SuicideFuel Math thread problem (official)

[Grim_Reaper]

Let V be a vector space with dim(V) = n < ∞. Suppose that L: V -> V is a linear map and that 0 is an eigenvalue of L.
Prove that L cannot be surjective.

If 0 is an eigenvalue, then there exists a non-zero vector in the kernel, which means it's nontrivial so dim(L(V)) < n. Therefore, L is not subjective.

 

[CountBleck]

If 0 is an eigenvalue, then there exists a non-zero vector in the kernel, which means it's nontrivial so dim(L(V)) < n. Therefore, L is not subjective.

hey good job. i know it wasnt the most challenging problem, but i just wanted to throw in a cute little algebra problem

 

[Ahnfeltia]

Chad and Pepe are playing with a pile of black, red and green stones.
When the game starts, there are k many black pieces (stone piles) on the game board
k could be any positive integer.

Players take turns. Chad took the first move. There are 2 options for a move:
(1) Place a red or green stone (from an infinite supply) on top of an existing stone pile. At the end of the round, there may not be two tiles of the same color in the stack.
A stack can contain up to three stones.
or
(2) Take out 2 stacks, but the top colors of the two stacks must be the same. Stack height doesn't matter.

If you can't take any action, you lose.
Determine whether either player can use a strategy to guarantee victory, and if so, explain how.
k=6

I think Chad can always force a win. First, Chad takes away 2 stacks. After that, he keeps mirroring Pepe's move. I didn't check super thoroughly, but I think this works.

 

[Ahnfeltia]

Consider the function f(n), such that f(0) = 0 and f(n) = f(⌊n/10⌋) + n - 10⌊n/10⌋, how many algorisms does the smallest positive integer m has, such that f(m) = 2015?

I'll just find the smallest m s.t. f(m) = 2015. Given that the algorithm is already specified, I don't really understand the "how many algorithms" part of the question.

For n < 10, f(n) = n
For 9 < n < 100, f(n) = ⌊n/10⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋
For 99 < n < 1000, f(n) = ⌊n/10⌋ - 9⌊⌊n/10⌋/10⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋ - 9⌊n/100⌋
For 999 < n < 10,000, f(n) = ⌊n/10⌋ - 9⌊⌊n/10⌋/10⌋ - 9⌊⌊n/10⌋/100⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋ - 9⌊n/100⌋ - 9⌊n/1000⌋

The pattern should be clear by now. If n has digits abcd with a > 0, then f(n) = abcd - 9(abc + ab + a) = a + b + c + d. At this stage, it should be easy to convince yourself that f is just the digit sum function.

The smallest number with digit sum 2015 is an 8 followed by 223 9s since 2015 = 223*9 + 8.

@Grim_Reaper @incelpardo99

 

[trying to ascend]

I'll just find the smallest m s.t. f(m) = 2015. Given that the algorithm is already specified, I don't really understand the "how many algorithms" part of the question.

For n < 10, f(n) = n
For 9 < n < 100, f(n) = ⌊n/10⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋
For 99 < n < 1000, f(n) = ⌊n/10⌋ - 9⌊⌊n/10⌋/10⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋ - 9⌊n/100⌋
For 999 < n < 10,000, f(n) = ⌊n/10⌋ - 9⌊⌊n/10⌋/10⌋ - 9⌊⌊n/10⌋/100⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋ - 9⌊n/100⌋ - 9⌊n/1000⌋

The pattern should be clear by now. If n has digits abcd with a > 0, then f(n) = abcd - 9(abc + ab + a) = a + b + c + d. At this stage, it should be easy to convince yourself that f is just the digit sum function.

The smallest number with digit sum 2015 is an 8 followed by 223 9s since 2015 = 223*9 + 8.

@Grim_Reaper @incelpardo99

Correct

 

[CountBleck]

Let A be an n x n matrix with complex entries. If trace(AB)=0 for every n x n matrix B, show that A=0.

 

[Ahnfeltia]

Let A be an n x n matrix with complex entries. If trace(AB)=0 for every n x n matrix B, show that A=0.

