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Chapter 6: Problem 19

Let \(\mathrm{T}\) and \(\mathrm{U}\) be positive definite operators on an inner product space V. Prove the following results. (a) \(\mathrm{T}+\mathrm{U}\) is positive definite. (b) If \(c>0\), then \(c \mathrm{~T}\) is positive definite. (c) \(\mathrm{T}^{-1}\) is positive definite. Visit goo.gl/cQch7i for a solution.

Short Answer

Expert verified
To prove the positive-definiteness of the given operators, we can follow these steps: (a) For T + U being positive definite, we can rewrite the inner product as 鉄(T+U)v, v鉄 = 鉄═v + Uv, v鉄 and use the linearity properties of inner products to get 鉄(T+U)v, v鉄 = 鉄═v, v鉄 + 鉄║v, v鉄. Since T and U are both positive definite, their sum will also be positive, proving that T + U is positive definite. (b) For cT being positive definite with c>0, we rewrite the inner product as 鉄(cT)v, v鉄 = 鉄╟(Tv), v鉄 and use the linearity properties of inner products to get 鉄(cT)v, v鉄 = c鉄═v, v鉄. Since c > 0 and T is positive definite, their product will also be positive, proving that cT is positive definite for c > 0. (c) To prove T鈦宦 is positive definite, we introduce a new vector w as Tv, so T鈦宦箇 = v. We then calculate the inner product 鉄(T^{-1})v, v鉄, which is equal to 鉄╳, v鉄. Using the Cauchy-Schwarz inequality, we find that 鉄(T^{-1})v, v鉄 is positive for any non-zero vector v, proving that T鈦宦 is positive definite.

Step by step solution

01

Part (a) - Proving T + U is Positive Definite

Recall that an operator T is positive definite if, for any non-zero vector v, the inner product 鉄═v, v鉄 is positive. To prove that T + U is positive definite, we need to show that 鉄(T+U)v, v鉄 > 0 for any non-zero vector v. Step 1: Rewrite the inner product We can rewrite the inner product of (T + U)v and v as: 鉄(T+U)v, v鉄 = 鉄═v + Uv, v鉄 Step 2: Using linearity properties of inner products Using linearity properties of inner products, we can simplify the above expression: 鉄═v + Uv, v鉄 = 鉄═v, v鉄 + 鉄║v, v鉄 Step 3: Positive-definiteness of T and U Since T and U are both positive definite operators, we know that 鉄═v, v鉄 > 0 and 鉄║v, v鉄 > 0. Step 4: Sum of positive values Thus, the sum of these two positive values, i.e., 鉄(T+U)v, v鉄 = 鉄═v, v鉄 + 鉄║v, v鉄, will also be positive. Hence, T + U is positive definite.
02

Part (b) - Proving cT is Positive Definite for c > 0

We need to prove that if c is a positive constant, then cT is also a positive definite operator. Now, let's show that 鉄(cT)v, v鉄 > 0 for any non-zero vector v. Step 1: Rewrite the inner product We can rewrite the inner product of (cT)v and v as: 鉄(cT)v, v鉄 = 鉄╟(Tv), v鉄 Step 2: Using linearity properties of inner products Using the scalar property of inner products, we can rewrite the expression: 鉄╟(Tv), v鉄 = c鉄═v, v鉄 Step 3: Positive-definiteness of T Since T is a positive definite operator, we know that 鉄═v, v鉄 > 0. Step 4: Product of positive values Since c > 0 and 鉄═v, v鉄 > 0, their product c鉄═v, v鉄 will also be positive. Hence, cT is positive definite for c > 0.
03

Part (c) - Proving T^{-1} is Positive Definite

To prove that the inverse of a positive definite operator T, denoted by T^{-1}, is also positive definite, we need to show that 鉄(T^{-1})v, v鉄 > 0 for any non-zero vector v. Step 1: Define w as Tv Let's introduce a new vector w such that w = Tv. Step 2: Inverse operator Since T^{-1} is the inverse of T, we have T^{-1}w = v. Step 3: Inner product of T^{-1}v and v Now, we need to find 鉄(T^{-1})v, v鉄, which is equivalent to 鉄╳, (T^{-1})v鉄. Since T^{-1}w = v, this is equal to 鉄╳, v鉄. Step 4: Positive-definiteness of T Using the positive-definiteness of T, we know that 鉄╳, w鉄 = 鉄═v, v鉄 > 0. Step 5: Cauchy-Schwarz inequality Using the Cauchy-Schwarz inequality, we have: |鉄╳, v鉄﹟^2 鈮 鉄╳, w鉄┾煥v,v鉄 Step 6: Compare 鉄╳, v鉄 to zero Since 鉄╳, w鉄 > 0 and 鉄╲, v鉄 > 0, this inequality implies: 0 < |鉄╳, v鉄﹟^2 鈮 鉄╳, w鉄 鉄╲,v鉄 As shown above, |鉄╳, v鉄﹟^2 > 0, which means that 鉄╳, v鉄 鈮 0. Step 7: Positive-definiteness of T^{-1} Since we know that 鉄╳, v鉄 = 鉄(T^{-1})v, v鉄, we can now conclude that 鉄(T^{-1})v, v鉄 is positive for any non-zero vector v, and thus, T^{-1} is positive definite.

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Key Concepts

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

Inner Product Space
An inner product space is a vector space equipped with an additional structure called an inner product. This inner product is a function that takes two vectors and returns a scalar, providing a way to measure angles and lengths. It's a foundational concept in mathematics and physics.

