Linear Independence and Linear Dependence, Ex 2 

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Very well explained... Best explain I found on RU-vid thanks... patrick

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Thanks patrickJMT. You made me understand linear independence in less than an hour :) Awesome video.

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Video was great and really helped a lot just wanted to point out you totally called that 7 a 4 when you first started at 0:23

great comment :) glad i could help you!

how did you choose -2? do you just pick an arbitrary number or was there a reason that you picked that?

*Patrick said 4, but meant 7.*

@Nicola72av nicola, you are very welcome : )

Super helpful!! Thank you for the videos!

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@PlanetEarthSupport happy that i could help out you wonderful people (most of you at least!) at least a little bit

hi, im lost with this one, hope you can help: The value(s) of λ such that the vectors v1 = (-3 - 2λ, 4, -4) and v2 = (4 - λ, 8, -8) are linearly dependent is (are)?? How do we start?

It is very helpful. Thank you so much

An 11 year old RU-vid video is more helpful than the whole unit tutors at Uni.

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@JayZeus7 getting there : ) want to talk about basis and linear transformations first ; )

If the vectors have zero as any of their components, a much easier solution exists. In this example it's obvious that a1 + 2a3 = 0 so you get a1 = -2a3 and 2a1 + a2 = 0 so you get a2 = -2a1 pick a3 to be = -1 a1 = 2, a2 = -2, a3 = -1 now check if these a1 a2 a3 values hold for all equations 2*4 + (-2)*1 + (-1)*7 =/= 0 so it's independent

This was extremely helpful

"We're having fun" ... sometimes I forget, so thanks :)

You learn it to me so good!!

the best tutor ever

lol nice. my teacher let us use a calculator on the 2nd test since we already proved ourselves on the 1st test (when reducing to echelon form).

for linear dependent/linear independent, can we interchange the row when doing row of reduction?

Because there is a 1 in that column. Meaning he can times that row (3) with the opposite coefficient of row 1 and 2, in this case -2 and 4 to zero them out.

not hard at all if you know how to get to reduced echelon form. you are awesome man thank you so much

Patrick, I have a question. In the beginning you said the making all the variables equal zero is the trivial solution. But you made the matrix for the system of equations into row echelon form. Would have leaving the matrix in ordinary echelon found a result that isn't trivial?

If you just do ordinary operations, with zeros only below the diagonal line and not above, and not making the diagonal line all 1's, it would still be a trivial solution, because any number x times a3 = 0 will force a3 to equal zero, thus making a1 and a2 zero as well. As long as the solution is the zero vector and there are no free variables you should get linear independence. Hope that was clear!

Great explanation.

very well explanation thank you

What is the matrix of the vectors turns out to be inconsistent, no solutions? You've covered; if consistent (unique or infinite solutions but not where a1, a2, ..., an=0), then LD ..and if consistent where a1, a2, ...,a3 must=0, then LID. What if inconsistent?

Thank you!!

thanks man,very nice tutorial ;)

great videos 👍

Thank you

thanks really useful

That was so helpful :,)

learned sooo much!

What if all the scalars like a1=0, a2=0 & a3=0 and all the vector matrices is also = 0. Will that be Linearly dependent or independent !?

Thanks Heaps :)

thank u brother

is there a rule for how many rows and columns you need to make zero or is this done until no other rows or columns can be made zero?

Thank you very much!!!!

extremely helpful!

thx for the explanation :)!

can i just stop at guassian row reduction(echelon form) method to know whether it is linearly independent/dependent.. or do i have to row reduce it the way you do....

so do i have to convert the matrix to ref or rref?

Very helpful to me this 9 year old video🥰

Can anyone tell me that at 5:33 how the alfa 1,2,3 becomes zero?

since the right side column is all zeroes, the system is a homogeneous system. Thus it is consistent always (has at least one solution).

is every linearly independent set all always equal to just zero?

So, if you get a non-trivial solution, does that mean that the system is automatically dependent?

So pretty much if you can achieve RREF it's Linearly independent and if you can only get REF its dependent?

The teachers now have projectors in every class... Not sure why they dont just play Patricks videos instead of wasting 3 classes trying to explain something he does in 5 minutes!

thank you in advance

Thanks

thank you

Hi patrick, How can you know you can zero out in the video 4:26.

So if a question asks give the number of linearly independent columns. What does that mean? Do I have to solve it all the way as you just did? If so does that mean all the columns are linearly independent? I'm confused on that. Could you please help me?

What would you do if the middle vector is all zero?

how do you know when to stop doing row reduction?

Is it wrong to say that if a set of vectors v1,v2,...,vn is linearly independent, then they span R^n ? Is or is it not always true?

wait so when you have it in echelon form and the last row is all 0s why doesnt that make a4 a free variable and you can let it equal an arbitrary constant there for it would be linearly dependent?

Thanks a lot

thank you boss.

I wish you are my professor.

Why can you just look away from the zero row?

So can someone explain to me why the bottom row being all zeros doesn't matter in this situation. I get that a1 = 0, a2 = 0, and a3 = 0. but wouldn't a bottom row of all zeros signify SOMETHING? anything?

thanks da!! :)

I know this is old, but I really need to know, how do you prove this with the definition. I am having a headache trying to find this answer AND NO ONE DOES IT!

I'm not clear with the logic of row that why we are multiplying and doing minus or plus

how does a1 equal zero in the first row?

thats very helpfulll..thanks a lott

"2,0,1,4"???

So, in other words, if you can reduce a system to reduced row echelon form, then that IS the only solution? I've kind of noticed that this happens often with square matrices, and I'm not sure if it's a coincidence or if I'm doing something wrong?

Which pen you are using?

what if the system has no solution ? for eg we have the last row as [ 0 0 0 | 2}

thank you............

thanks

Thanks, boss