Thanks for the kind words, 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 much, Adam
When calculating the second null space the third row should be [-24 36 -24]. You left out the negative sign but put it there during calculation so answer is still correct.
They are different. An eigenvector is any vector x that satisfies the equation Ax = Lx where L is a scalar we call the eigenvalue associated with the eigenvector x. Since there are many vectors that satisfy this equation (e.g. an infinite number of them), we might be interested in compactly describing the entire collection of vectors that satisfy the equation Ax = Lx. That is exactly what a basis does. Remember, basis is a collection of vectors. If we have a basis for the eigenspace, any linear combination of the vectors in the basis also satisfies the equation Ax = Lx. Since we can't write down EVERY vector that satisfies the equation, writing down the basis for the eigenspace is kind of the next best thing since we then know what form a solution must have (e.g. it can be written as some linear combination of vectors in the basis). I hope that helps. Make sure to check out my website adampanagos.org for lots of additional content that you might find helpful. Thanks, Adam
@@AdamPanagos thanks. just check what you said, all linear combination of eigenbasis is the (eigen)space/set satisfying the equation Ax=Lx(L is a scalar.)
Thanks for the kind words, 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 much, Adam
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
A basis for a set of vectors does not necessarily have to be an orthogonal basis. The vectors found in this example are indeed a basis for the eigenspace since all vectors in the eigenspace can be written as a linear combination of the vectors (that is essentially the definition of a basis). One could enforce an orthogonal constraint as well, but that was not required/performed in this particular video. Hope that helps. Adam
So wait after i find the associated eigenvectors for the eig. values, then i have shown the eigenspace? No need to write the solution? (i cant just leave it as you did). I'm taking Diff. Equations concurrently so i might be confusing methods or reasoning. It's pretty challenging to get used to
The number of vectors in the basis is the dimension of the space. So, if we find that a space can be represented with 3 vectors as the basis, the space has a dimension of 3. Hope that helps, Adam
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.
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
Yes, x3 is a free variable so we can choose any value we want. I selected x3 = 1. If you select x3 = -1 you'll get the answer you proposed. Hope that helps, Adam
This video helped me a fuck ton, I wish I could hug you, the whole x3 can be any value thing was so hard to grasp for me //Swedish mechanical engineerstudent
excellent vid. hey, so on that last eigenvector.... since we traditionally do NOT use fractions or decimal numbers in our choice, then by choosing for X2 to be = 2, then X1 = 3. so instead of an awkward (1.5, 1, 0), we can use (3, 2, 0). At least in my years of doing this, I've noticed the latter to be the trend. curious, was this NOT part of what you were taught?
It ends up being the same thing. We're solving an equation that's equal to zero. So, you can write the equation either way, since we can multiply an equation equal to zero by negative 1 on each sides without changing the solution. Hope that helps, Adam
@@AdamPanagos Thanks! THAT for sometime finally made sense to me out of all I've been reading about the topic. Nice and concise and makes logical sense too.
@@GaryTugan 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.
yes, x2 IS the free variable. which is why he solved for x1 in terms of x2. When a variable (x1 or x2 or x3, etc.) is the free variable, we then write the others in terms of that.
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
You're welcome, thanks for watching. Make sure to check out my website adampanagos.org where I have lots of other material you might find helpful. Thanks, Adam
Only two minutes in and I understood more things on this subject than in my actual classes keep it up .👍
Thanks for the kind words, 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 much, Adam
When calculating the second null space the third row should be [-24 36 -24]. You left out the negative sign but put it there during calculation so answer is still correct.
Yes, thanks so much. I just added an annotation to the video to note the mistake. Good catch!
I have to thank youtube for teaching me Linear Algebra
Hi Adam ! I like your videos very much followed them for my Linear Algebra course in year 1 of Engineering in Sweden.
Excellent, glad to hear that! Good luck!
Wonderful, simple explanation. Thank you!
Thanks!
My textbook does A-(lambda)i while you do (lambda)i -A, does that matter?
AKM Pros For comparison, mine is (lambda)i-A. IDK what yours is.
Mine does A-lambdaI too.
So it turns out they're both the same. It just one of those cases where they're equivalent by algebraic manipulation.
Actually, maybe it doesn't matter. Both are gotten from Av = λv anyways.
Jay G Sorry, I didn't see your reply before I commented again.
great video dude, you explained a pretty complex topic really well and made it simple
Thanks!
Thank you for this video. Was very well explained. I'm taking detailed notes for my upcoming quiz.
Thanks, good luck!
is there any difference between an eigenvector and a bases for the eigenspace or are they the same thing
They are different. An eigenvector is any vector x that satisfies the equation Ax = Lx where L is a scalar we call the eigenvalue associated with the eigenvector x.
Since there are many vectors that satisfy this equation (e.g. an infinite number of them), we might be interested in compactly describing the entire collection of vectors that satisfy the equation Ax = Lx. That is exactly what a basis does.
