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Chapter 6: Problem 4

(a) Complete the proof in Example 5 that ( 路, 路} is an inner product (the Frobenius inner product) on Mnxn(F). (b ) Use the Frobenius inner product to compute IIAII, IIBII, and (A, B) for $$ A=\left(\begin{array}{cc} 1 & 2+i \\ 3 & i \end{array}\right) \quad \text { and } \quad B=\left(\begin{array}{cc} 1+i & 0 \\ i & -i \end{array}\right) $$

Short Answer

Expert verified
The Frobenius norm for A, IIAII, is 4, the Frobenius norm for B, IIBII, is 1, and the Frobenius inner product between A and B, (A, B), is \(3 + 4i\).

Step by step solution

01

(a) Prove Frobenius inner product

Let's show that the Frobenius inner product satisfies the three properties of inner product: 1. Conjugate symmetry: \((X, Y) = trace(X^*Y) = \sum_{i=1}^n \sum_{j=1}^n (X^*)_{ij}Y_{ij} = \sum_{i=1}^n \sum_{j=1}^n \overline{X_{ji}}Y_{ij}\) Now, let's find \((Y, X)^*\): \((Y, X) = trace(Y^*X) = \sum_{i=1}^n \sum_{j=1}^n (Y^*)_{ij}X_{ij} = \sum_{i=1}^n \sum_{j=1}^n \overline{Y_{ji}}X_{ij}\) \((Y, X)^* = \overline{trace(Y^*X)} = \sum_{i=1}^n \sum_{j=1}^n \overline{\overline{Y_{ji}}X_{ij}} = \sum_{i=1}^n \sum_{j=1}^n \overline{Y_{ji}}Y_{ij}\) Thus, \((X, Y) = (Y, X)^*\) holds. 2. Linearity: \((aX + bY, Z) = trace((aX + bY)^* Z) = trace(aX^* Z + bY^* Z) = trace(aX^* Z) + trace(bY^* Z) = a \sum_{i=1}^n \sum_{j=1}^n (X^*)_{ij}Z_{ij} + b \sum_{i=1}^n \sum_{j=1}^n (Y^*)_{ij}Z_{ij} = a(X, Z) + b(Y, Z)\) Thus, \((aX + bY, Z) = a(X, Z) + b(Y, Z)\) holds. 3. Positive definiteness: \((X, X) = \text{trace}(X^*X) = \sum_{i=1}^n \sum_{j=1}^n (X^*)_{ij}X_{ij} = \sum_{i=1}^n \sum_{j=1}^n \overline{X_{ji}}X_{ij}\) Since \(X_{ij}\overline{X_{ji}}\) is always a non-negative real number, their sum is also non-negative. Thus, \((X, X) \geq 0\). Also, \((X, X) = 0\) implies that each term \(X_{ij}\overline{X_{ji}} = |X_{ij}|^2 = 0\), so X = 0. Hence, the Frobenius inner product satisfies all three properties, so it is an inner product on Mnxn(F).
02

(b) Compute inner product properties

Now, let's compute \((A, A)\), \((B, B)\), and \((A, B)\) for the given matrices A and B using the Frobenius inner product: Matrix A: $$ A=\left(\begin{array}{cc} 1 & 2+i \\ 3 & i \end{array}\right) $$ Matrix B: $$ B=\left(\begin{array}{cc} 1+i & 0 \\ i & -i \end{array}\right) $$ 1. Frobenius norm for A, IIAII: \((A, A) = trace(A^*A) = trace\left(\begin{array}{cc}1 & 3\\\overline{2+i} & \bar{i}\end{array}\right)\left(\begin{array}{cc}1 & 2+i\\3 & i\end{array}\right) = trace\left(\begin{array}{cc}10 & 1+5i\\1-5i & 6\end{array}\right) = 10 + 6 = 16\) IIAII = \(\sqrt{(A, A)} = \sqrt{16} = 4\) 2. Frobenius norm for B, IIBII: \((B, B) = trace(B^*B) = trace\left(\begin{array}{cc}\overline{1+i} & \bar{i}\\0 & i\end{array}\right)\left(\begin{array}{cc}1+i & 0\\i & -i\end{array}\right) = trace\left(\begin{array}{cc}3 & i-2\\-i & -2\end{array}\right) = 3 - 2 = 1\) IIBII = \(\sqrt{(B, B)} = \sqrt{1} = 1\) 3. Frobenius inner product for A and B, (A, B): \((A, B) = trace(A^*B) = trace\left(\begin{array}{cc}1 & 3\\\overline{2+i} & \bar{i}\end{array}\right)\left(\begin{array}{cc}1+i & 0\\i & -i\end{array}\right) = trace\left(\begin{array}{cc}1+3i & 3-3i\\-1+2i & 2+i\end{array}\right) = (1+3i) + (2+i) = 3 + 4i\) Hence, the Frobenius norm for A is 4, for B is 1, and the Frobenius inner product between A and B is \(3 + 4i\).

