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Chapter 2: Problem 11

Let \(V\) be a finite-dimensional vector space with ordered bases $\alpha, \beta\(, and \)\gamma$. (a) Prove that if \(Q\) and \(R\) are the change of coordinate matrices that change \(\alpha\)-coordinates into \(\beta\)-coordinates and \(\beta\)-coordinates into \(\gamma\)-coordinates, respectively, then \(R Q\) is the change of coordinate matrix that changes \(\alpha\)-coordinates into \(\gamma\)-coordinates. (b) Prove that if \(Q\) changes \(\alpha\)-coordinates into \(\beta\)-coordinates, then \(Q^{-1}\) changes \(\beta\)-coordinates into \(\alpha\)-coordinates.

Short Answer

Expert verified
In short, for part (a), we show that the matrix product \(RQ\) changes \(\alpha\)-coordinates into \(\gamma\)-coordinates by expressing \(P_\gamma\) in terms of \(P_\alpha\) and \(Q\), and then multiplying the matrices \(R\) and \(Q\): \(P_\gamma = (RQ) P_\alpha\). For part (b), we prove that if \(Q\) changes \(\alpha\)-coordinates into \(\beta\)-coordinates, then \(Q^{-1}\) changes \(\beta\)-coordinates into \(\alpha\)-coordinates by multiplying both sides by \(Q^{-1}\) on the left and using the definition of inverse matrices: \(Q^{-1} P_\beta = P_\alpha\).

Step by step solution

01

Definitions and Notations

Let's denote by \(P_\alpha\), \(P_\beta\), and \(P_\gamma\) the coordinate representations of a vector \(v\in V\) with respect to the bases \(\alpha\), \(\beta\), and \(\gamma\), respectively. Then, we have the following relations: - When applying the change of coordinate matrix \(Q\) from \(\alpha\)-coordinates to \(\beta\)-coordinates, we get: \(P_\beta = Q P_\alpha\). - When applying the change of coordinate matrix \(R\) from \(\beta\)-coordinates to \(\gamma\)-coordinates, we get: \(P_\gamma = R P_\beta\). Now we will prove parts (a) and (b) of the problem.
02

Proving Part (a)

We want to prove that the matrix product \(RQ\) changes \(\alpha\)-coordinates into \(\gamma\)-coordinates. 1. Express \(P_\gamma\) in terms of \(P_\alpha\) and \(Q\), using the relationship between \(P_\beta\) and \(P_\alpha\): \(P_\beta = Q P_\alpha\). So, we get \(P_\gamma = R (Q P_\alpha)\). 2. Multiply the matrices \(R\) and \(Q\): \(P_\gamma = (RQ) P_\alpha\). From step 2, we can see that the product \(R Q\) is the change of coordinate matrix that changes \(\alpha\)-coordinates into \(\gamma\)-coordinates, as required, and part (a) is proven.
03

Proving Part (b)

We want to prove that if \(Q\) changes \(\alpha\)-coordinates into \(\beta\)-coordinates, then \(Q^{-1}\) changes \(\beta\)-coordinates into \(\alpha\)-coordinates. 1. Since \(Q\) changes \(\alpha\)-coordinates to \(\beta\)-coordinates, we know that \(P_\beta = Q P_\alpha\). 2. Multiply both sides of the equation by \(Q^{-1}\) on the left: \(Q^{-1} P_\beta = Q^{-1}(Q P_\alpha)\). 3. Apply the definition of inverse matrices to simplify the equation: \(Q^{-1} P_\beta = (Q^{-1}Q) P_\alpha\). 4. Since \(Q^{-1}Q\) is the identity matrix, we have \(Q^{-1} P_\beta = I P_\alpha\). 5. For any vector \(u\) and identity matrix \(I\), we have \(Iu = u\). Thus, \(Q^{-1} P_\beta = P_\alpha\). From step 5, we can see that the inverse matrix \(Q^{-1}\) changes \(\beta\)-coordinates into \(\alpha\)-coordinates, as required, and part (b) is proven.

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Most popular questions from this chapter

A differential equation $$ y^{(n)}+a_{n-1} y^{(n-1)}+\cdots+a_{1} y^{(1)}+a_{0} y=x $$ is called a nonhomogeneous linear differential equation with constant coefficients if the \(a_{i}\) 's are constant and \(x\) is a function that is not identically zero.(a) Prove that for any \(x \in{C}^{\infty}\) there exists $y \in{C}^{\infty}\( such that \)y$ is a solution to the differential equation. Hint: Use Lemma 1 to Theorem \(2.32\) to show that for any polynomial \(p(t)\), the linear operator $p(\mathrm{D}): \mathrm{C}^{\infty} \rightarrow \mathrm{C}^{\infty}$ is onto. (b) Let \(V\) be the solution space for the homogeneous linear equation $$ y^{(n)}+a_{n-1} y^{(n-1)}+\cdots+a_{1} y^{(1)}+a_{0} y=0 . $$ Prove that if \(z\) is any solution to the associated nonhomogeneous linear differential equation, then the set of all solutions to the nonhomogeneous linear differential equation is $$ \\{z+y: y \in \mathrm{V}\\} . $$

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.

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 \(A(B C) . \quad[\text { Note that }(A B) C=A(B C) .]\)

Find the inverse of each of the following matrices (if it exists): \\[ A=\left[\begin{array}{ll} 7 & 4 \\ 5 & 3 \end{array}\right], \quad B=\left[\begin{array}{ll} 2 & 3 \\ 4 & 5 \end{array}\right], \quad C=\left[\begin{array}{rr} 4 & -6 \\ -2 & 3 \end{array}\right], \quad \quad D=\left[\begin{array}{ll} 5 & -2 \\ 6 & -3 \end{array}\right] \\]

Determine which of the following matrices are unitary: \(A=\left[\begin{array}{rr}i / 2 & -\sqrt{3} / 2 \\ \sqrt{3} / 2 & -i / 2\end{array}\right], \quad B=\frac{1}{2}\left[\begin{array}{cc}1+i & 1-i \\\ 1-i & 1+i\end{array}\right], \quad C=\frac{1}{2}\left[\begin{array}{ccc}1 & -i & -1+i \\ i & 1 & 1+i \\ 1+i & -1+i & 0\end{array}\right]\)
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