Question

A matrix \(A\) and vectors \(\vec{b}, \bar{u}\) and \(\vec{v}\) are given. Verify that \(A \vec{u}=\overrightarrow{0}, A \vec{v}=\vec{b}\) and \(A(\vec{u}+\vec{v})=\vec{b}\).$$ \begin{array}{l} A=\left[\begin{array}{ccc} 2 & -2 & -1 \\ -1 & 1 & -1 \\ -2 & 2 & -1 \end{array}\right] \\ \vec{b}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right], \vec{u}=\left[\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right], \vec{v}=\left[\begin{array}{c} 1 \\ 1 \\ -1 \end{array}\right] \end{array} $$

Short answer

The three conditions are verified: (1) \(A \vec{u} = \overrightarrow{0}\), (2) \(A \vec{v} = \vec{b}\), (3) \(A(\vec{u}+\vec{v}) = \vec{b}\).

Step-by-step solution

Calculate A\vec{u}

To verify that \( A \vec{u} = \overrightarrow{0} \), we need to multiply the matrix \( A \) by the vector \( \vec{u} \). Perform the matrix-vector multiplication:\[A \vec{u} = \begin{bmatrix} 2 & -2 & -1 \ -1 & 1 & -1 \ -2 & 2 & -1 \end{bmatrix} \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} = \begin{bmatrix} (2\cdot1) + (-2\cdot1) + (-1\cdot0) \ (-1\cdot1) + (1\cdot1) + (-1\cdot0) \ (-2\cdot1) + (2\cdot1) + (-1\cdot0) \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \].Thus, \( A \vec{u} = \overrightarrow{0} \) is confirmed.

Calculate A\vec{v}

Now, verify that \( A \vec{v} = \vec{b} \) by multiplying \( A \) by \( \vec{v} \):\[A \vec{v} = \begin{bmatrix} 2 & -2 & -1 \ -1 & 1 & -1 \ -2 & 2 & -1 \end{bmatrix} \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix} = \begin{bmatrix} (2\cdot1) + (-2\cdot1) + (-1\cdot(-1)) \ (-1\cdot1) + (1\cdot1) + (-1\cdot(-1)) \ (-2\cdot1) + (2\cdot1) + (-1\cdot(-1)) \end{bmatrix} = \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \].So, \( A \vec{v} = \vec{b} \) is confirmed.

Calculate A(\vec{u}+\vec{v})

To verify \( A(\vec{u}+\vec{v}) = \vec{b} \), first find \( \vec{u} + \vec{v} \):\[\vec{u} + \vec{v} = \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} + \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix} = \begin{bmatrix} 2 \ 2 \ -1 \end{bmatrix}\]Next, calculate the matrix-vector product:\[A(\vec{u} + \vec{v}) = \begin{bmatrix} 2 & -2 & -1 \ -1 & 1 & -1 \ -2 & 2 & -1 \end{bmatrix} \begin{bmatrix} 2 \ 2 \ -1 \end{bmatrix} = \begin{bmatrix} (2\cdot2) + (-2\cdot2) + (-1\cdot(-1)) \ (-1\cdot2) + (1\cdot2) + (-1\cdot(-1)) \ (-2\cdot2) + (2\cdot2) + (-1\cdot(-1)) \end{bmatrix} = \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \].Thus, \( A(\vec{u} + \vec{v}) = \vec{b} \) is confirmed.

Key concepts

Null Space

The null space of a matrix, also known as the kernel, is a fundamental concept in linear algebra. It consists of all the vectors that, when multiplied by the matrix, result in the zero vector. In simpler terms, it's the set of solutions to the equation \( A \vec{x} = \overrightarrow{0} \). This space is essential for understanding the behavior and properties of the matrix.

In the given exercise, we calculated \( A \vec{u} = \overrightarrow{0} \). This indicates that the vector \( \vec{u} \) is part of the null space of matrix \( A \). This concept is crucial because it helps reveal intrinsic properties of the transformation represented by \( A \).

The significance of null space lies in several key areas:

• Determining invertibility: If the null space contains only the zero vector, the matrix is invertible.
• Solutions to linear systems: It provides all potential solutions to homogeneous linear equations.
• Understanding linear transformations: It gives insights into how transformation affects input vectors.

Grasping the null space helps students comprehend deeper matrix properties and solve intricate linear algebra problems.

Vector Addition

Vector addition is the operation of adding two or more vectors together to produce another vector. This operation is fundamental in linear algebra and is used in numerous mathematical and practical applications. The basic rule for vector addition is that vectors are added component-wise.

In the exercise, we computed the addition of vectors \( \vec{u} \) and \( \vec{v} \) as follows: \[ \vec{u} + \vec{v} = \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} + \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix} = \begin{bmatrix} 2 \ 2 \ -1 \end{bmatrix} \].
Each corresponding component of the two vectors was added to form a new vector.

Reasons why vector addition is important include:

• Calculating resultant vectors: Often used in physics to combine forces or velocities.
• Understanding geometric transformations: Useful in translating points by vectors.
• Determining vector subspaces: Combined vectors can encompass spaces such as linear spans.

This simple operation lays the groundwork for more complex topics in vector algebra.

Linear Transformation

Linear transformation is a function between two vector spaces that preserves addition and scalar multiplication. This property implies that linear transformations map lines to lines (or points, in degenerate cases) and preserve the operations of addition and scalar multiplication.

In our example, matrix multiplication acts as a linear transformation. We see how matrix \( A \) transforms the vectors \( \vec{u} \) and \( \vec{v} \).

A linear transformation \( T \) from vector space \( V \) to vector space \( W \) can be written in a concise form as \( T(\vec{x}) = A \vec{x} \), where \( A \) is a matrix.

Key properties of linear transformations include:

• Preservation of vector addition: If \( T(\vec{x} + \vec{y}) = T(\vec{x}) + T(\vec{y}) \).
• Preservation of scalar multiplication: If \( T(c \cdot \vec{x}) = c \cdot T(\vec{x}) \), for scalar \( c \).
• Representation by matrices: Any linear transformation can be represented using a matrix.

Understanding linear transformations is essential because they form the backbone of numerous operations across mathematics and applied sciences, such as graphical manipulations, system solutions, and more.