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Linear Algebra Example Problems - Spanning Vectors #1
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- čas přidán 7. 03. 2015
- adampanagos.org
Course website: www.adampanago...
In this problem we work with the vectors v1 and v2 and determine if the set {v1, v2} spans R2.
If the set of vectors {v1,v2} spans R2, then ANY vector from R2 can be written as a linear combination of these vectors. To see if this is true, an arbitrary vector from R2 is selected an and an augmented matrix is constructed and solved.
If you enjoyed my videos please "Like", "Subscribe", and visit adampanagos.org to setup your member account to get access to downloadable slides, Matlab code, an exam archive with solutions, and exclusive members-only videos. Thanks for watching!
Been staring at my textbook for an hour and this video finally got it through my head. Thanks.
Wow.
Always had trouble fully understanding this concept until watching this video, I always wondered what it meant and not until I watched this video did I fully grasp the concept. Very well explained. Thank you!
Glad I could help, thanks!
Oh! very well explained. At last, I was able to get the problem and was able to solve it. Thank you so much! More videos to come.
Glad you were able to get it figured out. Best!
You are a legend for putting so much effort & time into making all these videos. The best thing about you is the format & presentation, whatever tool you are using it's great. It's so clear, precise and you explain so good. THANKS AGAIN SIR!
u just said it for me
Simple and glittering way of explaining.
This video is fantastic, I also love the fact that we don't have to sit around waiting for you to write things, I hate that in other math videos.
Glad you like the video, thanks!
Thank you Prof Panagos! This Linear Algebra series is great!! Really helpful! Would you please do more Advanced Algebra videos? I am a Math major student and I got to say watching your videos is so much fun!
Glad the videos have helped. I'm continuing to make more videos as I have time. Hopefully I'll get to some more abstract algebra concepts at some point in the future. Thanks for watching!
Best. Video. . . Explained well
Thanks!
Thanks for the useful examples. You made a lot of efforts for these examples. You are a great man.
Glad to have helped; thanks for watching.
Very well explained, sir😊
Pretty easy to understand!
...This video makes me happy, Adam. Using a not too difficult example, Spanning vectors are clearly explained! Thank you, Jan-W
I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
@@AdamPanagos ...Thank you for your kind reply and recommendation, I will definitely do that after watching several Linear Algebra videos of yours, as well as recommending your website and youtube channel to other friends interested in Linear Algebra or possible other topics! It really is a jungle of Mathematics on the internet, hence many thanks to you for providing more manageable structure in this... once again Adam, thank you very much! Jan-W
Best video on spanning ..thank you so much for this!
You're very welcome, thanks for watching!
this video is really helpful, just before my exams - thank you
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
Hello
The video is very beautiful
And one question please
Why do you this E2=_1/3E2
But you can do this
E2=3E1+E2
And continue the solution
Yes, what you suggest is fine as well. There are always MANY ways to do row reduction and still end up at the correct/same answer. What specific steps you perform is often just personal preference (similar to how people might simplify algebraic expressions in different steps but still end up at the same final answer). Hope that help!
Adam
@@AdamPanagos Well done, thank you
@@Becarefulthismath31.4 You're welcome, thanks for watching!
At 2:50, you said x1 = a times a-b over 3. Did you mean minus?
Yes, sorry for misspeaking. You are correct. "a MINUS the quantity......"
Thank U!!
You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.
Nice explanation 😀
Thank you for the kind words, I appreciate you watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
for two vectors to span a space they should be linearly independent , isn't it sir?
Thank you!
You're welcome. Glad I could help, thanks for watching.
Thank you sir for making my concepts perfect
Amazing! Thank U ! You saved my life!
It is clearly explained and I was able to understand span of a vector
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
Bes video for the vector span
Thanks sir. Great explanation 👍🏻👍🏻👍🏻
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org where I have a lot of other videos and resources available that you might find helpful. Thanks, Adam.
thanks
excellent video!
I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
@@AdamPanagos i love you
Da best
Thanks for the kind words! If you found the video useful make sure to check out my website adampanagos.org where I have a lot of other resources available and consider supporting development of new videos through www.patreon.com/agpanagos. Thanks for watching!
Adam
Hi
Can I ask, how did you compute your a and b
There isn't any computation for a and b. The values a and b are just arbitrary elements of the R2 vector [a; b]. They are variables; they aren't computed. We're seeing if for ANY values of a and b, we can write the given vectors as a linear combination to yield [a; b]. It turns out we can if we use the values of x1 and x2 computed in the video. So, no matter what a and b are, using x1 = a-(a-b)/3 and x2 = (a-b)/3 yields the linear combination x1*[1; 1] and x2*[1; -2] equal to [a; b].
Thank you so much i was impanicato ma mo tutto meglio daie
You're very welcome, thanks for watching.