Induction proves a statement by using the previous case to imply the next one. Consider the following

proof to demonstrate. We will show that for all positive integers n,

1 + 2+ + n =

(n(n + 1))/ 2

First, we show it holds for n = 1. For n = 1, the above statement becomes (1+1)/2 = 1, which is true. Now, we

will show that the previous result, for n, implies the next one, for n + 1. To do this, we assume the result

holds for n. That is, we assume

1 + 2+ + n = (n(n + 1))/ 2

This is called the Induction Hypothesis (IH). Next, we prove the next case for (n + 1) as follows:

1 + 2+ + n + (n + 1) = (1 + 2 + + n) + n + 1 = ((n(n + 1))/ 2 )+ n + 1 = ((n^2 + n)/2) + ((2n + 2)/2) = ((n^2 + 3n + 2)/2) = (((n + 1)(n + 2))/2)

Notice that we used the induction hypothesis in the first line. The rest is just algebra. Also, we recognize

1 + 2 + + n + (n + 1) = ((n+1)(n+2))/2 as the statement with n replaced by n + 1. So, because it holds for

n = 1, and each result implies the next one, it is true for all positive integers n. This is how induction works.

(Pi is the mathematical symbol)

3. Suppose that {v1, . . . , vn} is a basis of Rn, which are eigenvectors of both A and B. That is, say:

(i) (lambda 1, v1), . . . (lambda n, vn) are eigenpairs of A

(ii) (rho 1, v1), . . . (rho n, vn) are eigenpairs of B.

Show that AB = BA.

This is a linear algebra question, please solve it and show all the necessary steps to solve it and show all the assumptions and please solve it in a very clear handwriting or by typing it using a computer. Thanks a lot.