Remember, basis is a collection of vectors.
If we have a basis for the eigenspace, any linear combination of the vectors in the basis also satisfies the equation Ax = Lx. Since we can't write down EVERY vector that satisfies the equation, writing down the basis for the eigenspace is kind of the next best thing since we then know what form a solution must have (e.g. it can be written as some linear combination of vectors in the basis).
I hope that helps. Make sure to check out my website adampanagos.org for lots of additional content that you might find helpful. Thanks, Adam
@@AdamPanagos thanks. just check what you said, all linear combination of eigenbasis is the (eigen)space/set satisfying the equation Ax=Lx(L is a scalar.)
Very clear explanation, thanks!
Glad to help, thanks for watching!
Thanks my friend...you're a life saver
thanks man keep producing the work! It was really helpful
Thanks for the kind words, 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 much, Adam
thank you so much! MUCH better than my textbook:)
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
shouldnt their inner product to be equals to zero if they form a basis? is not zero, but 1. It seems that im missing something
A basis for a set of vectors does not necessarily have to be an orthogonal basis. The vectors found in this example are indeed a basis for the eigenspace since all vectors in the eigenspace can be written as a linear combination of the vectors (that is essentially the definition of a basis). One could enforce an orthogonal constraint as well, but that was not required/performed in this particular video. Hope that helps.
Adam
thank you!! It helped a lot, i had this question all day
So wait after i find the associated eigenvectors for the eig. values, then i have shown the eigenspace? No need to write the solution? (i cant just leave it as you did). I'm taking Diff. Equations concurrently so i might be confusing methods or reasoning. It's pretty challenging to get used to
how to find the dimensio correspond to the eigenspace?
help me senpaiiii
The number of vectors in the basis is the dimension of the space. So, if we find that a space can be represented with 3 vectors as the basis, the space has a dimension of 3. Hope that helps,
Adam
for the second λI-A last, on column 1 line 3, its supposed to be -24 right? then x1=x3
Great video! Many thanks
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.
Thank you
Clear and concise explanation! Thank you!
Does the same process apply if we have complex eigenvector and we want to create real basis from it?
So eigenbasis I.e basis of eigenvectors are just the resultant eigenvectors we get right ?!?
Thanks for the video. That was easy to understand. In the last part why we did row reduction? Can anyone can answer? Thanks
Thank you! It helped a lot.
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
can the null space of number two also be [1,0,-1] or is [-1,0,1] the only right answer?
Yes, x3 is a free variable so we can choose any value we want. I selected x3 = 1. If you select x3 = -1 you'll get the answer you proposed. Hope that helps,
Adam
This video helped me a fuck ton, I wish I could hug you, the whole x3 can be any value thing was so hard to grasp for me //Swedish mechanical engineerstudent
excellent vid. hey, so on that last eigenvector.... since we traditionally do NOT use fractions or decimal numbers in our choice, then by choosing for X2 to be = 2, then X1 = 3. so instead of an awkward (1.5, 1, 0), we can use (3, 2, 0). At least in my years of doing this, I've noticed the latter to be the trend. curious, was this NOT part of what you were taught?
Either is fine. The direction is all that matters. Hope that helps,
Adam
Not gonna lie, chief: ya went a little heavy on the snobby tone on this one x.x
@@jacmac225 was a straight up question. your sensibilities harmed by questions...?
Is it not supposed to be A-lambda*I?
With way is fine, you'll get the same answer since we're setting equal to zero. A-LI = 0 or LI-A = 0 are the same. Hope that helps,
Adam
Thank you sir for this amazing video!
Glad you liked it, thanks!
In my University's book it is A-(lambda)I.
It ends up being the same thing. We're solving an equation that's equal to zero. So, you can write the equation either way, since we can multiply an equation equal to zero by negative 1 on each sides without changing the solution. Hope that helps,
Adam
@@AdamPanagos helped me thanks
so eigenspace and eigenvectors are the same thing?
No, not quite. An eigenvector x is just a vector that satisfies Ax = Lx. The span of all eigenvectors is the eigenspace.
@@AdamPanagos Thanks! THAT for sometime finally made sense to me out of all I've been reading about the topic. Nice and concise and makes logical sense too.
@@GaryTugan 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.
In the final part, isn't x2 free and x3 = 1 since the columns are what correspond to each variable value? Otherwise, great stuff!
yes, x2 IS the free variable. which is why he solved for x1 in terms of x2. When a variable (x1 or x2 or x3, etc.) is the free variable, we then write the others in terms of that.
thank you I was losing my mind.
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
isnt it A-(lamda*x)
「ビデオサウンドは、私の想像を超えて、かなり良いです」、
Thanks!
thank youu
thxxx
You're welcome, thanks for watching. Make sure to check out my website adampanagos.org where I have lots of other material you might find helpful. Thanks,
Adam
I love you