Problem 12 - Differential Equations and Linear Algebra

Question

Determine orthogonal bases for rowspace( \(A\) ) and colspace( \(A\) ). $$A=\left[\begin{array}{ll} 1 & 5 \\ 2 & 4 \\ 3 & 3 \\ 4 & 2 \\ 5 & 1 \end{array}\right]$$

Step-by-Step Solution

Answer

The orthogonal basis for the row space of the given matrix A is: $$\left\{ \begin{bmatrix} 1 \\ 5 \end{bmatrix}, \begin{bmatrix} -6/13 \\ -18/13 \end{bmatrix} \right\}$$ And the orthogonal basis for the column space of the given matrix A is: $$\left\{ \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \end{bmatrix}, \begin{bmatrix} 2/11 \\ -2/11 \\ -6/11 \\ -10/11 \\ -14/11 \end{bmatrix} \right\}$$

Step 1: Find the bases for the row space and the column space of A

First, we need to find the row echelon form of the matrix A by applying Gaussian elimination. In this process, we perform elementary row operations to transform the matrix into a triangular form. $$A=\left[\begin{array}{ll} 1 & 5 \\\ 2 & 4 \\\ 3 & 3 \\\ 4 & 2 \\\ 5 & 1 \end{array}\right] \xrightarrow{R2-2R1} \left[\begin{array}{ll} 1 & 5 \\\ 0 & -6 \\\ 3 & 3 \\\ 4 & 2 \\\ 5 & 1 \end{array}\right] \xrightarrow{R3-3R1} \left[\begin{array}{ll} 1 & 5 \\\ 0 & -6 \\\ 0 & -12 \\\ 4 & 2 \\\ 5 & 1 \end{array}\right]$$ $$\xrightarrow{R4-4R1} \left[\begin{array}{ll} 1 & 5 \\\ 0 & -6 \\\ 0 & -12 \\\ 0 & -18 \\\ 5 & 1 \end{array}\right] \xrightarrow{R5-5R1} \left[\begin{array}{ll} 1 & 5 \\\ 0 & -6 \\\ 0 & -12 \\\ 0 & -18 \\\ 0 & -24 \end{array}\right]$$ Now, the matrix is in row echelon form. The non-zero rows will form the basis for the row space of A: $$Rowspace(A) = \ operatorname{span} \left\{ \begin{bmatrix} 1 \\ 5 \end{bmatrix}, \begin{bmatrix} 0 \\ -6 \end{bmatrix} \right\}$$ Similarly, we can extract the column space of A by writing down the spanning set formed by the non zero columns of A: $$Colspace(A) = \ operatorname{span} \left\{ \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \end{bmatrix}, \begin{bmatrix} 5 \\ 4 \\ 3 \\ 2 \\ 1 \end{bmatrix} \right\}$$

