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Apply the Gram-Schmidt orthonormalization process to transform the given basis for \(R^{n}\) into an orthonormal basis. Use the vectors in the order in which they are given. \(B=\\{(3,4,0,0),(-1,1,0,0),(2,1,0,-1),(0,1,1,0)\\}\)

Short Answer

Expert verified
The orthonormal basis after applying the Gram-Schmidt orthonormalization process to the original basis set is: \(\{(3/5,4/5,0),(-4/5,3/5,0),(2/5,-1/5,-2/\sqrt{5}), (0,0,1/\sqrt{5})\}\)

Step by step solution

01

Normalizing First Vector

The first vector (3,4,0,0) is normalized by dividing it by its magnitude. The magnitude of a vector \(v = (v_{1}, v_{2},...,v_{n})\) is given by \(\sqrt{v_{1}^{2} + v_{2}^{2} + ... + v_{n}^{2}}\). So, the normalized version of vector 1 is given by dividing every element of the vector by \( \sqrt{3^2+4^2} = 5\). Thus the first orthonormal vector, \( U_1 \), is \((3/5,4/5,0,0)\).
02

Orthogonalizing and Normalizing Second Vector

Subtract the projection of second vector (-1,1,0,0) onto the first orthonormal vector from the second vector to orthogonalize it. The projection of a vector \(a\) onto a vector \(u\) is given by \((a \cdot u) \cdot u\). It is necessary to normalize this resulting vector as done in Step 1. This gives us our second orthonormal vector.
03

Orthogonalizing and Normalizing Third Vector

Subtract the projections of the third vector onto all the previous orthonormal vectors from the third vector to orthogonalize it. Following these steps, calculate the third orthonormal vector. The method is similar as in Step 2.
04

Orthogonalizing and Normalizing Fourth Vector

Repeat the process for the fourth vector the same way we did in steps 2 and 3, by subtracting the projections of the fourth vector onto all the previous orthonormal vectors from the fourth vector to orthogonalize it. Then normalize the resulting vector. This gives us the last orthonormal vector to make up the orthonormal basis set.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Orthonormal Basis
Let's unravel the concept of an orthonormal basis, which is fundamental in the realm of linear algebra. Imagine each vector in an orthonormal basis as a rigorously orthogonal (at right angles) and normalized (of unit length) element within a vector space.

An orthonormal basis simplifies computations within the space such as when calculating projections, since vectors with these properties do not interfere with one another. In fact, the Gram-Schmidt process is one of the most celebrated methods for transforming a set of vectors into an orthonormal basis. This provides advantages for simplifying matrix operations and understanding geometrical and functional aspects of vector spaces.

The beauty of an orthonormal basis is that it allows us to represent other vectors in the space with unique combinations of the basis vectors, in the simplest form possible.
Vector Normalization
Now, let's touch on vector normalization. You have a vector, and you want to bring it down to size, so to speak. To normalize a vector means to scale it such that its length or magnitude is equal to 1, transforming it into what we call a unit vector.

To normalize a vector, you merely divide each component by the vector's magnitude. This process is pivotal in the early stages of the Gram-Schmidt orthonormalization method, as it guarantees that the 'unit' part of 'orthonormal' is satisfied. After normalizing the first vector in a set, subsequent vectors will then go through a process of orthogonalization before also being normalized themselves.

Converting to a Unit Vector

Imagine you have a vector \(v\). Its magnitude is given by the square root of the sum of the squares of its components. By dividing each component of \(v\) by its magnitude, you get the normalized vector or unit vector, often denoted by \(\hat{v}\).
Vector Projection
Vector projection is another pivotal operation that shows up in many linear algebra applications, particularly in the Gram-Schmidt process. When you project one vector onto another, essentially, you're casting a shadow of one vector onto the other, if you consider the light source to be coming from the direction of the vector being projected onto.

The projection takes the form of a scalar multiplication of the vector you’re projecting onto. We calculate it using the dot product, which reflects how much one vector extends in the direction of another. Once you have the dot product, you multiply the vector you’re projecting onto by the result, normalized by its magnitude squared to scale it appropriately.

