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Chapter 6: Problem 9

Let \(\mathrm{W}=\operatorname{span}(\\{(i, 0,1)\\})\) in \(\mathrm{C}^{3}\). Find orthonormal bases for \(\mathrm{W}\) and \(\mathrm{W}^{\perp}\).

Short Answer

Expert verified
Orthonormal basis for W: {(\(\frac{i}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\))} Orthonormal basis for W鈯�: {(\(\frac{1}{\sqrt{2}}, 0, \frac{-i}{\sqrt{2}}\)), (\(\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{i}{\sqrt{3}}\))}

Step by step solution

01

Normalize the given vector

The given vector in W is v = (i, 0, 1) and we need to normalize this vector. To do this, calculate its norm: \(\|v\| = \sqrt{|i|^2 + |0|^2 + |1|^2} = \sqrt{2}\) Now, divide the vector by its norm to get the orthonormal basis for W: \(\frac{1}{\sqrt{2}}(i, 0, 1) = (\frac{i}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})\)
02

Find an orthogonal basis for W鈯�

We need to find two linearly independent vectors that are orthogonal to the given vector (i, 0, 1). Let's call these two vectors u and v. To make sure these vectors are orthogonal to (i, 0, 1), we want their dot product to be zero: \((i, 0, 1) \cdot (u_1, u_2, u_3) = 0\) So, we need to find u = (u_1, u_2, u_3) such that: \((i, 0, 1) \cdot (u_1, u_2, u_3) = iu_1 + u_3 = 0\) Choose an arbitrary value for one of the components of u, for example, let's choose \(u_1 = 1\). Then, we have: \(iu_1 + u_3 = i + u_3 = 0\) which gives \(u_3 = -i\). Now, we have u = (1, u_2, -i). We can choose any value for u_2, for example, let it be 0. Thus, we get one vector u = (1, 0, -i). Now, we do the same thing for the second vector v = (v_1, v_2, v_3): \((i, 0, 1) \cdot (v_1, v_2, v_3) = iv_1 + v_3 = 0\) We can choose \(v_1 = -1\), so we have \(iv_1 + v_3 = -i + v_3 = 0\) and \(v_3 = i\). We can choose any value for v_2, for example, let it be 1. Thus, we get the second vector v = (-1, 1, i). So far, we have an orthogonal basis for W鈯�: {(1, 0, -i), (-1, 1, i)}.
03

Normalize the orthogonal basis for W鈯�

Now, we need to normalize the vectors (1, 0, -i) and (-1, 1, i) to get an orthonormal basis for W鈯�. First, let's normalize the vector (1, 0, -i): \(\|u\| = \sqrt{|1|^2 + |-i|^2} = \sqrt{2}\) Normalized vector u: \(\frac{1}{\sqrt{2}}(1, 0, -i) = (\frac{1}{\sqrt{2}}, 0, \frac{-i}{\sqrt{2}})\) Now, let's normalize the vector (-1, 1, i): \(\|v\| = \sqrt{|-1|^2 + |1|^2 + |i|^2} = \sqrt{3}\) Normalized vector v: \(\frac{1}{\sqrt{3}}(-1, 1, i) = (\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{i}{\sqrt{3}})\) Now we've found the orthonormal bases for W and W鈯�: Orthonormal basis for W: {(\(\frac{i}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\))} Orthonormal basis for W鈯�: {(\(\frac{1}{\sqrt{2}}, 0, \frac{-i}{\sqrt{2}}\)), (\(\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{i}{\sqrt{3}}\))}

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Key Concepts

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

Orthogonal Complement
The orthogonal complement of a vector space W, denoted as \(W^{\perp}\), consists of all vectors in the overall space that are perpendicular to every vector in W. If you imagine sticking a pencil upwards from a flat surface, any vectors lying on the surface are perpendicular to the pencil. This is a simple way to think about orthogonal complements.

In mathematical terms, for a vector space in \(\mathbb{C}^3\), if \(W = \text{span}\{(i, 0, 1)\}\), the task of finding \(W^{\perp}\) involves finding vectors \(u\) such that the dot product \((i, 0, 1) \cdot u = 0\). Here's how it's done:
  • By setting up the equation \(iu_1 + u_3 = 0\), you can solve for one component in terms of the others.
  • Choose arbitrary values for the remaining free variables to form a basis for \(W^{\perp}\).
In the given exercise, vectors \((1, 0, -i)\) and \((-1, 1, i)\) were found this way, forming an orthogonal basis for \(W^{\perp}\). After normalizing these vectors, we obtain the orthonormal basis.
Complex Vector Space
Complex vector spaces are similar to real vector spaces but include complex numbers as their elements, introducing a new dimension of mathematical exploration. When dealing with complex numbers, each element is a combination of a real part and an imaginary part, denoted generally as \(a + bi\).

