Linear Algebra Example Problems - Basis for an Eigenspace

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An eigenvector of a matrix is a vector v that satisfies Av = Lv. In other words, after A "acts on" the vector v, the result is the vector v just scaled by the constant number L.
In this example, we find the eigenvectors of a given 3x3 matrix. This is done by finding the null space of the matrix A-LI. The null space solution (A-LI)x = 0 always results in an infinite number of solutions for the vector x. As such, we find a basis for each one of these solutions, and thus the "basis for an eigenspace" terminology.
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[@donaldford2849]

Only two minutes in and I understood more things on this subject than in my actual classes keep it up .👍

[@AdamPanagos]

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

[@karandeepsinghchatha3872]

Hi Adam ! I like your videos very much followed them for my Linear Algebra course in year 1 of Engineering in Sweden.

[@AdamPanagos]

Excellent, glad to hear that! Good luck!

[@UifaleanAlexandru]

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?

[@marianne9317]

Wonderful, simple explanation. Thank you!

[@AdamPanagos]

Thanks!

[@MuhammadFaizan-br3cr]

Thanks my friend...you're a life saver

[@pyropower]

Thank you for this video. Was very well explained. I'm taking detailed notes for my upcoming quiz.

[@AdamPanagos]

Thanks, good luck!

[@ryanveyr9195]

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

[@ezrahinkley6132]

great video dude, you explained a pretty complex topic really well and made it simple

[@AdamPanagos]

Thanks!

[@malcolmcollinable]

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.

[@AdamPanagos]

Yes, thanks so much. I just added an annotation to the video to note the mistake. Good catch!

[@fahimimtiaz2153]

thanks man keep producing the work! It was really helpful

[@AdamPanagos]

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

[@mortuiambulante4683]

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

[@cristianadias8013]

Thank you sir for this amazing video!

[@AdamPanagos]

Glad you liked it, thanks!

[@sharonselvi6099]

for the second λI-A last, on column 1 line 3, its supposed to be -24 right? then x1=x3

[@shayandurrani3738]

Thanks for the video. That was easy to understand. In the last part why we did row reduction? Can anyone can answer? Thanks

[@gjennifer8887]

Very clear explanation, thanks!

[@AdamPanagos]

Glad to help, thanks for watching!

[@antoniaangelopoulou4269]

I have to thank youtube for teaching me Linear Algebra

[@elisesl7107]

thank you so much! MUCH better than my textbook:)

[@AdamPanagos]

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

[@dismasyves]

Thank you! It helped a lot.

[@AdamPanagos]

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

[@Kim-Yo-Jong123]

Thank you

[@oAbraksas]

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

[@AdamPanagos]

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

[@oAbraksas]

thank you!! It helped a lot, i had this question all day

[@mauricioperedo9657]

Great video! Many thanks

[@AdamPanagos]

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.

[@madhirajutrisha9240]

So eigenbasis I.e basis of eigenvectors are just the resultant eigenvectors we get right ?!?

[@supersonic174]

is there any difference between an eigenvector and a bases for the eigenspace or are they the same thing

[@AdamPanagos]

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

[@BoZhaoengineering]

@@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.)

[@GaryTugan]

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?

[@AdamPanagos]

Either is fine. The direction is all that matters. Hope that helps, Adam

[@jacmac225]

Not gonna lie, chief: ya went a little heavy on the snobby tone on this one x.x

[@GaryTugan]

@@jacmac225 was a straight up question. your sensibilities harmed by questions...?

[@PokeShadow77]

In the final part, isn't x2 free and x3 = 1 since the columns are what correspond to each variable value? Otherwise, great stuff!

[@GaryTugan]

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.

[@mutludogan5408]

thank youu

[@jamesa.646]

can the null space of number two also be [1,0,-1] or is [-1,0,1] the only right answer?

[@AdamPanagos]

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

[@dijahmasnawi6627]

how to find the dimensio correspond to the eigenspace?

[@dijahmasnawi6627]

help me senpaiiii

[@AdamPanagos]

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

[@Red_Zealot]

My textbook does A-(lambda)i while you do (lambda)i -A, does that matter?

[@Triplechoco52]

AKM Pros For comparison, mine is (lambda)i-A. IDK what yours is.

[@donedumi-leslie5304]

Mine does A-lambdaI too.

[@Triplechoco52]

So it turns out they're both the same. It just one of those cases where they're equivalent by algebraic manipulation.

[@donedumi-leslie5304]

Actually, maybe it doesn't matter. Both are gotten from Av = λv anyways.

[@donedumi-leslie5304]

Jay G Sorry, I didn't see your reply before I commented again.

[@derrick_rj.5035]

isnt it A-(lamda*x)

[@DAMfoxygrampa]

I love you

[@jonathanmilloway7345]

Is it not supposed to be A-lambda*I?

[@AdamPanagos]

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

[@naeemjamali1475]

In my University's book it is A-(lambda)I.

[@AdamPanagos]

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

[@poecoreface]

@@AdamPanagos helped me thanks

[@collincypret1983]

thank you I was losing my mind.

[@AdamPanagos]

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

[@sidmukhtadir6837]

so eigenspace and eigenvectors are the same thing?

[@AdamPanagos]

No, not quite. An eigenvector x is just a vector that satisfies Ax = Lx. The span of all eigenvectors is the eigenspace.

[@GaryTugan]

@@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.

[@AdamPanagos]

@@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.

[@roboter8024]

thxxx

[@AdamPanagos]

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

[@hfuytuyuyu6348]

「ビデオサウンドは、私の想像を超えて、かなり良いです」、

[@AdamPanagos]

Thanks!