Question

a. Show that \(\|\mathbf{x}+\mathbf{y}\|^{2}=\|\mathbf{x}\|^{2}+\|\mathbf{y}\|^{2}\) if and only if \(\mathbf{x}\) is orthogonal to \(\mathbf{y}\). b. If \(\mathbf{x}=\left[\begin{array}{l}1 \\ 1\end{array}\right], \mathbf{y}=\left[\begin{array}{l}1 \\ 0\end{array}\right]\) and \(\mathbf{z}=\left[\begin{array}{r}-2 \\ 3\end{array}\right],\) show that \(\|\mathbf{x}+\mathbf{y}+\mathbf{z}\|^{2}=\|\mathbf{x}\|^{2}+\|\mathbf{y}\|^{2}+\|\mathbf{z}\|^{2}\) but \(\mathbf{x} \cdot \mathbf{y} \neq 0, \mathbf{x} \cdot \mathbf{z} \neq 0,\) and \(\mathbf{y} \cdot \mathbf{z} \neq 0\)

Short answer

a. If orthogonal, then \(\mathbf{x} \cdot \mathbf{y} = 0\). b. The equation holds, but dot products are non-zero.

Step-by-step solution

Understanding the Expression

We need to show that \(\|\mathbf{x}+\mathbf{y}\|^2 = \|\mathbf{x}\|^2 + \|\mathbf{y}\|^2\) if and only if \(\mathbf{x}\) is orthogonal to \(\mathbf{y}\). Orthogonality means \(\mathbf{x} \cdot \mathbf{y} = 0\). This is an application of the Pythagorean theorem.

Defining Norm and Dot Product

Recall that for any vectors \(\mathbf{a}\) and \(\mathbf{b}\), we have \(\|\mathbf{a}\|^2 = \mathbf{a} \cdot \mathbf{a}\). So, \(\|\mathbf{x}+\mathbf{y}\|^2 = (\mathbf{x}+\mathbf{y}) \cdot (\mathbf{x}+\mathbf{y})\).

Expanding the Dot Product

Expand \((\mathbf{x} + \mathbf{y}) \cdot (\mathbf{x} + \mathbf{y})\):\[(\mathbf{x} + \mathbf{y}) \cdot (\mathbf{x} + \mathbf{y}) = \mathbf{x} \cdot \mathbf{x} + 2(\mathbf{x} \cdot \mathbf{y}) + \mathbf{y} \cdot \mathbf{y}\].

Simplify Using Orthogonality

For orthogonal vectors, \(\mathbf{x} \cdot \mathbf{y} = 0\), so:\[\|\mathbf{x}+\mathbf{y}\|^2 = \|\mathbf{x}\|^2 + \|\mathbf{y}\|^2\].The converse is true because if the expression holds, then \(\mathbf{x} \cdot \mathbf{y} = 0\) is implied.

Calculating for Given Vectors

Substitute \(\mathbf{x} = \begin{bmatrix} 1 \ 1 \end{bmatrix}\), \(\mathbf{y} = \begin{bmatrix} 1 \ 0 \end{bmatrix}\), \(\mathbf{z} = \begin{bmatrix} -2 \ 3 \end{bmatrix}\). Calculate \(\mathbf{x}+\mathbf{y}+\mathbf{z}\):\[\mathbf{x}+\mathbf{y}+\mathbf{z} = \begin{bmatrix} 1+1-2 \ 1+0+3 \end{bmatrix} = \begin{bmatrix} 0 \ 4 \end{bmatrix}\].

Calculating Norms

Find \(\|\mathbf{x}+\mathbf{y}+\mathbf{z}\|^2\):\[\|\begin{bmatrix} 0 \ 4 \end{bmatrix}\|^2 = 0^2 + 4^2 = 16\].Calculate individual norms: \[\|\mathbf{x}\|^2 = 2, \, \|\mathbf{y}\|^2 = 1, \, \|\mathbf{z}\|^2 = 13\].Sum: \(2 + 1 + 13 = 16\).

Verifying Dot Products

Calculate the dot products:\[\mathbf{x} \cdot \mathbf{y} = 1(1) + 1(0) = 1 \, (eq 0)\]\[\mathbf{x} \cdot \mathbf{z} = 1(-2) + 1(3) = 1 \, (eq 0)\]\[\mathbf{y} \cdot \mathbf{z} = 1(-2) + 0(3) = -2 \, (eq 0)\].None are zero, but \(\|\mathbf{x} + \mathbf{y} + \mathbf{z}\|^2 = \|\mathbf{x}\|^2 + \|\mathbf{y}\|^2 + \|\mathbf{z}\|^2\) holds.

Key concepts

Orthogonal Vectors

Orthogonal vectors are vectors that are at a "right angle" to each other. In mathematical terms, this means that their dot product is zero. For example, if you have two vectors, \( \mathbf{x} \) and \( \mathbf{y} \), they are orthogonal if \( \mathbf{x} \cdot \mathbf{y} = 0 \).
This concept is quite similar to the idea of perpendicular lines in geometry, but it applies to vectors of any dimension. The zero dot product ensures that where one vector ends, the direction to the other vector is entirely lateral, like two sides of a square meeting at a corner.
Orthogonality is an important feature in vector mathematics because it simplifies calculations and has meaningful interpretations in physics and engineering. You can think of orthogonal vectors as "independent" since they do not interact in a linear direction with one another.

Dot Product

The dot product (also known as the scalar product) is a way to multiply two vectors, resulting in a scalar (a single number). For vectors \( \mathbf{a} = [a_1, a_2] \) and \( \mathbf{b} = [b_1, b_2] \) in two-dimensional space, the dot product is calculated as \( \mathbf{a} \cdot \mathbf{b} = a_1 \cdot b_1 + a_2 \cdot b_2 \).
This operation is crucial because it provides a way to determine the degree of alignment between two vectors. If the dot product is zero, it indicates the vectors are orthogonal, and thus, independent of each other.

• It is a measure of similarity in direction.
• If both vectors are pointing in the same direction, the dot product will be maximum.
• If they are opposite, the dot product will be negative.
• If orthogonal, as mentioned, it will be zero.

This property is employed in various applications, such as calculating work done by a force in physics and in many algorithms in computer science.

Pythagorean Theorem

The Pythagorean theorem is a mathematical principle that applies to right triangles, but it also extends its utility into vector mathematics. For vectors, it states that the square of the norm of a vector sum is equal to the sum of the squares of the norms of the vectors, provided the vectors are orthogonal.
In formula terms:
If \( \mathbf{x} \) and \( \mathbf{y} \) are orthogonal, then \( \|\mathbf{x} + \mathbf{y}\|^2 = \|\mathbf{x}\|^2 + \|\mathbf{y}\|^2 \).
This is essentially the vector form of the Pythagorean theorem. It gets directly to the application of why we value orthogonal vectors: they allow us to easily compute resultant magnitudes. When vectors satisfy this condition, the geometry of the vectors forms a right triangle, where the resultant vector is the hypotenuse.

Vector Norm

The vector norm (or magnitude) is a measure of a vector's length. For a vector \( \mathbf{a} = [a_1, a_2] \), the vector norm is calculated using the formula \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2} \).
It gives an idea of "how much" of the vector is present, rather like the distance of a point from the origin in a coordinate system.

• Vector norms are non-negative; they are always zero or positive.
• A zero vector norm indicates a vector with no length.
• The norm is useful in normalization processes, where you make vectors unit length (length of one) for comparative purposes.

In any vector space, understanding the norm is fundamental. It lets you understand the size of elements within the space, which can further guide solutions in engineering, physics, and computer graphics.