Question

(a) Use the Gram-Schmidt process to find an orthonormal set of vectors out of \((1,-1,1),(-1,0,1)\), and \((2,-1,2)\). (b) Are these three vectors linearly independent? If not, find a zero linear combination of them by using part (a).

Short answer

The orthonormal set created from the given vectors using the Gram-Schmidt process is \(\{(1, -1, 1), (-1/3, -2/3, 2/3)\}\). The given vectors are not linearly independent, as the third vector \((2, -1, 2)\) can be expressed as a linear combination of the other two.

Step-by-step solution

Calculate the Orthogonal Vectors

Begin with the vector \(u_1 = (1,−1,1)\). The second vector \(v_2 = u_2 - proj_{u_1}u_2 = (-1,0,1) - [(u_1.u_2)/||u_1||^2] * u1 = (-1,0,1) - [(-2)/3]*(1, -1, 1)= (-1/3, -2/3, 2/3)\). The third vector \(v_3 = u_3 - proj_{u_1}u_3 - proj_{v_2}u3 = (2,-1,2) - [(u1.u3)/||u1||^2]*u1 - [(v2.u3)/||v2||^2]*v2 =0\).

Normalize the Orthogonal Vectors

Normalize the vectors obtained in the previous step to get an orthonormal set. We keep \(u_1\) as it is. We normalize \(-1/3, -2/3, 2/3\) by dividing it by its magnitude which is \(||(-1/3, -2/3, 2/3)|| = 1\) thus the normalized vector is \(-1/3, -2/3, 2/3\). Since \(v_3 = 0\), it does not contribute to the orthonormal basis.

Check for Linear Independence

Linear independence means that none of the vectors in the set can be expressed as a linear combination of the other vectors. In the context of the Gram-Schmidt process, if during the process a step is reached where \(v_i = 0\), then the vectors are not linearly independent. Here, \(v_3 = 0\) indicates that \((2,-1,2)\) is not linearly independent from \((1, -1, 1)\) and \((-1, 0, 1)\).

Find Zero Linear Combination

For linearly dependent vectors, some linear combination of them equals the zero vector. We calculate this from the formula for \(v_3\). The zero linear combination is found by rearranging the equation \(u_3−proj_{u1}u_3−proj_{v2}u3=0\) for \(u_3\). This gives \(u_3 = (-2/3) * u_1 +3 * v_2 = (2, -1, 2)\). This equation expresses \(u_3\) as a linear combination of \(u_1\) and \(u_2\), showing that the vectors are not independent.

Key concepts

Orthonormal Vectors

In the world of vectors, an orthonormal set of vectors is a powerful concept. It brings together the ideas of orthogonality and normalization. Here's how these ingredients make it work. Consider a set of vectors. This set is orthogonal if each pair of different vectors in the set is perpendicular to each other. In simpler terms, their dot product is zero.
But there's more to an orthonormal set than just being orthogonal. Each vector in the set also needs to have a magnitude or length of one. This brings in the normalization part of the equation.
Precise steps are taken to make vectors orthonormal. The Gram-Schmidt Process facilitates this transformation. The initial set of vectors is adjusted step-by-step, ensuring they are at right angles to each other and that each vector has a unit length.
This concept is immensely useful. For example, in computer graphics and data science, orthonormal sets allow us to simplify complex calculations and work more effectively with multi-dimensional data.

Linear Independence

The concept of linear independence is fundamental in algebra. It helps us understand the relationships between vectors. Linear independence means that no vector in a set can be created by adding or scaling other vectors in the set. In essence, the vectors are not entangled, each standing alone uniquely.
In practical terms, if you have a group of vectors, they are linearly independent if none of them can be written as a combination of others. If such a combination exists, they are dependent.
Here's a simple rule of thumb: If during the Gram-Schmidt process a zero vector appears, it indicates linear dependence. This was demonstrated in the exercise. The appearance of the zero vector here clearly revealed that one vector was a combination of others, showing dependence.

• Why is this important? Because in areas like machine learning or engineering, knowing whether vectors are independent or not can help optimize resources and reduce complexity.

Vector Projection

Vector projection is like shadow casting in mathematics. It projects the essence of one vector along another. When we project vector \(\mathbf{b}\) onto vector \(\mathbf{a}\), we obtain the component of \(\mathbf{b}\) that lies along \(\mathbf{a}\).
The formula is driven by the dot product: \( \text{proj}_{\mathbf{a}}\mathbf{b} = \frac{\mathbf{a} \,\cdot\, \mathbf{b}}{||\mathbf{a}||^2} \mathbf{a} \). This formula is vital in the Gram-Schmidt process.
As seen in the exercise, projection helps in adjusting vectors during the transformation to ensure orthogonality. Simply put, projection pulls vectors into alignment. They align parallel to a desired direction while maintaining proper perpendicularity with others.

• Understanding vector projection is key to effects in various domains, such as robotics for motion planning or in virtual reality environments to ensure objects behave in expected ways.