Visualizing Projection

Consider two vectors, \(a\) and \(b\). To project \(a\) onto \(b\), you’d compute \(\frac{a \cdot b}{\|b\|^2} b\), where \(a \cdot b\) is the dot product and \(\|b\|\) is the magnitude of \(b\). This gives the projected vector, which lies along \(b\) and represents \(a\)'s 'influence' in \(b\)'s direction.
Linear Algebra
Linear algebra is the backdrop against which the entire performance of the Gram-Schmidt orthonormalization takes place. It's a branch of mathematics that deals with vectors, vector spaces (also known as linear spaces), and linear transformations that move points in space without altering their structural integrity.

Within linear algebra, vectors can be manipulated in various ways to understand spaces and solve systems of linear equations. Techniques such as scaling (multiplying by a scalar), adding, and projecting vectors onto each other are the essence of linear algebra. These techniques let us explore the dimensions of any vector space and find sets of vectors that span or cover the entire space, such as bases, which can be made orthonormal for ease of calculations.
Magnitude of a Vector
The magnitude of a vector is essentially the measure of its length. In the journey of understanding vector spaces, you'll find this concept is not only central to the process of normalization but is widely applied across the entire field of linear algebra.

The magnitude is determined by taking the square root of the sum of the squares of each vector component. This is akin to calculating the hypotenuse of a right triangle in classic geometry, but extended into higher dimensions for vectors of n-components. The general formula is given as \(\|v\| = \sqrt{v_{1}^2 + v_{2}^2 + ... + v_{n}^2}\).

Applying Magnitude

For instance, take the vector \(v=(a,b,c)\). The magnitude of \(v\) is calculated by \(\sqrt{a^2 + b^2 + c^2}\). Understanding this will allow you to not just normalize vectors, but also helps in myriad ways, such as calculating distances between points in space and determining the modulus of vectors in physics.

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Most popular questions from this chapter

Let \(W\) be a subspace of the inner product space \(V\). Prove that the set \(W^{\perp}=\\{\mathbf{v} \in V:\langle\mathbf{v}, \mathbf{w}\rangle= 0 \text { for all } \mathbf{w} \in W\\}\) is a subspace of \(V\) Getting Started: To prove that \(W^{\perp}\) is a subspace of \(V,\) you must show that \(W^{\perp}\) is nonempty and that the closure conditions for a subspace hold (Theorem 4.5 ). (i) Find a vector in \(W^{\perp}\) to conclude that it is nonempty. (ii) To show the closure of \(W^{\perp}\) under addition, you need to show that \(\left\langle\mathbf{v}_{1}+\mathbf{v}_{2}, \mathbf{w}\right\rangle= 0\) for all \(\mathbf{w} \in W\) and for any \(\mathbf{v}_{1}, \mathbf{v}_{2} \in W^{\perp} .\) Use the properties of inner products and the fact that \(\left\langle\mathbf{v}_{1}, \mathbf{w}\right\rangle\) and \(\left\langle\mathbf{v}_{2}, \mathbf{w}\right\rangle\) are both zero to show this. (iii) To show closure under multiplication by a scalar, proceed as in part (ii). Use the properties of inner products and the condition of belonging to \(W^{\perp}\).

Determine whether each statement is true or false. If a statement is true, give a reason or cite an appropriate statement from the text. If a statement is false, provide an example that shows the statement is not true in all cases or cite an appropriate statement from the text. Prove that\(\|\mathbf{u}+\mathbf{v}\|^{2}+\|\mathbf{u}-\mathbf{v}\|^{2}=2\|\mathbf{u}\|^{2}+2\|\mathbf{v}\|^{2}\)for any vectors \(\mathbf{u}\) and \(\mathbf{v}\) in an inner product space \(V\)

Show that \(f\) and \(g\) are orthogonal in the inner product space \(C[a, b]\) with the inner product $$\langle f, g\rangle=\int_{a}^{b} f(x) g(x) d x$$ $$\begin{aligned}&C[0, \pi], \quad f(x)=1, \quad g(x)=\cos (2 n x)\\\&n=1,2,3, \dots\end{aligned}$$

Show that the volume \(V\) of a parallelepiped having \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) as adjacent edges is \(V=|\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w})|\)

Verify the Pythagorean Theorem for the vectors u and \(\mathbf{v}\). $$\mathbf{u}=(4,1,-5), \quad \mathbf{v}=(2,-3,1)$$

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