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Problem 320

# Let $\left(\mathrm{x}_{1}, \mathrm{x}_{2}, \ldots \mathrm{x}_{\mathrm{n}}\right),\left(\mathrm{y}_{1}, \mathrm{y}_{2}, \ldots, \mathrm{y}_{\mathrm{n}}\right),\left(\mathrm{z}_{1}, \mathrm{z}_{2}, \ldots \ldots \mathrm{z}_{\mathrm{n}}\right)$ be three vectors in $$\mathrm{R}^{\mathrm{n}}$$. Verify the following properties of the dot product using these vectors : a) $$\mathrm{X} \cdot \mathrm{X} \geq 0 ; \mathrm{X} \cdot \mathrm{X}=0$$ if and only if $$\mathrm{X}=0$$ b) $$X \cdot Y=Y \cdot X$$ c) $(\mathrm{X}+\mathrm{Y}) \cdot \mathrm{Z}=\mathrm{X} \cdot \mathrm{Y}+\mathrm{X} \cdot \mathrm{Z}$ d) $(\mathrm{CX}) \cdot \mathrm{Y}=\mathrm{X} \cdot(\mathrm{cY})=\mathrm{c}(\mathrm{X} \cdot \mathrm{Y})$

Expert verified
In summary, we have verified the properties of the dot product as follows: a) X.X ≥ 0; X.X = 0 if and only if X = 0: The dot product of a vector with itself is always non-negative, and is 0 if and only if the vector is the zero vector. b) X ∙ Y = Y ∙ X: The dot product is commutative, meaning it does not matter which vector we perform the operation on first. c) (X+Y) ∙ Z = X ∙ Z + Y ∙ Z: The dot product is distributive over vector addition, so we can perform the dot product separately on each vector and then add the results. d) (cX) ∙ Y = X ∙ (cY) = c(X ∙ Y): The dot product remains the same if we multiply one of the vectors by a scalar, and it equals the scalar multipled by the dot product of the original vectors.
See the step by step solution

## Step 1: Verify Property a) X.X ≥ 0; X.X = 0 if and only if X = 0

The dot product of a vector with itself is calculated as follows: $X \cdot X = x_1x_1 + x_2x_2 + ... + x_nx_n$ Notice that each term in the summation is a square of its corresponding component, which means that each term is non-negative. This means the dot product of a vector with itself must be greater than or equal to 0. Furthermore, if $$X \cdot X = 0$$, this implies that: $x_1x_1 + x_2x_2 + ... + x_nx_n = 0$ However, since squares of real numbers are non-negative, the only way this equation can be true is if each term is equal to 0: $x_1 = x_2 = ... = x_n = 0 \implies X = 0$ Thus, Property a) is verified.

## Step 2: Verify Property b) X ∙ Y = Y ∙ X

Let's compute both sides of the equation: $X \cdot Y = x_1y_1 + x_2y_2 + ... + x_ny_n$ $Y \cdot X = y_1x_1 + y_2x_2 + ... + y_nx_n$ Since multiplication is commutative, the terms for each respective side of the equation are the same. Therefore, Property b) is verified.

## Step 3: Verify Property c) (X+Y) ∙ Z = X ∙ Z + Y ∙ Z

Let's compute both sides of the equation: $(X+Y) \cdot Z = (x_1+y_1)z_1 + (x_2+y_2)z_2 + ... +(x_n+y_n)z_n$ $= x_1z_1 + x_2z_2 + ... + x_nz_n + y_1z_1 + y_2z_2 + ... + y_nz_n$ $X \cdot Z + Y \cdot Z = (x_1z_1 + x_2z_2 + ... + x_nz_n) + (y_1z_1 + y_2z_2 + ... + y_nz_n)$ As we can see, both sides of the equation are equal, thus Property c) is verified.

## Step 4: Verify Property d) (cX) ∙ Y = X ∙ (cY) = c(X ∙ Y)

Let's compute both sides of the equation: $(cX) \cdot Y = (cx_1)y_1 + (cx_2)y_2 + ... + (cx_n)y_n$ $= c(x_1y_1) + c(x_2y_2) + ... + c(x_ny_n)$ Next, compute $$X \cdot (cY)$$: $X \cdot (cY) = x_1(cy_1) + x_2(cy_2) + ... + x_n(cy_n)$ $= c(x_1y_1) + c(x_2y_2) + ... + c(x_ny_n)$ Now, compute c(X ∙ Y): $c(X \cdot Y) = c(x_1y_1 + x_2y_2 + ... + x_ny_n)$ $= c(x_1y_1) + c(x_2y_2) + ... + c(x_ny_n)$ We see that all three expressions are equal, therefore Property d) is verified.

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