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Chapter 7: Problem 37

Prove Theorem 7.12: Suppose \(E=\left\\{e_{i}\right\\}\) and \(E^{\prime}=\left\\{e_{i}^{\prime}\right\\}\) are orthonormal bases of \(V .\) Let \(P\) be the change-of-basis matrix from \(E\) to \(E^{\prime} .\) Then \(P\) is orthogonal. Suppose \\[e_{i}^{\prime}=b_{i 1} e_{1}+b_{i 2} e_{2}+\cdots+b_{i n} e_{n}, \quad i=1, \ldots, n\\] Using Problem \(7.18(\mathrm{b})\) and the fact that \(E^{\prime}\) is orthonormal, we get \\[\delta_{i j}=\left\langle e_{i}^{\prime}, e_{j}^{\prime}\right\rangle=b_{i 1} b_{j 1}+b_{i 2} b_{j 2}+\cdots+b_{i n} b_{j n}\\] Let \(B=\left[b_{i j}\right]\) be the matrix of the coefficients in (1). (Then \(P=B^{T}\) ) Suppose \(B B^{T}=\left[c_{i j}\right] .\) Then \\[c_{i j}=b_{i 1} b_{j 1}+b_{i 2} b_{j 2}+\cdots+b_{i n} b_{j n}\\] By (2) and (3), we have \(c_{i j}=\delta_{i j}\). Thus, \(B B^{T}=I\). Accordingly, \(B\) is orthogonal, and hence, \(P=B^{T}\) is orthogonal.

Short Answer

Expert verified
Since the product of matrix B and its transpose is the identity matrix, namely \(BB^T = I\), it follows that B is orthogonal. Therefore, the change-of-basis matrix P, which is the transpose of B, i.e., \(P = B^T\), is also orthogonal.

Step by step solution

01

Write the relation between E' and E using coefficients

Given the relationship between the orthonormal bases E and E', we can write it as: \(e_{i}^{\prime}=b_{i 1} e_{1}+b_{i 2} e_{2}+\cdots+b_{i n} e_{n}, \quad i=1,\ldots, n\) Here we are given n linear combinations, one for each of the n elements of E'.
02

Use Problem 7.18(b) and orthonormality

From Problem 7.18(b) and the fact that the bases E' are orthonormal, we obtain the following equation: \(\delta_{i j}=\left\langle e_{i}^{\prime}, e_{j}^{\prime}\right\rangle=b_{i 1} b_{j 1}+b_{i 2} b_{j 2}+\cdots+b_{i n} b_{j n}\) where \(\delta_{i j}\) is the Kronecker delta and is equal to 1 if i=j and 0 otherwise.
03

Define the matrix B and find its transpose

Define the matrix B with elements given by the coefficients from Step 1: \(B=\left[b_{i j}\right]\) Now, find the transpose of matrix B: \(P = B^T\) So, P is the change-of-basis matrix from E to E'.
04

Multiply matrix B and its transpose

Now, let's find the product of matrix B and its transpose, B^T: \(BB^T = \left[c_{i j}\right]\) So, we have: \(c_{i j}=b_{i 1} b_{j 1}+b_{i 2} b_{j 2}+\cdots+b_{i n} b_{j n}\)
05

Use the result from Step 2 and Step 4

From Step 2 and using the equation obtained in Step 4, we get: \(\delta_{i j}=c_{i j}\) This equation implies that the product BB^T is equal to the identity matrix I: \(BB^T = I\)
06

Conclusion

Since we found that the product of matrix B and its transpose is the identity matrix, we conclude that matrix B is orthogonal. And since P is the transpose of B, it follows that the change-of-basis matrix P is also orthogonal: \(P = B^T\) Hence, the theorem is proven.

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Key Concepts

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

Orthonormal Bases
An orthonormal basis is a set of vectors that are both mutually perpendicular (orthogonal) to each other and of unit length (normalized). When working with vector spaces, an orthonormal basis provides a simple and convenient framework for various calculations, including projections and transformations.

