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Chapter 11: Problem 87

Prove that \(|c \mathbf{v}|=|c||\mathbf{v}|,\) where \(c\) is a scalar and \(\mathbf{v}\) is a vector.

Short Answer

Expert verified
Question: Show that the magnitude of the product of a scalar and a vector is equal to the product of the magnitudes of the scalar and the vector (i.e., prove that $|c\mathbf{v}| = |c||\mathbf{v}|$).

Step by step solution

01

Define the absolute value of the scalar c and the magnitude of the vector v

The absolute value of a scalar \(c\), denoted as \(|c|\), is the non-negative value of the scalar. In other words, if \(c\) is positive, \(|c| = c\), and if \(c\) is negative, \(|c| = -c\). The magnitude of a vector \(\mathbf{v}\), denoted as \(|\mathbf{v}|\), is the length of the vector. If \(\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix}\), then the magnitude of \(\mathbf{v}\) is given by the formula $$|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + \cdots + v_n^2}.$$
02

Compute the product of the scalar and the vector

When we multiply a scalar \(c\) with a vector \(\mathbf{v}\), each element of the vector is multiplied by the scalar. Let \(\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix}\). Then, the product of \(c\) and \(\mathbf{v}\) is given by $$c \mathbf{v} = \begin{bmatrix} c v_1 \\ c v_2 \\ \vdots \\ c v_n \end{bmatrix}.$$
03

Calculate the magnitude of the product of the scalar and the vector

Next, we need to calculate the magnitude of the product of the scalar and the vector, \(|c \mathbf{v}|\). Since \(c \mathbf{v} = \begin{bmatrix} c v_1 \\ c v_2 \\ \vdots \\ c v_n \end{bmatrix}\), we have $$|c \mathbf{v}| = \sqrt{(cv_1)^2 + (cv_2)^2 + \cdots + (cv_n)^2}.$$ We can then factor out the square of the scalar \(c\), which simplifies the expression, $$|c \mathbf{v}| = \sqrt{c^2(v_1^2 + v_2^2 + \cdots + v_n^2)} = \sqrt{c^2 |\mathbf{v}|^2}.$$
04

Prove the desired property

Now that we have simplified the expression for \(|c \mathbf{v}|\), we can prove the desired property: $$|c \mathbf{v}| = \sqrt{c^2 |\mathbf{v}|^2} = \sqrt{c^2} \sqrt{|\mathbf{v}|^2} = |c| |\mathbf{v}|.$$ So, we have proven that \(|c \mathbf{v}|=|c||\mathbf{v}|,\) as desired.

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