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Chapter 5: Problem 11

Prove that the transpose of an orthogonal matrix is an orthogonal matrix.

Short Answer

Expert verified
We are given that A is an orthogonal matrix, i.e., A * A岬 = I. To prove A岬 is also orthogonal, we first find the transpose of A岬, denoted as (A岬)岬, which is equal to A. We need to show that A岬 * A = I. Since A is orthogonal, taking the transpose of A * A岬 = I, we obtain (A * A岬)岬 = I岬. Using the property (AB)岬 = B岬 * A岬, we rewrite it as (A岬)岬 * A岬 = I. Substituting (A岬)岬 = A, we have A * A岬 = I, proving that A岬 is orthogonal.

Step by step solution

01

Understanding Orthogonal Matrices

: A matrix A is orthogonal if its transpose A岬 is equal to its inverse A鈦宦. Mathematically, this can be written as: A * A岬 = A岬 * A = I where I is the identity matrix.
02

Finding the Transpose of the Transpose Matrix

: We are given that A is an orthogonal matrix, and we are asked to prove that A岬 is also orthogonal. To do this, let's first find the transpose of A岬, which can be denoted as (A岬)岬. Recall that when transposing a matrix, we essentially swap the rows and columns of the matrix. So, when finding the transpose of A岬, we are, in effect, undoing the transposition of A. Therefore, (A岬)岬 = A
03

Verifying the Orthogonality of A岬

: We now need to show that A岬 * A = I. If we can demonstrate that, then we have shown that A岬 is indeed orthogonal. Since A is orthogonal, we have: A * A岬 = I Taking the transpose of both sides of the equation, we obtain: (A * A岬)岬 = I岬 Since the transpose of the identity matrix is the identity matrix, we get: I岬 = I Now, using the property (AB)岬 = B岬 * A岬, we can rewrite the left side of our equation as: (A岬)岬 * A岬 = I But from Step 2, we know that (A岬)岬 = A, so substituting this in our equation, we get: A * A岬 = I Thus, we have proven that A岬 is, indeed, an orthogonal matrix.

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Most popular questions from this chapter

A (real or complex) scalar \(u\) is said to be an \(n\) th root of unity if \(u^{n}=1\) (a) Show that if \(u\) is an \(n\) th root of unity and \(u \neq 1\) then $$1+u++u^{2}+\cdots+u^{n-1}=0$$ \\[ \left[\text {Hint}: 1-u^{n}=(1-u)\left(1+u+u^{2}+\cdots+u^{n-1}\right)\right] \\] (b) Let \(\omega_{n}=e^{\frac{2 \pi i}{n}} .\) Use Euler's formula \(\left(e^{i \theta}=\right.\) \(\cos \theta+i \sin \theta)\) to show that \(\omega_{n}\) is an \(n\) th root of unity. (c) Show that if \(j\) and \(k\) are positive integers and if \(u=\omega_{n}^{j-1}\) and \(z=\omega_{n}^{-(k-1)},\) then \(u, z,\) and \(u z\) are all \(n\) th roots of unity.

Show that if \(U\) is an \(n \times n\) orthogonal matrix, then $$\mathbf{u}_{1} \mathbf{u}_{1}^{T}+\mathbf{u}_{2} \mathbf{u}_{2}^{T}+\cdots+\mathbf{u}_{n} \mathbf{u}_{n}^{T}=I$$

The functions \(\cos x\) and \(\sin x\) form an orthonormal \(\operatorname{set} \operatorname{in} C[-\pi, \pi] .\) If \(f(x)=3 \cos x+2 \sin x\) and \(g(x)=\cos x-\sin x\) use Corollary 5.5 .3 to determine the value of $$\langle f, g\rangle=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) g(x) d x$$

Let \(\mathbf{x}=(-1,-1,1,1)^{T}\) and \(\mathbf{y}=(1,1,5,-3)^{T}\) Show that \(\mathbf{x} \perp \mathbf{y}\). Calculate \(\|\mathbf{x}\|_{2},\|\mathbf{y}\|_{2},\|\mathbf{x}+\mathbf{y}\|_{2}\) and verify that the Pythagorean law holds.

Let \(P=A\left(A^{T} A\right)^{-1} A^{T},\) where \(A\) is an \(m \times n\) matrix of rank \(n\) (a) Show that \(P^{2}=P\) (b) Prove that \(P^{k}=P\) for \(k=1,2, \ldots\) (c) Show that \(P\) is symmetric. [Hint: If \(B\) is nonsingular, then \(\left.\left(B^{-1}\right)^{T}=\left(B^{T}\right)^{-1} .\right]\)
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