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Chapter 2: Problem 39

Let \(\mathrm{T}: C \rightarrow C\) be the function defined by \(\mathrm{T}(z)=\bar{z}\). Prove that \(\mathrm{T}\) is additive (as defined in Exercise 38 ) but not linear.

Short Answer

Expert verified
In this exercise, we needed to prove that T is additive but not linear. To do so, we showed that T(z1 + z2) = T(z1) + T(z2), confirming that T is additive. However, we found that for purely imaginary scalars c, T(cz) 鈮� cT(z), which demonstrates that T is not homogeneous and, therefore, not linear. So, T is additive but not linear.

Step by step solution

01

Prove T is Additive

To prove that T is additive, we must show that T(z1 + z2) = T(z1) + T(z2) for any two complex numbers z1 and z2. Let z1 = a1 + b1i and z2 = a2 + b2i be two arbitrary complex numbers in the complex plane, where a1, a2, b1, and b2 are real numbers. T(z1 + z2) = T((a1 + a2) + (b1 + b2)i) By definition, when z = x + yi (where x and y are real numbers), T(z) = (x - yi), the conjugate of z. We find the conjugates of the sum: T(z1 + z2) = (a1 + a2) - (b1 + b2)i Now, we separately find the conjugates of z1 and z2 and add them: T(z1) + T(z2) = (a1 - b1i) + (a2 - b2i) = (a1 + a2) - (b1 + b2)i Comparing the results of T(z1 + z2) and T(z1) + T(z2), we see that they are equal: T(z1 + z2) = T(z1) + T(z2) Thus, T is an additive function.
02

Prove T is not Linear (Show it's not Homogeneous)

To show that T is not linear, we must prove that it does not satisfy the homogeneity property: T(cz) 鈮� cT(z) for some complex number z and scalar c. Let z = x + yi, and let c be any non-zero scalar. First, let's find T(cz): T(cz) = T(c(x + yi)) = T(cx + cyi) Now, using the definition of T, we have: T(cz) = T(cx + cyi) = cx - cyi Next, we'll find cT(z): cT(z) = c(x - yi) = cx - cyi Comparing the results of T(cz) and cT(z), we see that they are equal: T(cz) = cT(z) From the results above, we see that T is homogeneous. However, T is not homogeneous when the scalar c is a purely imaginary number, i.e., c = bi, where b is a real number. Let's consider this case: T(cz) = T((bi)(x + yi)) = T(bxi - byi虏) = T(bxi + by) Now, using the definition of T, we have: T(cz) = T(bxi + by) = bx- yi - by Next, we'll find cT(z): cT(z) = (bi)(x - yi) = ix - i虏y = ix + y Comparing the results of T(cz) and cT(z), we see that they are not equal: T(cz) 鈮� cT(z) Thus, T is not homogeneous when the scalar c is a purely imaginary number. Since T satisfies the additivity property but is not homogeneous for some scalar c, we can conclude that T is additive but not linear.

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Key Concepts

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

Complex Numbers
Complex numbers are an extension of the real numbers and are fundamental in higher mathematics, including linear algebra. They come in the form of \(a + bi\), where \('a'\) and \('b'\) are real numbers, and \(i\) is the imaginary unit, satisfying the equation \(i^2 = -1\). The number \(a\) is called the 'real part', while \(b\) is the 'imaginary part'.

Complex numbers can be added, subtracted, multiplied, and divided, much like real numbers, but they also possess unique properties related to their two-dimensional nature. For example, the complex conjugate of a complex number \(z = a + bi\) is denoted as \(\bar{z} = a - bi\). As demonstrated in the original exercise, the function \(\mathrm{T}(z)\) maps a complex number to its conjugate, which preserves addition. Understanding complex numbers is vital for a range of in-depth mathematics and can model phenomena in physics and engineering.
Additive Functions
Additive functions are a class of functions that preserve addition across their inputs. In other words, if \(\mathrm{T}\) is an additive function, then for any two elements \(x\) and \(y\) in its domain, \(\mathrm{T}(x + y) = \mathrm{T}(x) + \mathrm{T}(y)\). This attribute is also known as the 'additivity property'.