They adhere to the same structural rules as real vector spaces. This means vector addition and scalar multiplication principles are maintained but expanded to include complex scalars:
  • Vectors are closed under addition and scalar multiplication with complex numbers.
  • There is a zero vector that acts as an additive identity.
  • Every vector has an additive inverse.
  • Distributive, associative, and commutative properties apply.
In the exercise, we are working within \(\mathbb{C}^3\), which means every vector has three complex components. This makes working with vectors here quite rich, as we account for both real and imaginary parts when performing mathematical operations.
Normalization
Normalization is a key process when developing orthonormal bases. It scales a vector to unit length (norm of 1), making it easier to work with and ensuring consistencies like directionality without magnitude influence.

To normalize a given vector, such as \((i, 0, 1)\):
  • Compute its norm: \(\|v\| = \sqrt{|i|^2 + |0|^2 + |1|^2} = \sqrt{2}\).
  • Divide each component of the vector by the norm to get \((\frac{i}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})\).
The beauty of normalization is that it transforms any vector into a unit vector while preserving its direction. This is essential when you want to establish an orthonormal set of vectors. Every vector in the orthonormal basis will have a norm of one, making calculations involving angles and lengths straightforward. In the process explained, normalization takes the orthogonal vectors found for \(W^{\perp}\) and adjusts their lengths, ensuring they become a standard set of building blocks for the vector space.

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Most popular questions from this chapter

Suppose that \(A\) and \(B\) are diagonalizable matrices. Prove or disprove that \(A\) is similar to \(B\) if and only if \(A\) and \(B\) are unitarily equivalent.

For a linear operator \(T\) on an inner product space \(V\), prove that \(T^{*} T=\) \(\mathrm{T}_{0}\) implies \(\mathrm{T}=\mathrm{T}_{0}\). Is the same result true if we assume that \(\mathrm{TT}^{*}=\mathrm{T}_{0}\) ?

Recall the moving space vehicle considered in the study of time contraction. Suppose that the vehicle is moving toward a fixed star located on the \(x\)-axis of \(S\) at a distance \(b\) units from the origin of \(S\). If the space vehicle moves toward the star at velocity \(v\), Earthlings (who remain "almost" stationary relative to \(S\) ) compute the time it takes for the vehicle to reach the star as \(t=b / v\). Due to the phenomenon of time contraction, the astronaut perceives a time span of \(t^{\prime}=t \sqrt{1-v^{2}}=(b / v) \sqrt{1-v^{2}}\). A paradox appears in that the astronaut perceives a time span inconsistent with a distance of \(b\) and a velocity of \(v\). The paradox is resolved by observing that the distance from the solar system to the star as measured by the astronaut is less than \(b\). (a) At time \(t\) (as measured on \(C\) ), the space-time coordinates of the star relative to \(S\) and \(C\) are $$ \left(\begin{array}{l} b \\ 0 \\ t \end{array}\right) \text {. } $$ (b) At time \(t\) (as measured on \(C\) ), the space-time coordinates of the star relative to \(S^{\prime}\) and \(C^{\prime}\) are $$ \left(\begin{array}{c} \frac{b-v t}{\sqrt{1-v^{2}}} \\ 0 \\ \frac{t-b v}{\sqrt{1-v^{2}}} \end{array}\right) $$ (c) For $$ x^{\prime}=\frac{b-t v}{\sqrt{1-v^{2}}} \quad \text { and } \quad t^{\prime}=\frac{t-b v}{\sqrt{1-v^{2}}}, $$ we have \(x^{\prime}=b \sqrt{1-v^{2}}-t^{\prime} v\). This result may be interpreted to mean that at time \(t^{\prime}\) as measured by the astronaut, the distance from the astronaut to the star, as measured by the astronaut, (see Figure 6.9) is $$ b \sqrt{1-v^{2}}-t^{\prime} v . $$ Assuming that the coordinate systems \(S\) and \(S^{\prime}\) and clocks \(C\) and \(C^{\prime}\) are as in the discussion of time contraction, prove the following results. (d) Conclude from the preceding equation that (1) the speed of the space vehicle relative to the star, as measured by the astronaut, is \(v\); (2) the distance from Earth to the star, as measured by the astronaut, is \(b \sqrt{1-v^{2}}\). Thus distances along the line of motion of the space vehicle appear to be contracted by a factor of \(\sqrt{1-v^{2}}\).

Let U be a linear operator on a finite-dimensional inner product space V. If \(\|\mathrm{U}(x)\|=\|x\|\) for all \(x\) in some orthonormal basis for \(\mathrm{V}\), must U be unitary? Justify your answer with a proof or a counterexample.

Let \(A\) be a normal matrix with an orthonormal basis of eigenvectors \(\beta=\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) and corresponding eigenvalues \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n} .\) Let \(V\) be the \(n \times n\) matrix whose columns are the vectors in \(\beta\). Prove that for each \(i\) there is a scalar \(\theta_{i}\) of absolute value 1 such that if \(U\) is the \(n \times n\) matrix with \(\theta_{i} v_{i}\) as column \(i\) and \(\Sigma\) is the diagonal matrix such that \(\Sigma_{i i}=\left|\lambda_{i}\right|\) for each \(i\), then \(U \Sigma V^{*}\) is a singular value decomposition of \(A\).
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