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Key Concepts

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

Linear Algebra
Linear algebra is a branch of mathematics that studies vectors, vector spaces, and linear transformations. It is essential in understanding both theoretical and applied mathematics. In particular, it provides tools for solving linear systems of equations.
Linear algebra focuses on structures such as ideations of matrices and determinants, providing a fundamental basis for various computations. Matrices and vectors are used extensively in computer graphics, engineering, physics, and more.
Understanding vectors within a space and the operations you can perform on them, like addition and scalar multiplication, is crucial. This helps in developing a solid foundation in both computational and theoretical aspects of mathematics.
Inner Product Spaces
An inner product space extends the notion of dot products of vectors to more general spaces. This space allows for defining angles and distances, which are not possible in broader vector spaces without such definitions.
Inner product spaces have specific properties, typically including linearity, conjugate symmetry, and positive-definiteness. These properties collectively enable more advanced discussions and problems in mathematics and physics.
For instance, the Frobenius inner product used in matrix operations helps define the angle and distance between two matrices. This makes generalizing concepts like the length of a vector or the angle between vectors in higher-dimension spaces possible.
Matrix Operations
Matrix operations are the foundation of many computations in linear algebra. Addition, multiplication, and finding determinants of matrices are routine operations.
The Frobenius inner product is particularly useful for operations involving matrices. It simplifies computing matrix norms and products which are essential in various applications like machine learning, signal processing, and optimization algorithms.
When performing these operations, matrices interact similar to vectors but with more complex rules due to their two-dimensional nature. Criteria such as row and column alignment dictate these operations, guiding rules for existence or otherwise of results like the matrix product.
Conjugate Symmetry
Conjugate symmetry is a fundamental property for inner products especially when dealing with complex numbers. It ensures that the inner product of two complex matrices or vectors behaves predictably, allowing for meaningful geometric and algebraic interpretations.
This property essentially states that the inner product of two elements must equal the complex conjugate of the inner product in the reverse order. Mathematically, for matrices or complex numbers, this is reflected as \((X, Y) = \overline{(Y, X)}\).
Conjugate symmetry facilitates many mathematical proofs and derivations, providing a necessary foundation for determining orthogonality and other similar properties in spaces that include complex elements.

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Most popular questions from this chapter

Let \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{W}\) be a linear transformation, where \(\mathrm{V}\) and \(\mathrm{W}\) are finitedimensional inner product spaces. Prove that \(\left(R\left(T^{*}\right)\right)^{\perp}=N(T)\), using the preceding definition.

This exercise provides a converse to Exercise 20. Let \(V\) be a finitedimensional inner product space with inner product \(\langle\cdot, \cdot \cdot\rangle\), and let \(\langle\cdot, \cdot\rangle^{\prime}\) be any other inner product on V. (a) Prove that there exists a unique linear operator \(T\) on \(V\) such that \(\langle x, y\rangle^{\prime}=\langle\mathrm{T}(x), y\rangle\) for all \(x\) and \(y\) in \(\mathrm{V}\). Hint: Let \(\beta=\) \(\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) be an orthonormal basis for \(\mathrm{V}\) with respect to \(\langle\cdot \cdot \cdot\rangle\), and define a matrix \(A\) by \(A_{i j}=\left\langle v_{j}, v_{i}\right\rangle^{\prime}\) for all \(i\) and \(j\). Let \(\mathrm{T}\) be the unique linear operator on \(\mathrm{V}\) such that \([\mathrm{T}]_{\beta}=A\). (b) Prove that the operator \(T\) of (a) is positive definite with respect to both inner products.

Let \(W\) be a finite-dimensional subspace of an inner product space \(V\). Show that if \(\mathrm{T}\) is the orthogonal projection of \(\mathrm{V}\) on \(\mathrm{W}\), then \(\mathrm{I}-\mathrm{T}\) is the orthogonal projection of \(\mathrm{V}\) on \(\mathrm{W}^{\perp}\).

(a) Bessel's Inequality. Let \(\mathrm{V}\) be an inner product space, and let \(S=\) \(\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) be an orthonormal subset of \(\mathrm{V}\). Prove that for any \(x \in \mathrm{V}\) we have $$ \|x\|^{2} \geq \sum_{i=1}^{n}\left|\left\langle x, v_{i}\right\rangle\right|^{2} . $$ Hint: Apply Theorem \(6.6\) to \(x \in \mathrm{V}\) and \(\mathrm{W}=\operatorname{span}(S)\). Then use Exercise 10 of Section 6.1. (b) In the context of (a), prove that Bessel's inequality is an equality if and only if \(x \in \operatorname{span}(S)\).

Determine which of the mappings that follow are bilinear forms. Justify your answers. (a) Let \(\mathrm{V}=\mathrm{C}[0,1]\) be the space of continuous real-valued functions on the closed interval \([0,1]\). For \(f, g \in \mathrm{V}\), define $$ H(f, g)=\int_{0}^{1} f(t) g(t) d t . $$ (b) Let \(\mathrm{V}\) be a vector space over \(F\), and let \(J \in \mathcal{B}(\mathrm{V})\) be nonzero. Define \(H: \mathrm{V} \times \mathrm{V} \rightarrow F\) by $$ H(x, y)=[J(x, y)]^{2} \quad \text { for all } x, y \in \mathrm{V} . $$ (c) Define \(H: R \times R \rightarrow R\) by \(H\left(t_{1}, t_{2}\right)=t_{1}+2 t_{2}\). (d) Consider the vectors of \(\mathrm{R}^{2}\) as column vectors, and let \(H: \mathrm{R}^{2} \rightarrow R\) be the function defined by \(H(x, y)=\operatorname{det}(x, y)\), the determinant of the \(2 \times 2\) matrix with columns \(x\) and \(y\). (e) Let \(\mathrm{V}\) be a real inner product space, and let \(H: \mathrm{V} \times \mathrm{V} \rightarrow R\) be the function defined by \(H(x, y)=\langle x, y\rangle\) for \(x, y \in \mathrm{V}\). (f) Let \(\mathrm{V}\) be a complex inner product space, and let \(H: \mathrm{V} \times \mathrm{V} \rightarrow C\) be the function defined by \(H(x, y)=\langle x, y\rangle\) for \(x, y \in \mathrm{V}\).
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