/*! 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 16 Let \(B\) be an \(n \times n\) i... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 16

Let \(B\) be an \(n \times n\) invertible matrix. Define $\Phi: M_{n \times n}(F) \rightarrow M_{n \times n}(F)\( by \)\Phi(A)=B^{-1} A B\(. Prove that \)\Phi$ is an isomorphism.

Short Answer

Expert verified
To prove that the given linear transformation \(\Phi(A) = B^{-1} A B\) is an isomorphism, we need to show that it is both injective and surjective. It is straightforward to verify that \(\Phi\) is linear. To prove injectivity, if we have \(\Phi(A_1) = \Phi(A_2)\), then it implies that \(A_1 = A_2\). To prove surjectivity, for any matrix \(C\), we can find a matrix \(A\) such that \(\Phi(A) = C\). Therefore, since \(\Phi\) is linear, injective, and surjective, it is indeed an isomorphism.

Step by step solution

01

Prove that \(\Phi\) is linear

To show that \(\Phi\) is linear, we need to verify that for any matrices \(A_1, A_2 \in M_{n \times n}(F)\) and any scalar \(k \in F\), the following two properties hold: 1. \(\Phi(A_1 + A_2) = \Phi(A_1) + \Phi(A_2)\) 2. \(\Phi(kA_1) = k\Phi(A_1)\) Let's check both properties: 1. \(\Phi(A_1 + A_2) = B^{-1}(A_1 + A_2)B = B^{-1}A_1B + B^{-1}A_2B = \Phi(A_1) + \Phi(A_2)\) 2. \(\Phi(kA_1) = B^{-1}(kA_1)B = kB^{-1}A_1B = k\Phi(A_1)\) Since both properties hold, \(\Phi\) is linear.
02

Prove that \(\Phi\) is injective

To show that \(\Phi\) is injective, we need to show that for any matrices \(A_1, A_2 \in M_{n \times n}(F)\), if \(\Phi(A_1) = \Phi(A_2)\), then \(A_1 = A_2\). Suppose \(\Phi(A_1) = \Phi(A_2)\). Then, we have \(B^{-1}A_1B = B^{-1}A_2B\). Multiply both sides by \(B\) on the left and \(B^{-1}\) on the right: \(A_1B = A_2B\) Since \(B^{-1}\) exists, we can multiply both sides by \(B^{-1}\) on the right: \(A_1 = A_2\) Thus, if \(\Phi(A_1) = \Phi(A_2)\), then \(A_1 = A_2\), which proves that \(\Phi\) is injective.
03

Prove that \(\Phi\) is surjective

To show that \(\Phi\) is surjective, we need to prove that for any matrix \(C \in M_{n \times n}(F)\), there exists a matrix \(A \in M_{n \times n}(F)\) such that \(\Phi(A) = C\). Let \(C \in M_{n \times n}(F)\). Define the matrix \(A = BCB^{-1}\). Then, we have: \(\Phi(A) = B^{-1}(BCB^{-1})B = (B^{-1}B)C(B^{-1}B) = C\) Thus, for any matrix \(C\), there exists a matrix \(A\) such that \(\Phi(A) = C\), which proves that \(\Phi\) is surjective.
04

Conclusion

Since \(\Phi\) is a linear transformation that is both injective and surjective, \(\Phi\) is an isomorphism.

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Most popular questions from this chapter

Repeat Example 7 with the polynomial \(p(x)=1+x+2 x^{2}+x^{3}\).

Suppose \(A\) and \(B\) are orthogonal matrices. Show that \(A^{T}, A^{-1}, A B\) are also orthogonal.

Let \(B=\left[\begin{array}{lll}1 & 8 & 5 \\ 0 & 9 & 5 \\ 0 & 0 & 4\end{array}\right] .\) Find a triangular matrix \(A\) with positive diagonal entries such that \(A^{2}=B\)

Let \(V\) be a finite-dimensional vector space with the ordered basis \(\beta\). Prove that \(\psi(\beta)=\beta^{* *}\), where \(\psi\) is defined in Theorem \(2.26\).

Let \(V\) be the solution space of an \(n\) th-order homogeneous linear differential equation with constant coefficients. Fix \(t_{0} \in R\), and define a mapping \(\Phi: V \rightarrow \mathrm{C}^{n}\) by $$ \Phi(x)=\left(\begin{array}{c} x\left(t_{0}\right) \\ x^{\prime}\left(t_{0}\right) \\ \vdots \\ x^{(n-1)}\left(t_{0}\right) \end{array}\right) \quad \text { for each } x \in \mathrm{V} . $$ (a) Prove that \(\Phi\) is linear and its null space is the zero subspace of \(\mathrm{V}\). Deduce that \(\Phi\) is an isomorphism. Hint: Use Exercise 14 . (b) Prove the following: For any \(n\) th-order homogeneous linear differential equation with constant coefficients, any \(t_{0} \in R\), and any complex numbers \(c_{0}, c_{1}, \ldots, c_{n-1}\) (not necessarily distinct), there exists exactly one solution, \(x\), to the given differential equation such that \(x\left(t_{0}\right)=c_{0}\) and \(x^{(k)}\left(t_{0}\right)=c_{k}\) for $k=1,2, \ldots, n-1$.
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