Question

Given \(\mathbf{V}=2 \mathbf{i}-3 \mathrm{j}, \mathbf{U}=\mathbf{i}+2 \mathbf{k}\), write the dyadics \(\mathbf{U} \mathbf{V}\) and \(\mathbf{V U}\) in dyadic and in matrix form. Find \(\mathbf{W}=(\mathrm{UV}) \cdot \mathbf{U}\) and write this equation in matrix form.

Short answer

The dyadic products in matrix form are \( \mathbf{U} \mathbf{V} = \begin{bmatrix} 2 & -3 & 0 \ 0 & 0 & 0 \ 4 & -6 & 0 \end{bmatrix} \) and \( \mathbf{V} \mathbf{U} = \begin{bmatrix} 2 & 0 & 4 \ -3 & 0 & -6 \ 0 & 0 & 0 \end{bmatrix} \). The resulting vector is \(\mathbf{W} = \begin{bmatrix} 2 \ 0 \ 4 \end{bmatrix} \)

Step-by-step solution

- Define V and U

Identify and define the vectors: \( \mathbf{V}=2 \mathbf{i}-3 \mathbf{j} \ \mathbf{U}=\mathbf{i}+2 \mathbf{k} \)

- Calculate the dyadic product UV

The dyadic product \(\mathbf{U} \mathbf{V}\) is given by replacing each component of \( \mathbf{U} \) into a scalar multiple of \( \mathbf{V} \)\( \mathbf{U} \mathbf{V} = ( \mathbf{i} + 2 \mathbf{k} )( 2 \mathbf{i} - 3 \mathbf{j} ) \)Which results in: \( \mathbf{U} \mathbf{V} = 2 \mathbf{i} \mathbf{i} - 3 \mathbf{i} \mathbf{j} + 4 \mathbf{k} \mathbf{i} - 6 \mathbf{k} \mathbf{j} \)

- Convert UV dyadics to matrix form

Convert the dyadic product to matrix form by aligning the basis vectors: \[ \mathbf{U} \mathbf{V} = \begin{bmatrix} 2 & -3 & 0 \ 0 & 0 & 0 \ 4 & -6 & 0 \end{bmatrix} \]

- Calculate the dyadic product VU

The dyadic product \( \mathbf{V} \mathbf{U} \) is given by replacing each component of \(\mathbf{V} \) into a scalar multiple of \( \mathbf{U} \)\( \mathbf{V} \mathbf{U} = ( 2 \mathbf{i} - 3 \mathbf{j} )( \mathbf{i} + 2 \mathbf{k} ) \)This results in:\( \mathbf{V} \mathbf{U} = 2 \mathbf{i} \mathbf{i} + 4 \mathbf{i} \mathbf{k} - 3 \mathbf{j} \mathbf{i} - 6 \mathbf{j} \mathbf{k} \)

- Convert VU dyadics to matrix form

Convert the dyadic product to matrix form by aligning the basis vectors: \[ \mathbf{V} \mathbf{U} = \begin{bmatrix} 2 & 0 & 4 \ -3 & 0 & -6 \ 0 & 0 & 0 \end{bmatrix} \]

- Calculate W

Calculate \( \mathbf{W} \) using the product \( (\mathbf{UV}) \cdot \mathbf{U} \)First, write \( \mathbf{U} \) in column vector form: \( \mathbf{U} = \begin{bmatrix} 1 \ 0 \ 2 \end{bmatrix} \)Next, multiply the matrix form of \( \mathbf{UV} \) with \( \mathbf{U} \):\[ \mathbf{W} = \begin{bmatrix} 2 & -3 & 0 \ 0 & 0 & 0 \ 4 & -6 & 0 \end{bmatrix} \begin{bmatrix} 1 \ 0 \ 2 \end{bmatrix} \]

- Perform matrix multiplication

Multiply the matrices to obtain: \[ \mathbf{W} = \begin{bmatrix} 2 \cdot 1 + (-3) \cdot 0 + 0 \cdot 2 \ 0 \cdot 1 + 0 \cdot 0 + 0 \cdot 2 \ 4 \cdot 1 + (-6) \cdot 0 + 0 \cdot 2 \end{bmatrix} = \begin{bmatrix} 2 \ 0 \ 4 \end{bmatrix} \]

Key concepts

Vector Algebra

In vector algebra, vectors are mathematical entities that have both magnitude and direction. Vectors are usually written in terms of basis vectors. For this exercise, vectors \(\textbf{V}\) and \(\textbf{U}\) are given as \(\textbf{V} = 2 \mathbf{i} - 3 \mathbf{j}\) and \(\textbf{U} = \mathbf{i} + 2 \mathbf{k}\). These basis vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are unit vectors along the x, y, and z directions, respectively.

One very interesting operation involving vectors is the **dyadic product**. The dyadic product of vectors \(\textbf{U}\) and \(\textbf{V}\) represents a tensor (a matrix in this case) formed by all combinations of the basis vectors. In this exercise, we explored how to compute and represent these dyadic products as matrices.

Another key concept is when we multiply the resulting matrix (from the dyadic product) by another vector to get another vector. This process involves matrix multiplication, which we will discuss in the next section.

Matrix Multiplication

Matrix multiplication is fundamental in linear algebra and is used to perform various operations on matrices and vectors. It involves multiplying elements of rows from one matrix with columns of another and summing the products.

For the matrix forms of the dyadic products computed in this exercise: \[ \textbf{U V} = \begin{bmatrix} 2 & -3 & 0 \ 0 & 0 & 0 \ 4 & -6 & 0 \ \end{bmatrix} \] \[ \textbf{V U} = \begin{bmatrix} 2 & 0 & 4 \ -3 & 0 & -6 \ 0 & 0 & 0 \ \end{bmatrix} \], we multiply them by the vector \(\textbf{U}\) in column form to obtain \(\textbf{W}\).

For example, multiplying the dyadic product \( \textbf{U V}\) with \(\textbf{U} = \begin{bmatrix} 1 \ 0 \ 2 \ \end{bmatrix}\) involves calculating each element in the resulting matrix. The result is \[ \textbf{W} = \begin{bmatrix} 2 \ 0 \ 4 \ \end{bmatrix} \].

Understanding the computation of matrix multiplication helps us manipulate and understand more complex structures in linear algebra, leading to deeper insights in vector transformations and applications.

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, vector spaces, linear transformations, and matrices. Linear algebra forms the foundation for many areas of mathematics and engineering.

In this exercise, we explored linear algebra concepts by working with vectors and their dyadic products. The dyadic product is a specific type of linear transformation that maps two vectors into a matrix.

Understanding how these dyadic products work, and being able to convert between vector algebra and matrix representations, is key in applying linear algebra to practical problems. For example:

• **Mechanics**: analyzing forces and movements in engineering.
• **Computer Graphics**: transforming shapes and animations.
• **Quantum Mechanics**: representing quantum states.

In essence, mastering the principles of linear algebra helps tackle various scientific and real-world challenges with mathematical precision.