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Chapter 8: Problem 15

Let \(D\) be a 2 -linear, alternating function. Show that \(D(A, B)=-D(B, A)\)

Short Answer

Expert verified
Using the 2-linearity property of the given alternating function \(D\), we have \(D(A + B, A + B) = D(A, B) + D(B, A)\). Since \(D\) is alternating, we also know that \(D(A + B, A + B) = -D(A + B, A + B)\). Comparing these equations and applying that \(D\) is alternating, we deduce that \(D(A, B) = -D(B, A)\).

Step by step solution

01

Understand the problem

We're given a 2-linear, alternating function D and we need to show that D(A, B) = -D(B, A). Let's recall the definitions of a 2-linear function and an alternating function. A function D is 2-linear if it is linear with respect to each of its arguments, meaning that for any scalars a, b and any matrices P, Q, R: 1. D(aP + bQ, R) = aD(P, R) + bD(Q, R) 2. D(P, aQ + bR) = aD(P, Q) + bD(P, R) A function D is alternating if for any matrices P, Q: 1. D(P, Q) = -D(Q, P) 2. D(P, P) = 0 (since D(P, P) = -D(P, P)) Now let's use these properties to show that D satisfies the alternating property.
02

Use 2-linearity and apply it to swap the arguments

To show that D(A, B) = -D(B, A), we need to swap the order of the arguments A and B. We can use the 2-linearity property of D to alter the arguments and eventually show that swapping the arguments results in the change of sign. Using the 2-linearity property, we can write: 1. D(A + B, A + B) = D(A, A) + D(A, B) + D(B, A) + D(B, B) As D is an alternating function, we know that D(A, A) = D(B, B) = 0. Thus, 1. D(A + B, A + B) = D(A, B) + D(B, A)
03

Show that D(A, B) = -D(B, A)

Now, as D is an alternating function, we also have: 2. D(A + B, A + B) = -D(A + B, A + B) Now we compare equation 1 and equation 2: D(A + B, A + B) = D(A, B) + D(B, A) -D(A + B, A + B) = D(A, B) + D(B, A) Since the left sides are equal, the right sides must also be equal: D(A, B) + D(B, A) = - D(A + B, A + B) But we also know that D(A + B, A + B) = D(A, B) + D(B, A) from equation 1. Plugging this into the equation above, we get: D(A, B) + D(B, A) = - (D(A, B) + D(B, A)) This shows that D(A, B) = -D(B, A), which is the required proof that D is an alternating function.

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Most popular questions from this chapter

Consider a permutation \(\sigma=j_{1} j_{2} \ldots j_{n} .\) Let \(\left\\{e_{i}\right\\}\) be the usual basis of \(K^{n},\) and let \(A\) be the matrix whose \(i\) th row is \(e_{j_{i}}\left[\text { i.c., } A=\left(e_{j_{1}}, e_{j_{2}}, \ldots, e_{j_{n}}\right)\right] .\) Show that \(|A|=\operatorname{sgn} \sigma\)

Let \(\mathbf{D}: V \rightarrow V\) be the differential operator; that is, \(\mathbf{D}(f(t))=d f / d t .\) Find \(\operatorname{det}(\mathbf{D})\) if \(V\) is the vector space of (a) \(\left\\{1, t, \ldots, t^{5}\right\\}\) functions with the following bases: (b) \(\left\\{e^{t}, e^{2 t}, e^{3 t}\right\\}, \quad(\mathrm{c})\\{\sin t, \cos t\\}\)

Let \(\tau \in S_{n} .\) Show that \(\tau \circ \sigma\) runs through \(S_{n}\) as \(\sigma\) runs through \(S_{n},\) that is, \(S_{n}=\left\\{\tau \circ \sigma: \sigma \in S_{n}\right\\}\)

Let \(\sigma=24513\) and \(\tau=41352\) be permutations in \(S_{5} .\) Find \((\mathrm{a}) \quad \tau \circ \sigma,(\mathrm{b}) \quad \sigma^{-1}\)

Let \(A=\left[\begin{array}{rrrr}1 & 2 & 3 & 2 \\ 1 & 0 & -2 & 3 \\ 3 & -1 & 2 & 5 \\ 4 & -3 & 0 & -1\end{array}\right]\) and \(B=\left[\begin{array}{rrrr}1 & 3 & -1 & 5 \\ 2 & -3 & 1 & 4 \\ 0 & -5 & 2 & 1 \\ 3 & 0 & 5 & -2\end{array}\right] .\) Find the minor and the signed minor corresponding to the following submatrices: \(\left.\begin{array}{lll}\text { (a) } & A(1,4 ; & 3,4\end{array}\right)\) (b) \(\quad B(1,4 ; \quad 3,4)\) \(\begin{array}{llll}\text { (c) } & A(2,3 ; & 2,4),(\mathrm{d}) & B(2,3 ; & 2,4)\end{array}\)
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