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Chapter 6: Problem 22

Let \(Q\) be an orthogonal matrix with an eigenvalue \(\lambda_{1}=1\) and let \(\mathbf{x}\) be an eigenvector belonging to \(\lambda_{1}\) Show that \(\mathbf{x}\) is also an eigenvector of \(Q^{T}\)

Short Answer

Expert verified
We can show that the eigenvector \(\mathbf{x}\) of the orthogonal matrix \(Q\) with eigenvalue \(\lambda_{1} = 1\) is also an eigenvector of \(Q^T\). Since \(Q\) is orthogonal, we know that \(Q^T = Q^{-1}\). We have: \(Q \mathbf{x} = \mathbf{x}\) By multiplying \(Q^T\) on both sides, we get: \(Q^T (Q \mathbf{x}) = Q^Tx\) Substituting \(Q^T\) with \(Q^{-1}\), we have: \(Q^{-1} (Q \mathbf{x}) = Q^T \mathbf{x}\) Since \(Q\) is orthogonal, \(Q^{-1}Q = I\) (the identity matrix): \(I \mathbf{x} = Q^T \mathbf{x}\) That implies \(\mathbf{x} = Q^T \mathbf{x}\). Therefore, \(\mathbf{x}\) is an eigenvector of \(Q^T\) with the same eigenvalue 1.

Step by step solution

01

Write the definition of an eigenvector and eigenvalue

Let's begin by recalling the definition of an eigenvalue and eigenvector. For a given matrix Q and a non-zero vector x, 位 is an eigenvalue and x is its associated eigenvector if: Q x = 位 x In our case, 位鈧� = 1, so: Q x = x
02

Show that Q^T x is equal to x using the fact that Q is orthogonal

As mentioned in the analysis, since Q is orthogonal, we know that Q^T = Q鈦宦�. Now, let's multiply Q^T on both sides of the expression Q x = x: Q^T (Q x) = Q^T x Since Q^T = Q鈦宦�, we can substitute Q^T with Q鈦宦�: Q鈦宦� (Q x) = Q^T x Now, since Q is orthogonal, we know that Q鈦宦� Q = I (the identity matrix): I x = Q^T x The identity matrix multiplied by any vector is the vector itself: x = Q^T x
03

Verify that x is also an eigenvector of Q^T with eigenvalue 1

We have shown in step 2 that Q^T x = x. Since Q^T x = 1 * x, we can now conclude that x is an eigenvector of the transpose Q^T with eigenvalue 位 = 1. Hence, we have proven that if x is an eigenvector of an orthogonal matrix Q with eigenvalue 位鈧� = 1, then x is also an eigenvector of its transpose Q^T with the same eigenvalue 1.

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Most popular questions from this chapter

Let \(A\) be a symmetric positive definite \(n \times n\) matrix and let \(S\) be a nonsingular \(n \times n\) matrix. Show that \(S^{T} A S\) is positive definite

Transform the \(n\) th-order equation \\[ y^{(n)}=a_{0} y+a_{1} y^{\prime}+\cdots+a_{n-1} y^{(n-1)} \\] into a system of first-order equations by setting \(y_{1}=y\) and \(y_{j}=y_{j-1}^{\prime}\) for \(j=2, \ldots, n .\) Determine the characteristic polynomial of the coefficient matrix of this system.

Let \(A\) be an \(m \times n\) matrix of rank \(n\) with singular value decomposition \(U \Sigma V^{T}\). Let \(\Sigma^{+}\) denote the \(n \times m\) matrix $$\left(\begin{array}{ccccc} \frac{1}{\sigma_{1}} & & & \\ & \frac{1}{\sigma_{2}} & & \\ & & \ddots & \\ & & & \frac{1}{\sigma_{n}} \end{array}\right)$$ and define \(A^{+}=V \Sigma^{+} U^{T} .\) Show that \(\hat{\mathbf{x}}=A^{+} \mathbf{b}\) satisfies the normal equations \(A^{T} A \mathbf{x}=A^{T} \mathbf{b}\)

It follows from Exercise 14 that, for a diagonalizable matrix, the number of nonzero eigenvalues (counted according to multiplicity) equals the rank of the matrix. Give an example of a defective matrix whose rank is not equal to the number of nonzero eigenvalues.

Let \(A\) be a Hermitian matrix and let \(\mathbf{x}\) be a vector in \(\mathbb{C}^{n} .\) Show that if \(c=\mathbf{x} A \mathbf{x}^{H},\) then \(c\) is real.
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