Step 2: Apply the Gram-Schmidt process to orthogonalize both bases

We can now proceed to find the orthogonal bases for both row space and column space using the Gram-Schmidt process. First, for the row space: Let \(v_1 = \begin{bmatrix}1 \\ 5\end{bmatrix}\) and \(v_2 = \begin{bmatrix}0 \\ -6\end{bmatrix}\) Then use Gram-Schmidt process to find orthogonal vectors: $$u_1 = v_1 = \begin{bmatrix}1 \\ 5\end{bmatrix}$$ $$u_2 = v_2 - proj_{u_1}(v_2) = \begin{bmatrix}0 \\ -6\end{bmatrix} - \frac{v_2 \cdot u_1}{u_1 \cdot u_1} u_1 = \begin{bmatrix}0 \\ -6\end{bmatrix} - \frac{-30}{26} \begin{bmatrix}1 \\ 5\end{bmatrix} = \begin{bmatrix}0 \\ -6\end{bmatrix} - \frac{30}{26} \begin{bmatrix}1 \\ 5\end{bmatrix} = \begin{bmatrix} -6/13 \\ -18/13 \end{bmatrix}$$ So, the orthogonal basis for the row space is given by: $$Orthogonal\_row\_space = \left\{ \begin{bmatrix} 1 \\ 5 \end{bmatrix}, \begin{bmatrix} -6/13 \\ -18/13 \end{bmatrix} \right\}$$ Now, for the column space: Let \(w_1 = \begin{bmatrix}1 \\ 2 \\ 3 \\ 4 \\ 5\end{bmatrix}\) and \(w_2 = \begin{bmatrix}5 \\ 4 \\ 3 \\ 2 \\ 1\end{bmatrix}\) Then use Gram-Schmidt process to find orthogonal vectors: $$u_1 = w_1 = \begin{bmatrix}1 \\ 2 \\ 3 \\ 4 \\ 5\end{bmatrix}$$ $$u_2 = w_2 - proj_{u_1}(w_2) = \begin{bmatrix}5 \\ 4 \\ 3 \\ 2 \\ 1\end{bmatrix} - \frac{w_2 \cdot u_1}{u_1 \cdot u_1} u_1 = \begin{bmatrix}5 \\ 4 \\ 3 \\ 2 \\ 1\end{bmatrix} - \frac{35}{55} \begin{bmatrix}1 \\ 2 \\ 3 \\ 4 \\ 5 \end{bmatrix} = \begin{bmatrix}5 \\ 4 \\ 3 \\ 2 \\ 1\end{bmatrix} - \frac{7}{11} \begin{bmatrix}1 \\ 2 \\ 3 \\ 4 \\ 5\end{bmatrix} = \begin{bmatrix}2/11 \\ -2/11 \\ -6/11 \\ -10/11 \\ -14/11\end{bmatrix}$$ So, the orthogonal basis for the column space is given by: $$Orthogonal\_col\_space = \left\{ \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \end{bmatrix}, \begin{bmatrix} 2/11 \\ -2/11 \\ -6/11 \\ -10/11 \\ -14/11 \end{bmatrix} \right\}$$ Now we have orthogonal bases for both row space and column space of the given matrix A.

Key Concepts

Row Space

The row space of a matrix is the set of all possible linear combinations of its row vectors. This means it consists of all the vectors that you can form by adding and scaling the rows of the matrix.
This space gives you insight into the dimensionality of the solutions and the relationships between the data rows.

• To find a basis for the row space, you typically use row operations to simplify the matrix into a row echelon form.
• The non-zero rows in this echelon form generate the row space.

In our original example, after performing Gaussian elimination, the matrix is transformed into row-echelon form. From this, the basis for the row space is derived from the non-zero rows, which are linearly independent and span the row space.

Column Space

The column space of a matrix is composed of all the linear combinations of its column vectors. It essentially represents the output space of the linear transformation that the matrix describes when it acts on input vectors.
This space is critical for understanding the possible outputs of systems modeled by the matrix.

• To determine the column space, you typically identify columns that are linear combinations of others using the matrix's row echelon form.
• The pivot columns (those corresponding to leading ones in row echelon form) are used as a basis for the column space.

In the example, the original matrix's columns are used to form the column space, showing which vectors linearly combine to form the space.

Gaussian Elimination

Gaussian Elimination is a method used to simplify matrices and solve systems of linear equations. It involves performing a series of elementary row operations to transform a given matrix into row echelon form (and optionally into reduced row echelon form).

• The primary goal is to make the matrix easier to work with, especially when finding solutions to systems of equations.
• It systematically zeros out entries below each leading pivot position in a column.

In the step-by-step solution, Gaussian Elimination is crucial to changing the matrix into a form where the bases for the row and column spaces can be clearly identified. This method is particularly helpful for understanding and visualizing the structure of linear transformations.

Gram-Schmidt Process

The Gram-Schmidt Process is an algorithm used to transform a set of linearly independent vectors into an orthogonal set. This is especially useful in many mathematical applications, including finding orthogonal bases for vector spaces.

• It works by taking a set of vectors and orthogonalizing them one by one, ensuring each new vector is orthogonal to the ones that came before it.
• The process involves subtracting off components already represented by previous vectors in the orthogonal set.

In our example, after determining initial bases for both the row and column spaces, the Gram-Schmidt Process is applied to convert these into orthogonal bases. This ensures that each vector is independent, with no overlap, making calculations involving these vectors simpler and more robust.