Question

(a) Use the Gram-Schmidt process to find an orthonormal set of vectors out of \((1,-1,2),(-2,1,-1)\), and \((-1,-1,4)\). (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 from the Gram-Schmidt process is \{ u, v2, w2 \}. They are not linearly independent because there exists a nonzero linear combination that results in the zero vector, specifically -u1 + v1 = 0.

Step-by-step solution

Implementing Gram-Schmidt procedure

We start with the Gram-Schmidt process to find an orthogonal basis. The process involves projecting vectors onto others and subtracting the projections from the original vectors. Here are the steps involved in the process, using \( u,v,w \) as the given vectors, \( u1, v1, w1 \) as the orthogonal basis and \( v2, w2 \) as the orthonormal basis: \[u1 = u = (1,-1,2),\]\[v1 = v - \(\frac {v \cdot u1} {u1 \cdot u1}\) u1 = (-2,1,-1) - \(\frac {(1\cdot -2) - (1\cdot1) + (2\cdot -1)} {(1 \cdot 1) + (-1 \cdot -1) + (2 \cdot 2)}\) (1, -1, 2) = (-1,0,-1),\]\[w1 = w - \(\frac {w \cdot u1} {u1 \cdot u1}\) u1 - \(\frac {w \cdot v1} {v1 \cdot v1}\) v1 = (-1,-1,4) - \(\frac {(-1\cdot1) + (-1\cdot-1) + (4\cdot2)} {(1 \cdot 1) + (-1 \cdot -1) + (2 \cdot 2)}\) u1 - \(\frac {(-1\cdot-1) + (-1\cdot0) + (4\cdot -1)} {(-1 \cdot -1) + (0 \cdot 0) + (-1 \cdot -1)}\) v1 = (-1,-1,4) - 3u1 +1v1.

Normalizing the vectors

Next, we normalize these vectors to the length 1. This involves dividing each vector by its magnitude,\[v2 = \(\frac{v1}{\lVert v1 \rVert}\) = \(\frac{(-1,0,-1)}{\sqrt{(-1)^2 + 0 + (-1)^2}}\) = \(-\frac {1}{\sqrt{2}}\) (1, 0, 1),\]\[w2 = \(\frac{w1}{\lVert w1 \rVert}\) = \(\frac{-1u1 + 1v1}{\sqrt{(-1u1 + 1v1) \cdot (-1u1 + 1v1)}}\) = 0.]

Checking for linear independence

We can conclude that the original set of vectors \((u,v,w)\) is linearly dependent as there is a trivial solution when we try to solve for the scalar coefficients in the following linear combination:\[au+ bv + cw = 0,\]in which \(a, b, c\) are the scalar coefficients, not all equal to zero. Using the orthonormal vectors completed in the previous steps, this relationship can be easily expressed as:\[-1u1 +1v1 = 0,\]which is a nonzero linear combination of the original vectors, thereby demonstrating they are linearly dependent.

Key concepts

Orthonormal Vectors

The concept of orthonormal vectors is fundamental in linear algebra. It combines two important properties: orthogonality and normalization. Vectors are orthogonal if their dot product is zero, meaning they are at a right angle to each other. When they are also normalized, they are reduced to unit length (magnitude of 1).
Creating an orthonormal set from a given set of vectors involves processes like the Gram-Schmidt process. This task is essential in fields like computer graphics, physics, and statistics.
In our original problem, the Gram-Schmidt process was used to transform a vector set into an orthonormal set. By projecting each vector onto the others and adjusting them, we ensure they do not overlap, making them orthogonal. Finally, by dividing each orthogonal vector by its own magnitude, the set becomes orthonormal.
This allows for simplified mathematical operations, as calculations involving orthonormal vectors often become more straightforward.

Linear Independence

Linear independence describes a scenario where no vector in a set can be formed as a linear combination of the others. If the vectors are linearly independent, it means none of them can be expressed as a straightforward mix of any subset of the others. This idea is crucial for understanding the dimensions of vector spaces.
The concept holds importance in various applications, such as in determining the rank of a matrix or finding solutions to a system of linear equations. Checking for linear independence can be done by finding scalars that satisfy the equation:

• If the only solution to a combination like au + bv + cw = 0 is when all the scalars are zero, the set is independent.

In the given problem, the attempt to express the vectors using orthonormal transformations revealed that they were dependent, as a non-trivial combination was found.

Vector Normalization

Vector normalization is the process of scaling a vector so that it has a length of one. This is done without changing its direction. It's an important operation because many algorithms in data science and programming require vectors of unit length for simplification, like finding unit vectors for directions in physics or machine learning.
The normalization process involves dividing each component of the vector by its magnitude. The magnitude (or length) of a vector \( \mathbf{v} = (x, y, z) \) is given by the square root of the sum of the squares of its components: \( \lVert \mathbf{v} \rVert = \sqrt{x^2 + y^2 + z^2} \).
Once calculated, the normalized vector becomes:

• \( \frac{\mathbf{v}}{\lVert \mathbf{v} \rVert} \)

In the problem, normalization followed the step of forming orthogonal vectors, ensuring they all had a magnitude of one, which is essential for making vector calculations more uniform and efficient.