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

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)

1. Suppose Pi : Rn --> Rn is a linear map that satisfies Pi^2 = Pi, or equivalently,

Pi(Pi(x)) = Pi(x) for all x in Rn. (x here is a vector)

Note: You may NOT assume Pi is invertible. In fact, any projection map Pi satisfies Pi^2 = Pi, and P=I is

the only invertible one. This problem will show that Pi^2 = Pi implies Pi is a projection map.

(a) Show that ker(Pi) and range(Pi) are complements in Rn.

(b) Show that Pi is the projection onto range(Pi) with respect to ker(Pi).

(c) Show that if lambda is eigenvalue of P, then lambda = 0 or 1.

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.