1. Give an example, with brief justification, of each of the...

1. Give an example, with brief justification, of each of the following. (a) A linear operator $L: \mathbb{R}^2 \to \mathbb{R}^2$ that has exactly one fixed point. (b) An isometry $f: \mathbb{R}^2 \to \mathbb{R}^2$ that is not a linear operator on $\mathbb{R}^2$. (c) A linear operator $L: \mathbb{R}^2 \to \mathbb{R}^2$ that is not an isometry.

Transcript

00:01 Hello everyone, in this problem we are given with the operator t which is a mapping from h to h which is an isometry if this isometry if norm of tf is equal to norm of f for all f belongs to h.

00:23 So, with this property we are given the first part we need to prove that if t is an isometry then we need to prove that tf tg is equal to fg for all fg belongs to h.

00:37 So, given here t is an isometry so we have that norm of tx will be equal to norm of x.

00:49 So, now tx tx can be written as x x.

00:55 So, now t asterix tx x can be written as i x x as t asterix tx value is i.

01:10 So, now applying this for tf tg so we have now taking t asterix to tf so it will be t asterix of tf tg.

01:23 So, we can write this value to be i of fg.

01:28 So, multiplying this we have the value of fg.

01:31 So, with this we have first prove the first part of tf tg is equal to fg.

01:36 So, now let us move on to the second part of the problem where we need to prove that if t is an isometric then t asterix t is equal to i.

01:46 We need to prove this.

01:47 So, we are given that t is isometry so we can write t to be norm of tf the whole square will be equal to norm of f the whole square.

02:01 So, it will be tf tf can be written as f f.

02:07 So, now taking t asterix to first tf so we get t asterix of tf tf it will be i f f.

02:19 So, with this we can say that the value of t asterix t is to be i.

02:24 So, this is one required proof here.

02:28 Now let us move on to the next part of the problem.

02:31 The next part if t is surjective and isometry then we need to prove that t t asterix to be equal to i.

02:40 So, here we know that t asterix t to be equal to i and t is surjective since t is clearly a bijection.

03:03 So, with this we can say that the value of t inverse exists.

03:08 So, therefore t asterix t multiplied by t inverse we get this value as i i multiplied by t inverse.

03:19 So, this is nothing but t inverse.

03:21 So, now multiplying this t asterix t with t inverse so t t gets cancelled.

03:31 So, now we get t asterix to be equal to t inverse.

03:34 Now multiplying t on both sides we get t t asterix to be equal to t multiplied by t inverse we will be getting it as inverse.

03:45 So, i so which implies that t is unitary.

03:52 So, this is the proof of this part...

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