In the context of the exercise, the set \(E\) and \(E'\) are considered orthonormal bases of the vector space \(V\). This means every vector in the space can be represented uniquely as a linear combination of the basis vectors. Additionally, due to their orthonormal nature, the inner product \(\langle e_i, e_j \rangle\) is 0 if \(i eq j\) and 1 if \(i = j\), satisfying the conditions needed for an orthonormal set.
Kronecker Delta
The Kronecker delta function \(\delta_{ij}\) is a simple yet powerful tool used in mathematics to concisely express the orthogonality property of vectors in an orthonormal basis. It is defined as being equal to 1 if the indices \(i\) and \(j\) are the same and 0 otherwise.

In our problem, the Kronecker delta is used to describe the relationship between \(e'_i\) and \(e'_j\), the vectors of the new basis \(E'\). If we take the inner product of any two distinct basis vectors, according to the orthogonality property, we get zero. If the vectors are the same, the inner product is one. This is critical because this simple rule helps us quickly identify when two vectors are orthogonal and also underpins the concept of an orthogonal change-of-basis matrix, like the one described in the exercise (matrix \(P\)).
Transpose of a Matrix
The transpose of a matrix \(B\) denoted \(B^T\), is achieved by flipping the matrix over its diagonal. This means that the rows of \(B\) become the columns of \(B^T\) and vice versa. In many mathematical operations involving matrices, particularly those related to inner products and orthogonality, transposes play a critical role.

In the given exercise, the transpose of the matrix \(B\) is used to form the change-of-basis matrix \(P\), which converts coordinates from the original orthonormal basis \(E\) to the new basis \(E'\). The significance of \(P\) being the transpose is fundamental to proving that the matrix is orthogonal. As shown in the steps, this orthogonality of \(P\) implies \(P\) satisfies the condition \(P P^T = I\), where \(I\) is the identity matrix, which is precisely what is observed in the final step of the solution, confirming our original assertion.

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Most popular questions from this chapter

Consider the subspace \(U\) of \(\mathbf{R}^{4}\) spanned by the vectors: \\[v_{1}=(1,1,1,1), \quad v_{2}=(1,1,2,4), \quad v_{3}=(1,2,-4,-3)\\] Find (a) an orthogonal basis of \(U ;\) (b) an orthonormal basis of \(U\) (a) Use the Gram-Schmidt algorithm. Begin by setting \(w_{1}=u=(1,1,1,1)\). Next find \\[v_{2}-\frac{\left\langle v_{2}, w_{1}\right\rangle}{\left\langle w_{1}, w_{1}\right\rangle} w_{1}=(1,1,2,4)-\frac{8}{4}(1,1,1,1)=(-1,-1,0,2)\\] Set \(w_{2}=(-1,-1,0,2) .\) Then find \\[\begin{aligned} v_{3}-\frac{\left\langle v_{3}, w_{1}\right\rangle}{\left\langle w_{1}, w_{1}\right\rangle} w_{1}-\frac{\left\langle v_{3}, w_{2}\right\rangle}{\left\langle w_{2}, w_{2}\right\rangle} w_{2} &=(1,2,-4,-3)-\frac{(-4)}{4}(1,1,1,1)-\frac{(-9)}{6}(-1,-1,0,2) \\ &=\left(\frac{1}{2}, \frac{3}{2},-3,1\right) \end{aligned}\\] Clear fractions to obtain \(w_{3}=(1,3,-6,2) .\) Then \(w_{1}, w_{2}, w_{3}\) form an orthogonal basis of \(U\) (b) Normalize the orthogonal basis consisting of \(w_{1}, w_{2}, w_{3}\). Because \(\left\|w_{1}\right\|^{2}=4,\left\|w_{2}\right\|^{2}=6,\) and \(\left\|w_{3}\right\|^{2}=50,\) the following vectors form an orthonormal basis of \(U:\) \\[u_{1}=\frac{1}{2}(1,1,1,1), \quad u_{2}=\frac{1}{\sqrt{6}}(-1,-1,0,2), \quad u_{3}=\frac{1}{5 \sqrt{2}}(1,3,-6,2)\\]

Let \(\mathbf{M}=\mathbf{M}_{2,2}\) with inner product \(\langle A, B\rangle=\operatorname{tr}\left(B^{T} A\right) .\) Find an orthogonal basis for the orthogonal complement of (a) diagonal matrices, (b) symmetric matrices.