Some properties of an inner product include:
  • Linearity: The inner product is linear in its first argument. For vectors \(u, v, w\) and scalar \(c\), \(\langle cu+v, w\rangle = c\langle u, w\rangle + \langle v, w\rangle\).
  • Symmetry: It satisfies \(\langle u, v\rangle = \overline{\langle v, u\rangle}\), where the bar denotes complex conjugation.
  • Positive-Definiteness: For any vector \(v\), \(\langle v, v\rangle \geq 0\), and it's zero only if \(v\) is the zero vector.
These properties ensure that inner product spaces generalize Euclidean spaces and form the basis for understanding more complex structures like positive definite operators. Inner product spaces help us analyze the behavior of operators, making them fundamental in the study of linear mappings.
Linear Operators
Linear operators are functions that map vectors from one vector space to another while preserving the operations of vector addition and scalar multiplication. They play a crucial role in various mathematical areas, offering a simplified way to understand complex transformations.

To qualify as linear, an operator \(T\) must satisfy two main properties:
  • Additivity: \(T(u + v) = T(u) + T(v)\) for all vectors \(u\) and \(v\).
  • Homogeneity: \(T(cu) = cT(u)\) for any scalar \(c\) and vector \(u\).
Linear operators can represent various transformations such as rotations, reflections, and scalings. A specialized type of linear operator, the positive definite operator, has special properties in an inner product space:
  • For any non-zero vector \(v\), the inner product \(\langle Tv, v\rangle\) is strictly positive.
This means positive definite operators preserve a certain amount of energy or distance in a system, making them indispensable tools in physics and engineering.
Inverse Operator
An inverse operator reverses the effect of another operator, akin to how division undoes multiplication. Given a linear operator \(T\), the inverse operator \(T^{-1}\) satisfies \(T(T^{-1}(v)) = T^{-1}(T(v)) = v\) for vector \(v\). This concept is essential in solving systems of linear equations and understanding functions.Finding an inverse is only possible if \(T\) is bijective (one-to-one and onto), meaning it maps every element in its domain to a unique element in its range.

For positive definite operators, the inverse is particularly interesting:
  • If \(\langle Tv, v\rangle > 0\) for all non-zero \(v\), then \(T^{-1}\) is also positive definite.
  • This property ensures that the positive definiteness of the transformation is preserved, even when the operation is reversed.
Inverse operators are not just theoretical constructs; they are key in practical computations like those in computer graphics and optimization, where reverting transformations accurately is critical.

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Most popular questions from this chapter

This exercise relates the singular values of a well-behaved linear operator or matrix to its eigenvalues. (a) Let \(\mathrm{T}\) be a normal linear operator on an \(n\)-dimensional inner product space with eigenvalues \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\). Prove that the singular values of \(\mathrm{T}\) are \(\left|\lambda_{1}\right|,\left|\lambda_{2}\right|, \ldots,\left|\lambda_{n}\right|\). (b) State and prove a result for matrices analogous to (a).

24\. Let \(T\) be a linear operator on a real inner product space \(V\), and define \(H: \mathrm{V} \times \mathrm{V} \rightarrow R\) by \(H(x, y)=\langle x, \mathrm{~T}(y)\rangle\) for all \(x, y \in \mathrm{V}\). (a) Prove that \(H\) is a bilinear form. (b) Prove that \(H\) is symmetric if and only if \(\mathrm{T}\) is self-adjoint. (c) What properties must \(\mathrm{T}\) have for \(H\) to be an inner product on \(\mathrm{V}\) ? (d) Explain why \(H\) may fail to be a bilinear form if \(\mathrm{V}\) is a complex inner product space.

Let \(V\) be a finite-dimensional inner product space, and let \(T\) and \(U\) be self-adjoint operators on \(V\) such that \(T\) is positive definite. Prove that both TU and UT are diagonalizable linear operators that have only real eigenvalues. Hint: Show that UT is self-adjoint with respect to the inner product \(\langle x, y\rangle^{\prime}=\langle\mathrm{T}(x), y\rangle\). To show that TU is self-adjoint, repeat the argument with \(\mathrm{T}^{-1}\) in place of \(\mathrm{T}\).

Let \(H: \mathrm{R}^{2} \times \mathrm{R}^{2} \rightarrow R\) be the function defined by $$ H\left(\left(\begin{array}{l} a_{1} \\ a_{2} \end{array}\right),\left(\begin{array}{l} b_{1} \\ b_{2} \end{array}\right)\right)=a_{1} b_{2}+a_{2} b_{1} \quad \text { for }\left(\begin{array}{l} a_{1} \\ a_{2} \end{array}\right),\left(\begin{array}{l} b_{1} \\ b_{2} \end{array}\right) \in \mathrm{R}^{2} $$ (a) Prove that \(H\) is a bilinear form. (b) Find the \(2 \times 2\) matrix \(A\) such that \(H(x, y)=x^{t} A y\) for all \(x, y \in \mathrm{R}^{2} .\)

Let \(S\) be the set of all \(\left(t_{1}, t_{2}, t_{3}\right) \in \mathrm{R}^{3}\) for which $$ 3 t_{1}^{2}+3 t_{2}^{2}+3 t_{3}^{2}-2 t_{1} t_{3}+2 \sqrt{2}\left(t_{1}+t_{3}\right)+1=0 $$ Find an orthonormal basis \(\beta\) for \(\mathrm{R}^{3}\) for which the equation relating the coordinates of points of \(\mathcal{S}\) relative to \(\beta\) is simpler. Describe \(\mathcal{S}\) geometrically.
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