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Chapter 7: Problem 99

A complex matrix \(A\) is unitary if it is invertible and \(A^{-1}=A^{H}\). Alternatively, \(A\) is unitary if its rows (columns) form an orthonormal set of vectors (relative to the usual inner product of \(\mathbf{C}^{n}\) ). Find a unitary matrix whose first row is: (a) a multiple of \((1,1-i)\); (b) a multiple of \(\left(\frac{1}{2}, \frac{1}{2} i, \frac{1}{2}-\frac{1}{2} i\right)\).

Short Answer

Expert verified
For case (a), the unitary matrix is: \(A = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1 - i}{\sqrt{3}} \\ \frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}\) For case (b), the unitary matrix is: \(A = \begin{pmatrix} \frac{2}{\sqrt{3}}\left(\frac{1}{2}\right) & \frac{2}{\sqrt{3}}\left(\frac{1}{2}i\right) & \frac{2}{\sqrt{3}}\left(\frac{1}{2}-\frac{1}{2}i\right) \\ -\frac{1}{2} & -\frac{1}{2}i & \frac{1}{2}+\frac{1}{2}i \\ -\frac{1}{2} & \frac{1}{2}-\frac{1}{2}i & -\frac{1}{2}i \end{pmatrix}\)

Step by step solution

01

Normalizing the first row vector

We have the given vector \((1,1-i)\) as a multiple of the first row. First, we need to normalize this vector: Vector: \(v_1 = (1, 1-i)\) Normalized vector: \(u_1 = \frac{v_1}{\|v_1\|}\) Magnitude of \(v_1\): \(\|v_1\| = \sqrt{|1|^2 + |1-i|^2} = \sqrt{1 + 2} = \sqrt{3}\) Normalized vector: \(u_1 = \frac{1}{\sqrt{3}}(1, 1-i)\)
02

Finding an orthogonal vector

Now we need to find another vector \(v_2\) orthogonal to the first: By definition of orthogonality: \(\langle u_1, v_2 \rangle = 0\) Let \(v_2 = (x, y)\). Then, \(\frac{1}{\sqrt{3}}\left\langle (1, 1-i), (x, y) \right\rangle = 0\) Plug in the values and simplify: \(\frac{1}{\sqrt{3}}(x + y - iy) = 0\) The above equation should hold true for any values of \(x\) and \(y\), so look for any pairs that would satisfy the equation. After testing different values, we can choose \(y = 1\) and \(x = i\). Then, \(v_2 = (i, 1)\).
03

Normalizing the second row vector

Normalize the vector \(v_2\) to get the second row of the matrix: Magnitude of \(v_2\): \(\|v_2\| = \sqrt{|i|^2 + |1|^2} = \sqrt{2}\) Normalized vector: \(u_2 = \frac{v_2}{\|v_2\|} = \frac{1}{\sqrt{2}}(i, 1)\)
04

Forming the unitary matrix

We have both normalized rows, so we form the unitary matrix: \(A = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1 - i}{\sqrt{3}} \\ \frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}\) Now we have the unitary matrix for case (a). #Case (b)#
05

Normalizing the first row vector

We have the given vector \(\left(\frac{1}{2}, \frac{1}{2} i, \frac{1}{2}-\frac{1}{2}i\right)\) as a multiple of the first row. First, we need to normalize this vector: Vector: \(v_1 = \left(\frac{1}{2}, \frac{1}{2}i, \frac{1}{2}-\frac{1}{2}i\right)\) Normalized vector: \(u_1 = \frac{v_1}{\|v_1\|}\) Magnitude of \(v_1\): \(\|v_1\| = \sqrt{\frac{1}{4} + \frac{1}{4} + \frac{1}{4}} = \frac{\sqrt{3}}{2}\) Normalized vector: \(u_1 = \frac{2}{\sqrt{3}}\left(\frac{1}{2}, \frac{1}{2} i, \frac{1}{2}-\frac{1}{2} i\right)\)
06

Finding orthogonal vectors with Gram-Schmidt process

Now, we will use the Gram-Schmidt process and find two additional orthogonal (and normalized) vectors to form a unitary matrix: 1) Start with two independent vectors, say \((1,0,0)\) and \((0,1,0)\). 2) Take the projection of those vectors onto \(u_1\) and subtract it to make them orthogonal: \(v_2 = (1,0,0) - \langle(1,0,0), u_1\rangle u_1\) \(v_3 = (0,1,0) - \langle(0,1,0), u_1\rangle u_1\) 3) Normalize the orthogonal vectors: \(u_2 = \frac{v_2}{\| v_2 \|}\) \(u_3 = \frac{v_3}{\| v_3 \|}\) After calculating the vectors, we get: \(u_2 = \left(-\frac{1}{2}, -\frac{1}{2}i, \frac{1}{2}+\frac{1}{2}i\right)\) \(u_3 = \left(-\frac{1}{2}, \frac{1}{2}-\frac{1}{2}i, -\frac{1}{2}i\right)\)
07

