Illustrative vector calculus
Componentwise addition and scalar multiplication of vectors.
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Study Illustrative vector calculus #
Addition/Subtraction of vectors
Definition 1. The addition/subtraction of two vectors in \(\mathbb{R}^n\) is done componentwisely. Formally: Consider two vectors \[\mathbf{v}=\begin{pmatrix} v_1\\v_2\\\vdots\\v_n\end{pmatrix},~ \mathbf{w}=\begin{pmatrix} w_1\\w_2\\\vdots\\w_n\end{pmatrix} \in \mathbb{R}^n\,.\] Then we define \[\mathbf{v} + \mathbf{w} := \begin{pmatrix} v_1+w_1\\v_2+w_2\\\vdots\\v_n+w_n\end{pmatrix} ~~~~ \text{and} ~~~~ \mathbf{v} - \mathbf{w} := \begin{pmatrix} v_1-w_1\\v_2-w_2\\\vdots\\v_n-w_n\end{pmatrix}\, .\]
The addition of vectors corresponds to a composition of movements that are described by the vectors.
Scalar multiplication of vectors
Definition 2. The scalar multiplication of vectors in \(\mathbb{R}^n\) is done componentwisely. Formally: Consider any real number \(\lambda\in\mathbb{R}\) and any vector \[\mathbf{v}=\begin{pmatrix} v_1\\v_2\\\vdots\\v_n\end{pmatrix} \in \mathbb{R}^n\, .\] Then we define \[\lambda\cdot \mathbf{v} := \begin{pmatrix} \lambda \cdot v_1\\ \lambda \cdot v_2\\\vdots\\ \lambda \cdot v_n\end{pmatrix}\, .\]
Example 3. Consider the vector \(\mathbf{v}=\begin{pmatrix} 1\\2 \end{pmatrix}\). We determine the following three vectors: \[\text{(i)} -1\cdot \mathbf{v} = \begin{pmatrix}-1\\-2\end{pmatrix} \qquad\quad \text{(ii) } 2\cdot \mathbf{v} = \begin{pmatrix}2\\4\end{pmatrix} \qquad\quad \text{(iii) } \frac{1}{2}\cdot \mathbf{v} = \begin{pmatrix}\tfrac{1}{2}\\1\end{pmatrix}\]
The scalar multiplication of a vector with a constant \(\alpha\) corresponds to a stretching by the factor \(|\alpha|\), while the direction is switched if \(\alpha\) is negative.
Rules of calculation
Let \(\mathbf{u},\mathbf{v},\mathbf{w}\in \mathbb{R}^n\) be vectors and \(\alpha,\beta\in\mathbb{R}\) be numbers. Then the following hold:
(i) Commutativity: | \(\mathbf{v}+\mathbf{w} = \mathbf{w} + \mathbf{v}\) | |
(ii) Associativity: | \((\mathbf{u} + \mathbf{v}) +\mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})\) | |
(iii) Distributivity I: | \(\alpha \cdot (\mathbf{u} + \mathbf{v}) = \alpha\cdot \mathbf{u} + \alpha\cdot \mathbf{v}\) | |
(iv) Distributivity II: | \((\alpha + \beta)\cdot \mathbf{v} = \alpha\cdot \mathbf{v} + \beta\cdot \mathbf{v}\) |
Example 4.
\(3\cdot \begin{pmatrix} 13 \\ -12 \end{pmatrix} - 2\cdot \begin{pmatrix} 5\\ -4\end{pmatrix} = \begin{pmatrix} 39 \\ -36 \end{pmatrix} - \begin{pmatrix} 10\\ -8\end{pmatrix} = \begin{pmatrix} 29 \\ -28 \end{pmatrix}\)
\(2 \cdot \left( \begin{pmatrix} 3 \\ 2 \\-1 \end{pmatrix} + \begin{pmatrix} 0\\5\\-5 \end{pmatrix}\right) - \begin{pmatrix} 6\\-2\\2 \end{pmatrix} = 2\cdot \begin{pmatrix} 3 \\ 7 \\-6 \end{pmatrix} - \begin{pmatrix} 6\\-2\\2 \end{pmatrix} = \begin{pmatrix} 0 \\ 16 \\-14 \end{pmatrix}\)
Exercise 5. Find a vector \(\mathbf{x}\) which satisfies the given equation. \[\text{(i)} ~ 2\mathbf{x} - 3\cdot \begin{pmatrix} 1\\-1\\0 \end{pmatrix} = 4 \mathbf{x} + \begin{pmatrix} 1\\ 7\\ 2 \end{pmatrix} \hspace{1cm} \text{(ii)} ~ 2\cdot \left( \begin{pmatrix} 3\\1\\1 \end{pmatrix} - \mathbf{x} \right) = \mathbf{x} + \begin{pmatrix} 0\\3\\1\end{pmatrix}\]
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