Suppose a_ij != 0. Let B = PD where P is a permutation matrix which will swap columns i & j of A and D is a diagonal matrix with d_ii = 1 and d_kk = 0 for k != i. Then AB will be nil save for column i, whose diagonal element will be a_ij. However, a_ij = tr(AB) = 0 is now a contradiction.

 

[CountBleck]

Suppose a_ij != 0. Let B = PD where P is a permutation matrix which will swap columns i & j of A and D is a diagonal matrix with d_ii = 1 and d_kk = 0 for k != i. Then AB will be nil save for column i, whose diagonal element will be a_ij. However, a_ij = tr(AB) = 0 is now a contradiction.

good job ! an elegant solution too. that problem is an old favorite of mine

 

[General Alek]

These problems make me feel retarded

 

[Grim_Reaper]

I'll just find the smallest m s.t. f(m) = 2015. Given that the algorithm is already specified, I don't really understand the "how many algorithms" part of the question.

For n < 10, f(n) = n
For 9 < n < 100, f(n) = ⌊n/10⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋
For 99 < n < 1000, f(n) = ⌊n/10⌋ - 9⌊⌊n/10⌋/10⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋ - 9⌊n/100⌋
For 999 < n < 10,000, f(n) = ⌊n/10⌋ - 9⌊⌊n/10⌋/10⌋ - 9⌊⌊n/10⌋/100⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋ - 9⌊n/100⌋ - 9⌊n/1000⌋

The pattern should be clear by now. If n has digits abcd with a > 0, then f(n) = abcd - 9(abc + ab + a) = a + b + c + d. At this stage, it should be easy to convince yourself that f is just the digit sum function.

The smallest number with digit sum 2015 is an 8 followed by 223 9s since 2015 = 223*9 + 8.

@Grim_Reaper @incelpardo99

Correct

How many algorithms are there though?

 

[Ahnfeltia]

How many algorithms are there though?

Given that the algorithm is already specified, I don't really understand the "how many algorithms" part of the question.

 

[Ahnfeltia]

good job ! an elegant solution too. that problem is an old favorite of mine

It's a nice problem. I'll be assisting with linear algebra this quarter. Maybe I'll try to sneak in this problem.

 

[CountBleck]

It's a nice problem. I'll be assisting with linear algebra this quarter. Maybe I'll try to sneak in this problem.

damn youre living my dream

 

[Ahnfeltia]

damn youre living my dream

I hope you'll get there then. I also hope it'll bring you more joy than it does me. I'm currently getting a PhD but I don't see myself continuing in academia.

 

[CountBleck]

I hope you'll get there then. I also hope it'll bring you more joy than it does me. I'm currently getting a PhD but I don't see myself continuing in academia.

yea for now as a 3rd year undergrad academia seems like the move. but i imagine that may change once i actually see how brutal it is. i'll still try it out tho. i got nothing to lose

 

[Med Amine]

when you go asian, you never fail an equation

 

[trying to ascend]

How many algorithms are there though?

224

 

[Grim_Reaper]

224

Amazing

 

[Ahnfeltia]

yea for now as a 3rd year undergrad academia seems like the move. but i imagine that may change once i actually see how brutal it is. i'll still try it out tho. i got nothing to lose

My thoughts exactly at the time

 

[IronsideCel]

I hate IQ moggers.
If you answered a correct question in this thread, you must be a sexhaver.

 

[Ahnfeltia]

I hate IQ moggers.
If you answered a correct question in this thread, you must be a sexhaver.

I'm too busy doing math to have sex

 

[Grim_Reaper]

I hate IQ moggers.
If you answered a correct question in this thread, you must be a sexhaver.

@Ahnfeltia probably has foid students asking him to give them better grades, and in-exchange they have sex with him.

 

[IronsideCel]

@Ahnfeltia probably has foid students asking him to give them better grades, and in-exchange they have sex with them.

 

[Ahnfeltia]

@Ahnfeltia probably has foid students asking him to give them better grades, and in-exchange they have sex with him.

only happened thrice thus far

 

[Grim_Reaper]

only happened thrice thus far

@IronsideCel

 

[IronsideCel]

@IronsideCel

 

[FakeFakecel]

Man this thread reminded me of how fucking much I hate functions

 

[im done]