: \(\left\{ \begin{bmatrix}-4 \\ 0 \\ 6 \\ -2\end{bmatrix}, \begin{bmatrix}0 \\ 10 \\ 5 \\ 5\end{bmatrix}\right\}\), Dimension = 2 - Basis for row space: \(\left\{ \begin{bmatrix}1 & 0 & -\frac{3}{4}\end{bmatrix}, \begin{bmatrix}0 & 1 & \frac{13}{10}\end{bmatrix}\right\}\), Dimension = 2 - Basis for null space: \(\left\{\begin{bmatrix}\frac{3}{4} \\ -\frac{13}{10} \\ 1\end{bmatrix}\right\}\), Dimension = 1

Key Concepts

Basis

In linear algebra, a **basis** is a set of vectors that are linearly independent and span a vector space. Understanding the concept of a basis is crucial because it provides a way of describing the entire vector space using a minimal number of vectors.

• A set of vectors \( \{ \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \} \) is considered to be a basis if no vector in the set can be written as a linear combination of the others.
• The basis vectors can generate other vectors in the space through linear combinations.

In practical terms, for a given matrix, when we find a basis for its row space or column space, it helps us understand the linear structure of the matrices, such as which directions the matrix transformation affects. This is a foundational concept for determining the dimension of these spaces.

Dimension

The **dimension** of a vector space is the number of vectors in its basis, essentially telling us how many degrees of freedom or directions the space has.

• If a vector space has dimension \( n \), every basis for that space will contain exactly \( n \) vectors.
• Dimension can also be seen as a measure of the "size" or "capacity" of the space.

In the context of the row and column spaces of a matrix, the dimension will tell us about the rank of the matrix. For instance, from our problem, both the row space and column space for our matrix have a dimension of 2, indicating that both the row vectors and column vectors can span a two-dimensional space. Understanding dimension is crucial for determining things like rank and the solvability of matrix equations.

Row Space

The **row space** of a matrix is the set of all possible linear combinations of its row vectors. It provides insight into the linear relationships between the rows of a matrix.

• To find the row space, convert the matrix to row-reduced echelon form (RREF). The non-zero rows of the RREF give the basis for the row space.
• In our solution, the non-zero rows \( \begin{bmatrix}1 & 0 & -\frac{3}{4}\end{bmatrix}, \begin{bmatrix}0 & 1 & \frac{13}{10}\end{bmatrix} \)provide a basis for the row space.

This particular basis allows us to express any row in the matrix as a combination of these basis rows. The dimension of the row space is also known as the row rank of the matrix.

Column Space

The **column space** of a matrix, also known as the range, is the set of all possible linear combinations of its column vectors. It tells us about the solution set the matrix can produce or "map" to when applied to vectors.

• To determine the basis for the column space, identify the pivot columns in the RREF of the matrix, and take the corresponding original columns from the initial matrix.
• In our example, the pivot columns are the first and second ones in the RREF, thus forming the column basis vectors \( \begin{bmatrix}-4 \ 0 \ 6 \ -2\end{bmatrix}, \begin{bmatrix}0 \ 10 \ 5 \ 5\end{bmatrix} \).

This basis indicates which directions the transformations associated with the matrix can take. The dimension of the column space tells us how many linearly independent directions the columns of the matrix span, which is also known as the column rank of the matrix.

Null Space

The **null space** of a matrix is the set of all vectors that result in the zero vector when multiplied by the matrix. It helps us understand the solutions to the homogeneous equation \( A\mathbf{x} = \mathbf{0} \).

• To find a basis for the null space, solve the equation \( A\mathbf{x} = \mathbf{0} \) using the RREF of the matrix.
• In our case, the solution \( \begin{bmatrix}x_1 \ x_2 \ x_3\end{bmatrix} = k \begin{bmatrix}\frac{3}{4} \ -\frac{13}{10} \ 1\end{bmatrix} \)leads to a basis vector \( \begin{bmatrix}\frac{3}{4} \ -\frac{13}{10} \ 1\end{bmatrix} \).

This basis reflects the degrees of freedom in the matrix, with a dimension equal to the number of free variables. The null space is essential in understanding the rankings and dependencies within a linear system.