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Chapter 6: Problem 4

Let \(A\) be a nonsingular matrix and let \(\lambda\) be an eigenvalue of \(A .\) Show that \(1 / \lambda\) is an eigenvalue of \(A^{-1}\)

Short Answer

Expert verified
In summary, we showed that if \(Av = \lambda v\), then \((1/\lambda) v = A^{-1}v\), demonstrating that if \(\lambda\) is an eigenvalue of a nonsingular matrix \(A\), then \(1/\lambda\) is an eigenvalue of \(A^{-1}\).

Step by step solution

01

Definition of an eigenvalue and eigenvector

Recall the definition of an eigenvalue and a corresponding eigenvector. Given a square matrix A and a scalar 位, 位 is an eigenvalue of A if there exists a non-zero vector v such that Av = 位v.
02

Write the given equation in terms of A^(-1)

We are given that A is nonsingular and 位 is an eigenvalue of A. This means there exists a nonzero vector v such that: Av = 位v Now, we want to write this equation in terms of A^(-1). To do so, multiply both sides of the equation by A^(-1): A^(-1)Av = A^(-1)位v
03

Use the property that A^(-1)A = I

We know that the product of the inverse of a matrix and the matrix itself is the identity matrix I (A^(-1)A = I). So, we can simplify the left side of the equation: Iv = A^(-1)位v Since Iv = v, we now have: v = A^(-1)位v
04

Divide both sides by 位

Now, we want to show that A^(-1)v = (1/位)v. To do so, divide both sides of the equation by 位 (since 位 is nonzero, we can safely divide by it): (1/位)v = 位^(-1)A^(-1)位v
05

Demonstrate that 1/位 is an eigenvalue of A^(-1)

By rearranging the terms, we now have: A^(-1)v = (1/位)v The above equation shows that if Av=位v, then A^(-1)v=(1/位)v. This demonstrates that if 位 is an eigenvalue of A, then 1/位 is an eigenvalue of A^(-1), as required.

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Most popular questions from this chapter

Let \(A\) be a diagonalizable matrix whose eigenvalues are all either 1 or \(-1 .\) Show that \(A^{-1}=A\)

Let \(A\) be a symmetric positive definite matrix and let \(Q\) be an orthogonal diagonalizing matrix. Use the factorization \(A=Q D Q^{T}\) to find a nonsingular matrix \(B\) such that \(B^{T} B=A\)

Show that if \(A\) is symmetric positive definite, then \(\operatorname{det}(A)>0 .\) Give an example of a \(2 \times 2\) matrix with positive determinant that is not positive definite.

For each of the following matrices, compute the determinants of all the leading principal submatrices and use them to determine whether the matrix is positive definite: (a) \(\left(\begin{array}{rr}2 & -1 \\ -1 & 2\end{array}\right)\) (b) \(\left(\begin{array}{ll}3 & 4 \\ 4 & 2\end{array}\right)\) (c) \(\left(\begin{array}{rrr}6 & 4 & -2 \\ 4 & 5 & 3 \\ -2 & 3 & 6\end{array}\right)\) (d) \(\left(\begin{array}{rrr}4 & 2 & 1 \\ 2 & 3 & -2 \\ 1 & -2 & 5\end{array}\right)\)

Show that if a matrix \(U\) is both unitary and Hermitian, then any eigenvalue of \(U\) must equal either 1 or -1
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