Forming the unitary matrix

Now we combine the three normalized vectors as rows of the unitary matrix: \(A = \begin{pmatrix} \frac{2}{\sqrt{3}}\left(\frac{1}{2}\right) & \frac{2}{\sqrt{3}}\left(\frac{1}{2}i\right) & \frac{2}{\sqrt{3}}\left(\frac{1}{2}-\frac{1}{2}i\right) \\ -\frac{1}{2} & -\frac{1}{2}i & \frac{1}{2}+\frac{1}{2}i \\ -\frac{1}{2} & \frac{1}{2}-\frac{1}{2}i & -\frac{1}{2}i \end{pmatrix}\) Now we have the unitary matrix for case (b).

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Key Concepts

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

Orthonormal Vectors
In mathematics, especially in the study of linear algebra, orthonormal vectors play a critical role. These vectors are not just any set of vectors. They have two specific properties that make them particularly valuable in simplifying complex problems:
- **Orthogonal**: This means that the vectors are perpendicular to each other. For two vectors to be orthogonal, their dot product must equal zero. - **Normalized**: A vector is normalized if its length, or magnitude, equals one.
When a set of vectors not only maintains orthogonality but also normalization, they form an orthonormal set. This concept is critical when dealing with matrices, as forming an orthonormal basis can greatly simplify many mathematical tasks. Examples of such applications include simplifying complex calculations and ensuring that transformations preserve vector lengths and angles.
Gram-Schmidt Process
The Gram-Schmidt process is a method used to convert a non-orthogonal set of vectors into an orthogonal鈥攐r even better鈥攐rthonormal set. This procedure is essential in linear algebra as it builds an orthonormal basis for a subspace starting from a given set of linearly independent vectors. It works as follows:
1. **Start with a set of vectors**: Begin with any set of linearly independent vectors. 2. **Orthogonalize**: Adjust each vector in the set so they remain orthogonal. This involves subtracting projections of the current vector along previously adjusted vectors. 3. **Normalize**: After orthogonalizing the vectors, adjust their lengths to one.
The benefit of the Gram-Schmidt process is its ability to ensure that the new orthonormal basis retains the original space's dimension and characteristics. Using this process is particularly useful when creating a unitary matrix, as its rows or columns are required to be orthonormal.
Complex Matrix
A complex matrix at its core is a matrix where the elements are complex numbers. Each complex number can be expressed in the form a + bi, where a and b are real numbers, and i is the imaginary unit where i^2 = -1. Complex matrices are not just limited to real values but include imaginary components, which deeply enhance their application in many fields like engineering, physics, and computer science.
  • **Unitary Matrix**: A unitary matrix is a type of complex matrix that holds a special property. It is its own inverse's conjugate transpose, meaning when you multiply it by its conjugate transpose, you get the identity matrix. This property ensures stability in transformations, preserving vector norms and angles.
  • **Applications**: These matrices are vital in quantum mechanics, where preserving probabilities is crucial, leading to their frequent appearance in calculations involving quantum states.
Complex matrices and their properties, such as unitarity, are foundational in advanced mathematical analysis and problem-solving, bridging the gap between theoretical concepts and real-world applications.

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Most popular questions from this chapter

Prove Theorem 7.2: The norm in an inner product space \(V\) satisfies (a) \(\left[\mathrm{N}_{1}\right]\|v\| \geq 0 ;\) and \(\|v\|=0\) if and only if \(v=0\). (b) \(\left[\mathrm{N}_{2}\right]\|k v\|=|k|\|v\|\). (c) \(\left[\mathrm{N}_{3}\right]\|u+v\| \leq\|u\|+\|v\|\).

Let \(V\) be a complex inner product space. Verify the relation \\[\left\langle u, a v_{1}+b v_{2}\right\rangle=\bar{a}\left\langle u, v_{1}\right\rangle+\bar{b}\left\langle u, v_{2}\right\rangle\\]

Prove (a) \(\|\cdot\|_{1}\) is a norm on \(\mathbf{R}^{n}\). (b) \(\|\cdot\|_{\infty}\) is a norm on \(\mathbf{R}^{n}\).

Verify that the following is an inner product on \(\mathbf{R}^{2}\), where \(u=\left(x_{1}, x_{2}\right)\) and \(v=\left(y_{1}, y_{2}\right):\) $$ f(u, v)=x_{1} y_{1}-2 x_{1} y_{2}-2 x_{2} y_{1}+5 x_{2} y_{2} $$

Find the matrix \(A\) that represents the usual inner product on \(\mathbf{R}^{2}\) relative to each of the following bases: (a) \(\left\\{v_{1}=(1,4), \quad v_{2}=(2,-3)\right\\}\) (b) \(\left\\{w_{1}=(1,-3), \quad w_{2}=(6,2)\right\\}\).
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