Mathematical physics - writing proofs

[Hypercube]

Hi there!

So for about a month, I've been self-studying mathematical physics using Mathematical Physics by Hassani. It's a big change for me, after only Stewart calculus, Boyce DE and some linear algebra course, but it's been loads of fun! Now, writing proofs is something fairly new to me, and I think it would be good if someone more experienced goes through and confirms whether what I write is correct, or at the very least good enough.

1. Homework Statement

Homework Equations

L(ℂ, ℂ) refers to the set of endomorphisms on ℂ.

The Attempt at a Solution

Step 1. Analyse and reiterate the question in your own words.

Prove that all endomorphisms on ℂ produce constant-multiple of the input vector.Step 2. Attempt the proof.

Let T be an endomorphism on ℂ such that T(a) = b, and assume that b ≠ αa. Since the range of T is a subspace of ℂ, b must also be in ℂ. This leads to contradiction because ℂ now has a number of linearly independent vectors that exceeds its dimension (1). Hence, b=αa.===================Is this proof any good? I have spent quite a bit of time on it, and it is the best I could come up with.

Thanks

 

[fresh_42]

Hi there!

So for about a month, I've been self-studying mathematical physics using Mathematical Physics by Hassani. It's a big change for me, after only Stewart calculus, Boyce DE and some linear algebra course, but it's been loads of fun! Now, writing proofs is something fairly new to me, and I think it would be good if someone more experienced goes through and confirms whether what I write is correct, or at the very least good enough.

1. Homework Statement

View attachment 110742

Homework Equations

L(ℂ, ℂ) refers to the set of endomorphisms on ℂ.

The Attempt at a Solution

Step 1. Analyse and reiterate the question in your own words.

Prove that all endomorphisms on ℂ produce constant-multiple of the input vector.Step 2. Attempt the proof.

Let T be an endomorphism on ℂ such that T(a) = b, and assume that b ≠ αa. Since the range of T is a subspace of ℂ, b must also be in ℂ. This leads to contradiction because ℂ now has a number of linearly independent vectors that exceeds its dimension (1). Hence, b=αa.===================Is this proof any good? I have spent quite a bit of time on it, and it is the best I could come up with.

Thanks

Looks good. I only would spent some thoughts on the cases: Why is ##\alpha_a## the same for all ##a##? You only proved it for a single one. They could all be different. And it could be, that ##b=0##, then linear dependence is automatically true and the dimension argument breaks down. Could there as well exist ##b\neq 0## (for a different ##a##)?

 

[Hypercube]

This never even occurred to me. Indeed, what if a belongs to ker(T)? I will spend some more time on this and see if I can modify the answer to encompass those cases as well. Your input is very much appreciated, thank you! Now I know I'm on the right track.