91Ó°ÊÓ

An eigenvector, x, of matrix A is a nonzero vector such that the product of A and x results in the vector scaled by some scalar λ. In other words, the following equality holds: \[Ax = λx\] The scalar λ is called the eigenvalue.

We need to show that the property holds for m = 1. Since \(A^1 = A\), we have: \[A^1x = Ax\] From the definition of eigenvector and eigenvalue, we know that \(Ax = λx\). So, the property holds for m = 1.

Assume that the property holds for some positive integer m, that is: \[A^mx = λ^mx\] We will show that the property also holds for m + 1.

We want to show that: \[A^{m+1}x = λ^{m+1}x\] Now, let's compute \(A^{m+1}x\): \[A^{m+1}x = A(A^mx)\] From our inductive hypothesis, we know that \(A^mx = λ^mx\), so we can substitute this into the equation: \[A^{m+1}x = A(λ^mx)\] Now, we apply the property of scalar multiplication: \[A^{m+1}x = λ^m(Ax)\] From the definition of eigenvector and eigenvalue, we know that \(Ax = λx\), so we can substitute this into the equation: \[A^{m+1}x = λ^m(λx)\] Finally, we combine the repeated scalar multiplication to get: \[A^{m+1}x = λ^{m+1}x\] Thus, the property holds for m + 1. Since we have shown that the property holds for m = 1 (base case) and for m + 1 (inductive step), by mathematical induction, the property holds for all positive integers m. Therefore, \(A^mx\) is an eigenvector belonging to λ for all \(m = 1, 2, \ldots\).