In the context of the problem, \(\mathrm{T}\) is shown to be additive by observing that the conjugate of the sum of two complex numbers is equal to the sum of their conjugates. This property is key when dealing with linear transformations and is often one of the first steps in determining whether a function is linear. Ensuring that students understand additivity can help them tackle more complex properties of functions like linearity.
Homogeneity Property
The homogeneity property is one of the two conditions a function must satisfy to be considered linear; the other is the additivity property. A function \(\mathrm{T}\) is said to have the homogeneity property if for any scalar \(c\) and any element \(z\) in its domain, \(\mathrm{T}(cz) = c\mathrm{T}(z)\). Essentially, multiplying the input by a scalar \(c\) should have the same effect as multiplying the output by \(c\).

However, the function described in the exercise, \(\mathrm{T}(z) \to \bar{z}\), is not homogeneous when the scalar is purely imaginary, as the multiplication of a complex number by a purely imaginary scalar does not commute with the function \(\mathrm{T}\). This breaks one of the two crucial conditions for linearity, hence \(\mathrm{T}\) is not linear even though it is additive. Understanding this subtle difference is essential for students learning about linear transformations in linear algebra.

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Most popular questions from this chapter

Let $$ \mathrm{V}=\left\\{\left(\begin{array}{cc} a & a+b \\ 0 & c \end{array}\right): a, b, c \in F\right\\} . $$ Construct an isomorphism from \(V\) to \(F^{3}\).

Let \(T\) be the linear operator on \(R^{2}\) defined by $$ \mathrm{T}\left(\begin{array}{l} a \\ b \end{array}\right)=\left(\begin{array}{l} 2 a+b \\ a-3 b \end{array}\right) $$ let \(\beta\) be the standard ordered basis for \(\mathrm{R}^{2}\), and let $$ \beta^{\prime}=\left\\{\left(\begin{array}{l} 1 \\ 1 \end{array}\right),\left(\begin{array}{l} 1 \\ 2 \end{array}\right)\right\\} . $$ Use Theorem \(2.23\) and the fact that $$ \left(\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right)^{-1}=\left(\begin{array}{rr} 2 & -1 \\ -1 & 1 \end{array}\right) $$ to find \([\mathrm{T}]_{\beta^{\prime}}\).

Let \(\mathrm{V}\) be a finite-dimensional vector space over a field \(F\), and let \(\beta=\) \(\left\\{x_{1}, x_{2}, \ldots, x_{n}\right\\}\) be an ordered basis for \(\mathrm{V}\). Let \(Q\) be an \(n \times n\) invertible matrix with entries from \(F\). Define $$ x_{j}^{\prime}=\sum_{i=1}^{n} Q_{i j} x_{i} \quad \text { for } 1 \leq j \leq n, $$ and set \(\beta^{\prime}=\left\\{x_{1}^{\prime}, x_{2}^{\prime}, \ldots, x_{n}^{\prime}\right\\} .\) Prove that \(\beta^{\prime}\) is a basis for \(\mathrm{V}\) and hence that \(Q\) is the change of coordinate matrix changing \(\beta^{\prime}\)-coordinates into \(\beta\)-coordinates. Visit goo.gl/vsxsGH for a solution.

Let \(A\) and \(B\) be matrices for which the product matrix \(A B\) is defined, and let \(u_{j}\) and \(v_{j}\) denote the \(j\) th columns of \(A B\) and \(B\), respectively. If \(v_{p}=c_{1} v_{j_{1}}+c_{2} v_{j_{2}}+\cdots+c_{k} v_{j_{k}}\) for some scalars \(c_{1}, c_{2}, \ldots c_{k}\), prove that \(u_{p}=c_{1} u_{j_{1}}+c_{2} u_{j_{2}}+\cdots+c_{k} u_{j_{k}} .\) Visit goo.gl/sRpves for a solution.

Prove that the matrix $$ A=\left(\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right) $$ corresponds to a dominance relation. Use Exercise 22 to determine which persons dominate [are dominated by] each of the others within two stages.
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