How to Determine if a Set of Vectors is Linearly Independent [Passing Linear Algebra]

You see if you can find nonzero weights when writing the zero vector as a linear combination of the vectors in the set.
Interesting Theorem : If a set of vectors contains the zero vector, that set is automatically linearly independent. Why is this? Continue reading:
Think about the linear dependence relation: you pick zeros for all the weights for all the nonzero vectors in the set, but then for the zero vector you can pick ANY NUMBER for its weight and you will get the zero vector as a linear combination. In doing this, you've written a linear combo of the vectors in the set that equals the zero vector, and that one weight was nonzero, right? So, you've done it. The set is linearly dependent bc you can write a linear dependence relation.
Skip the background info: 2:09
Skip to how to eyeball: 6:14

[@LyndseyNguyen116]

4 years later and you're still helping people... ur a legend, thanks man!

[@Winklemanus]

I can tell the guy who does these is handsome.

[@stemsupport3645]

love u graham

[@simmzlingsmthiyane1809]

Focus

[@odgarig8601]

My boy thirsting in linear algebra comments section. Y'all wild.

[@coohjay]

i love how the voice cracking got more and more intense as the video went on, great vid btw :)

[@angelahuynh5074]

it's probably cause he's so passionate about math

[@Quiyum]

thought I was the only one lmfao

[@quinn4856]

I watched this video the night before my first linear algebra exam and now I'm watching it again the night before my linear algebra final lol

[@michael654]

I watched this video when I was 5 years old. Now I won the Nobel prize in math.

[@rohhanbhardwaj]

@@michael654 whattttttttttt

[@abbycash]

THANK YOU. Very helpful and easy to understand

[@GF-wf8rt]

Very good video. You don’t explain it using annoying theorem lingo. Quick and easy ways to check a span. Thanks!

[@angelahuynh5074]

We just learned about linear dependence this week and your explanations make it easier to understand :-)

[@estherabass7389]

thank you so much! i love how straightforward this was!

[@spookybrojo7186]

This really breaks it down, thanks!

[@Gabriel-dm2ru]

THANK YOU YOU ARE A LIFE SAVER

[@thienphan1764]

Really appreciate the videos, can't thank you enough 🙏🙏

[@huitzilopochtli2872]

Thank you so much!

[@danielprive6]

Archer's teaching Linear Algebra now

[@stepbystepscience]

Your definition at the beginning of the video was very good, most other sources make it way too complicated.

[@babi9500]

2:00 man after so many vids you're the only one who answered my question thanks sm

[@charliefleming3859]

great stuff

[@definitelynottigerwhitten5865]

Thanks brofessor!

[@williamhogrider4136]

Thnx man ;)

[@GB2G]

Thank you

[@yusufmoola6471]

or u can find the determinant and see that it equals zero so its linear dependent.

[@toddchavez4610]

My go to method

[@respectpartii6302]

only works in square matrices, so this is a most general method presented in the video.

[@xeeshankhan725]

You are going to help me pass my Exam >> XD

[@Sam-iy4in]

king

[@drealslimshady]

what if there are more elements in each vector than vectors (ex. a 4x3 matrix) would that also mean it is automatically linearly dependent?

[@Lucas-yh5zz]

Hi! not necessarily, for example if you had these two vectors: (1,3,5,7) and (2,6,10,14), you can see they are not linearly independant as the second one is clearly two times the first one.
pd:I know it's been 8 months, so you probably don't need the answer anymore, but other people may.

[@liveforkillarpit]

@@Lucas-yh5zz its after 1year helps me . thanks

[@ok-sj7bx]

What if there are more variables in the vectors than the matrix has columns?
For example vectors v_1 = (1,2,3) and vector v_2 (2,3,4) would come out (just play along) and have one free variable but that would still mean they have two "stable" variables? Which technically would be enough to form a plane? Or would it still shrink in size and form a line, since that would still mean there is a solution other than 0 for the coefficients c_1*v_1+c_2*v_2=0-vector.

[@ok-sj7bx]

Nvm literally the last comment responded to this.

[@cyrrender]

nah jit was goin thru puberty in this video 😭 jokes aside thx for the great explanation!

[@crazygreekpvp6635]

GOAT

[@Coolpanda77454]

question: according to the last example, is the opposite also guaranteed? If we have more entries in each vector than the number of total vectors, does that make it linearly independent?

[@monsieurLDN]

No only if every column has a pivot.
Pets say that you have a set of 2 vectors with 3 entries. If the two vectors are multiples (which is the same as overlapping) the set would be lineary dependent

[@ok-sj7bx]

@@monsieurLDN Lol check the newest comment, I literally asked the exact same question and then read one comment down and you answered my question. Thanks!

[@GenuinePeacefulTimes]

having only a trivial solution defines linear independence. you said we don't want that at the 3:00 mark. lmao you are wrong

[@thabelangncube7539]

can you show us how you went about the row reduction, am kinda getting some weird values

[@gattinocurioso6852]

mee to

[@cmdrbobert9862]

I don't really grasp how you know what you're doing to row by using c1, c2, and c3.

[@thehamsterarmy2380]

He's solving for the pivots variables in each row by using the columns in that row

[@matteos.3438]

your comment about the theorem that states if a set contains the zero vector makes it linearly independent is incorrect - it's the other way around because if one of the vectors were zero then we would have a dependence relation

[@monsieurLDN]

Doesn't it depend? you can only use the trivial solution of 0 to represent a 0-vector

[@monsieurLDN]

Nvm

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[@healthstellar8701]

5:54 nice

[@chrisdeluisa6648]

9:00

[@ASD-6132]

@@chrisdeluisa6648 6:09

[@SpeaksYourWord]

1:54 wtf is that supposed to mean

[@user-vw8ow3pm6u]

Okay so why did my professor make it so complicated for no reason

[@jameshindle4000]

Phattest gg's

[@amb.jworld3127]

Are you teaching those who are Learning it for the first time or are you lecturing a pro😒.

[@PURPLECUT]

some people know how to teach some don't and this guy does not.

[@thehamsterarmy2380]

honestly it made a lot of sense

[@EigenA]

I thought he did a good job. Maybe we should look at ourselves before commenting on the qualities of others.