Tricky Dot Products (2011 Putnam A4)

[@LetsSolveMathProblems]

I sincerely apologize for such a long wait. Although I will probably go into hibernation again in the fall once the school kicks in, I plan to post quite regularly throughout the rest of the summer, starting a series on basic theoretical linear algebra in the near future. I cannot give enough thanks to everyone who supported the channel while I was gone, and to those who sent me comments wishing me the best! I am excited to interact with our amazing community once again. =) As stated at the end of the video, the extra challenge problem is to prove or disprove whether the problem statement holds if and only if the same can be said about columns. Feel free to comment proposed solutions here!

[@Jan-vz5ge]

Really looking forward to your series on theoretical linear algebra 👍🏼

[@benjamingiribonimonteiro9393]

No need to apologize! Focus totally in your studies, this will be the best moment of your life in that sense, we totally understand and support you! I can't wait to see your theoretical linear algebra series, this is a very beautiful topic! My best wishes for you =)

[@VaradMahashabde]

Books before views Understandable, have a great semester!

[@bryanmoreno8925]

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[@alpe6127]

It's nice to hear from you again! :)

[@LetsSolveMathProblems]

It's nice to hear from you and other viewers again! :)

[@Vampianist3]

Bro two things for you have whooshed up sooo many levels since you started uni: your rate of speech, and the level of mathematics you use to tackle problems. Marvellous job!

[@redaassal8733]

The king is back , m so happy to hear that voice and that accent again !

[@pichass9337]

THE KING RETURNS!!!

[@charleswoodard8478]

Fast favorite. This gives me great insight as to the framework of approach for problem solving. Exquisite.

[@Ethan-mj6wy]

The absolute chief of clean and easy to understand explanations, back to business 👑

[@ayushbanerjee1187]

Probably the best math youtube channel! Please continue creating videos:)

[@hurotowel6539]

The wait has been long! It is great to see you posting videos again.

[@mariochavez3834]

See your video in my feed is the best feeling that I have in months

[@swotted]

YOU'RE BACK!

[@yaseengarehmohammadlou9349]

Very interesting. Thanks for solving this problem.

[@MuPrimeMath]

Hey!!

[@newkid9807]

Pu Prime Math

[@kaziaburousan166]

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[@matron9936]

He is back, back again

[@yashvangala]

The metaphysical being with a voice( and a white board) has returned!

[@atomiccompiler9495]

We missed you!

[@shambosaha9727]

The Return of the King

[@gardening_vibes]

Hey glad that you are back!

[@aswinibanerjee6261]

When are you resuming weekly math challenge ?

[@LetsSolveMathProblems]

After much debate, I have decided to officially call the end of the Weekly Math Challenge as it is, and pursue new projects in KZbin (such as the upcoming series on theoretical linear algebra) and other academic studies in personal life. I express my gratitude and apologies to you and other viewers who participated in the challenge.

[@aadityajha7502]

For challenge, wacth this math puzzle and try to solve it without watching solution.kzbin.info/www/bejne/o36XdGiFo7N8n5o

[@alejandrocasco8318]

Oh my god he's back

[@yoyokojo651]

He’s back!!!!

[@521Undertaker]

Welcome back!

[@yassinezaoui4555]

Welcome back 😊

[@Sciophile]

Another argument, as requested for the n even case. Since the columns sum to 0, we have that the columns are linearly dependent. Since row rank = column rank, we have a linear dependence between the rows. Now consider one such linear dependence relation. V_i1+V_i2+..+V_ik=0 where 1≤i_1≤i_2≤...≤i_k≤n. If k is odd, k is smaller than n so consider V_p where p=/=i_k doesn't appear on the list. Dotting V_p with the equation, we get 1=0 (since the left hand side is the sum of an odd number of 1s). So we conclude that k must be even; so dotting it instead with V_i1, we get 1=0 again; neither case works. (note in the n=odd case, we get a sum of an odd number of terms where we can't 'choose an extra one' we're also forced into the maximum length linear dependence relation by the V_p argument)

[@LetsSolveMathProblems]

That is an astonishingly clean solution! This is a very minor remark, but I believe your solution could use a small fix: We should consider 1≤i_1≤i_2≤...≤i_k≤n for n even (so k and n are not necessarily the same), and perform casework on the parity of k. If k is odd, there exists v_p with p != i_j for any j, and we can take the dot product for contradiction; if not, take dot product with v_{i_1} for contradiction.

[@Sciophile]

​@@LetsSolveMathProblems yes, that's what I meant and then I went and used awkward notation lol. I will fix it.

[@maxjackson6616]

Have you ever taken the putnam? How did you do?

[@alainrogez8485]

What does "dot product" mean?

[@VaradMahashabde]

Let'sSolveQuarantineBoredom! BTW, my solution for the extra question : I think this is actually trivially answered by the arguments used in the video. We saw that a matrix A fulfills this property iff A*A^t = [all 1] - I_n. Clearly, RHS is a symmetric matrix, and so A^t is also a solution. Which means that the row vectors of A^t, which are also the column vectors of A, fulfill this property. QED

[@LetsSolveMathProblems]

Unless I am missing something evident, I believe the extra question requires more work: Realize that when you take the transpose of A*A^t, you obtain A*A^t, so we do not immediately get information about the rows of A^t in the process. Also, consider multiplying a matrix {{1, 0}, {1, 1}} by {{0, 1}, {1, 1}}. The product is symmetric, but multiplied the other way, it is not.

[@VaradMahashabde]

@@LetsSolveMathProblems I done goofed

[@dkravitz78]

Much easier solution: Let J be matrix of all ones and j be vector of all ones. We want to know if there's a matrix A such that At*A=J+I mod 2 If n is odd then A=J-I works, as At*A=(J-I)^2= J^2-2IJ+I^2= nJ - 2J + I = (mod 2) J+I With the last equality because n is odd. If n is even, say At*A=J+I mod 2 But A*j=0 because each row sum must be even. Multiply both sides times j, get 0 =(J+I)j = (n+1)j = (mod 2) j This is a contradiction.

[@oliverseaman1671]

What a g

[@alejandrodeharo9509]

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