Let U be a linear operator on a finite-dimensional vector space V. Prove the following results. (a) $\mathrm{N}(\mathrm{U}) \subseteq \mathrm{N}\left(\mathrm{U}^{2}\right) \subseteq \cdots \subseteq \mathrm{N}\left(\mathrm{U}^{k}\right) \subseteq \mathrm{N}\left(\mathrm{U}^{k+1}\right) \subseteq \cdots .$ (b) If \(\operatorname{rank}\left(U^{m}\right)=\operatorname{rank}\left(U^{m+1}\right)\) for some positive integer \(m\), then \(\operatorname{rank}\left(U^{m}\right)=\operatorname{rank}\left(U^{k}\right)\) for any positive integer \(k \geq m\). (c) If \(\operatorname{rank}\left(U^{m}\right)=\operatorname{rank}\left(U^{m+1}\right)\) for some positive integer \(m\), then \(\mathrm{N}\left(\mathrm{U}^{m}\right)=\mathrm{N}\left(\mathrm{U}^{k}\right)\) for any positive integer \(k \geq m\). (d) Let \(T\) be a linear operator on \(V\), and let \(\lambda\) be an eigenvalue of \(T\). Prove that if $\operatorname{rank}\left((T-\lambda I)^{m}\right)=\operatorname{rank}\left((T-\lambda I)^{m+1}\right)$ for some integer \(m\), then $\mathrm{K}_{\lambda}=\mathrm{N}\left((\mathrm{T}-\lambda \mathrm{I})^{m}\right)$. (e) Second Test for Diagonalizability. Let \(\mathrm{T}\) be a linear operator on \(V\) whose characteristic polynomial splits, and let $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\( be the distinct eigenvalues of \)\mathrm{T}$. Then \(\mathrm{T}\) is diagonalizable if and only if $\operatorname{rank}\left(\mathrm{T}-\lambda_{i} \mathrm{I}\right)=\operatorname{rank}\left(\left(\mathrm{T}-\lambda_{i} \mathrm{I}\right)^{2}\right)\( for \)1 \leq i \leq k$. (f) Use (e) to obtain a simpler proof of Exercise 24 of Section 5.4: If \(\mathrm{T}\) is a diagonalizable linear operator on a finite-dimensional vector space \(\mathrm{V}\) and \(\mathrm{W}\) is a \(\mathrm{T}\)-invariant subspace of \(\mathrm{V}\), then \(\mathrm{T}_{\mathrm{W}}\) is diagonalizable.

Consider the vector space \(\mathbf{P}(t)\) with inner product \(\langle f, g\rangle=\int_{0}^{1} f(t) g(t) d t .\) Apply the GramSchmidt algorithm to the set \(\left\\{1, t, t^{2}\right\\}\) to obtain an orthogonal set \(\left\\{f_{0}, f_{1}, f_{2}\right\\}\) with integer coefficients. First set \(f_{0}=1 .\) Then find \\[t-\frac{\langle t, 1\rangle}{\langle 1,1\rangle} \cdot 1=t-\frac{\frac{1}{2}}{1} \cdot 1=t-\frac{1}{2}\\] Clear fractions to obtain \(f_{1}=2 t-1 .\) Then find \\[t^{2}-\frac{\left\langle t^{2}, 1\right\rangle}{\langle 1,1\rangle}(1)-\frac{\left\langle t^{2}, 2 t-1\right\rangle}{\langle 2 t-1,2 t-1\rangle}(2 t-1)=t^{2}-\frac{\frac{1}{3}}{1}(1)-\frac{\frac{1}{6}}{\frac{1}{3}}(2 t-1)=t^{2}-t+\frac{1}{6}\\] Clear fractions to obtain \(f_{2}=6 t^{2}-6 t+1 .\) Thus, \(\left\\{1,2 t-1,6 t^{2}-6 t+1\right\\}\) is the required orthogonal set.

Let \(U\) and \(W\) be subspaces of a finite-dimensional inner product space \(V\). Show that (a) \(\quad(U+W)^{\perp}=U^{\perp} \cap W^{\perp}\) (b) \(\quad(U \cap W)^{\perp}=U^{\perp}+W^{\perp}\)
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