/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 Let \(f(x)\) and \(g(x)\) be pol... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 37

Let \(f(x)\) and \(g(x)\) be polynomials and let \(A\) be a square matrix. Prove (a) \((f+g)(A)=f(A)+g(A)\) (b) \((f \cdot g)(A)=f(A) g(A)\) (c) \(f(A) g(A)=g(A) f(A)\)

Short Answer

Expert verified
By defining the given polynomials f(x) and g(x) and investigating their properties when acting on a square matrix A, we proved the following: (a) \((f+g)(A) = f(A) + g(A)\) (b) \((f \cdot g)(A)=f(A) g(A)\) (c) \(f(A) g(A) = g(A) f(A)\)

Step by step solution

01

Define polynomial functions

Let f(x) and g(x) be polynomials of degree n and m, respectively: \(f(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n\) \(g(x) = b_0 + b_1x + b_2x^2 + ... + b_mx^m\)
02

Prove (f+g)(A) = f(A) + g(A)

Objective: Prove that applying the sum of the polynomial functions to matrix A is equal to the sum of applying polynomial functions to A separately. Let h(x) = f(x) + g(x), then: \(h(x) = (a_0 + b_0) + (a_1 + b_1) x + (a_2 + b_2) x^2 + ... \) Now, let's find h(A), f(A), and g(A): \(h(A) = (a_0 + b_0)I + (a_1 + b_1)A + (a_2 + b_2)A^2 + ... \) \(f(A) = a_0I + a_1A + a_2A^2 + ... + a_nA^n \) \(g(A) = b_0I + b_1A + b_2A^2 + ... + b_mA^m \) Now, sum f(A) and g(A): \(f(A) + g(A) = (a_0I + b_0I) + (a_1A + b_1A) + (a_2A^2 + b_2A^2) + ... \) Comparing the results of h(A) and f(A) + g(A), we can see that: \((f+g)(A) = f(A) + g(A)\)
03

Prove (f*g)(A) = f(A) * g(A)

Objective: Prove that applying the product of polynomial functions to matrix A is equal to the product of applying the polynomial functions to A separately. Let p(x) = f(x) * g(x), then: \(p(x) = (a_0b_0) + (a_0b_1x + a_1b_0x) + ...\) Now, let's find f(A) * g(A) and p(A): \(f(A) * g(A) = (a_0I + a_1A + a_2A^2 + ...)(b_0I + b_1A + b_2A^2 + ...)\) By multiplying the matrices, we can observe that the resulting expression matches p(A). Thus, we have: \((f \cdot g)(A) = f(A) g(A)\)
04

Prove f(A) * g(A) = g(A) * f(A)

Objective: Prove that the order of matrix multiplication does not affect the resulting matrix for these polynomial functions. Using the fact that matrix exponentiation commutes with itself, i.e., \(A^nA^m = A^mA^n,\) we can prove that f(A) * g(A) = g(A) * f(A): \(f(A)g(A)=(a_0I + a_1A + a_2A^2 + ...)(b_0I + b_1A + b_2A^2 + ...)\) \(g(A)f(A)=(b_0I + b_1A + b_2A^2 + ...)(a_0I + a_1A + a_2A^2 + ...)\) Since the operations are associative and the order of matrix multiplication (for these polynomial functions) does not affect the resulting matrix, we can see that f(A) * g(A) = g(A) * f(A).

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Most popular questions from this chapter

Prove the converse of Exercise 8: If \(A\) and \(B\) are each \(m \times n\) matrices with entries from a field \(F\), and if there exist invertible $m \times m\( and \)n \times n\( matrices \)P\( and \)Q$, respectively, such that \(B=P^{-1} A Q\), then there exist an \(n\)-dimensional vector space \(\mathrm{V}\) and an \(m\)-dimensional vector space \(\mathrm{W}\) (both over \(F\) ), ordered bases \(\beta\) and \(\beta^{\prime}\) for \(\mathbf{V}\) and \(\gamma\) and \(\gamma^{\prime}\) for \(\mathbf{W}\), and a linear transformation $\mathrm{T}: \mathrm{V} \rightarrow \mathrm{W}$ such that $$ A=[\mathrm{T}]_{\beta}^{\gamma} \text { and } B=[\mathrm{T}]_{\beta^{\prime}}^{\gamma^{\prime}} . $$ Hints: Let $\mathrm{V}=\mathrm{F}^{n}, \mathrm{~W}=\mathrm{F}^{m}, \mathrm{~T}=\mathrm{L}_{A}\(, and \)\beta\( and \)\gamma$ be the standard ordered bases for \(\mathrm{F}^{n}\) and \(\mathrm{F}^{m}\), respectively. Now apply the results of Exercise 13 to obtain ordered bases \(\beta^{\prime}\) and \(\gamma^{\prime}\) from \(\beta\) and \(\gamma\) via \(Q\) and \(P\), respectively.

Prove that "is similar to" is an equivalence relation on $\mathrm{M}_{n \times n}(F)$.

Determine which of the following matrices are normal: \(A=\left[\begin{array}{cc}3+4 i & 1 \\ i & 2+3 i\end{array}\right]\) and \(B=\left[\begin{array}{cc}1 & 0 \\ 1-i & i\end{array}\right]\)

Prove Theorem 2.3: \(\quad(i)(A+B)^{T}=A^{T}+B^{T}\)Prove Theorem \(2.3: \quad\) (i) \((A+B)^{T}=A^{T}+B^{T}\), (ii) \(\left(A^{T}\right)^{T}=A\) (iii) \((k A)^{T}=k A^{T}.\)

Refer to the following matrices: $$A=\left[\begin{array}{rr} 1 & 2 \\ 3 & -4 \end{array}\right], \quad B=\left[\begin{array}{rr} 5 & 0 \\ -6 & 7 \end{array}\right], \quad C=\left[\begin{array}{rrr} 1 & -3 & 4 \\ 2 & 6 & -5 \end{array}\right], \quad D=\left[\begin{array}{rrr} 3 & 7 & -1 \\ 4 & -8 & 9 \end{array}\right]$$ Find (a) \(A B\) and \((A B) C\) (b) \(B C\) and \(A(B C) . \quad[\text { Note that }(A B) C=A(B C) .]\)
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