22. Diagonalization and Powers of A

[@bigfrankgaming2423]

This man single handedly saved my university algebra course, my teacher was just reading notes, he's actually expalining in a very clear manner.

[@BaranwalAYUSH]

This man has really reignited my passion for mathematics. Thank You Professor Strang for such amazing lectures.

[@charmenk]

Good professor with good old blackboard and white chalk teaching method. This is way better than all the fancy powerpoints that many teachers use now a days.

[@9888565407]

hey did ya benefit from these lectures

[@yusufkavcakar8744]

@@9888565407 yeah ı did

[@jiayuanwang1498]

I can't agree more!!!

[@kayfouroneseven]

this is a bazillion times more straightforward and clear than the lectures i pay for at my university. :( I appreciate this being online

[@bsmichael9570]

He tells it like a story. It’s like he’s taking us all on a journey. You can’t wait to see the next episode.

[@jollysan3228]

I agree. > Just one small correction at 32:30: It should have been S * LAMBDA^100 * c instead of LAMBDA^100 * S * c.

[@Slogan6418]

thank you

[@ozzyfromspace]

The sad thing was, a few moments later he was struggling to explain things because even though he hadn't pinned down the error, he someone knew that something wasn't quite right. But he obviously had the core idea nailed

[@alexandresoaresdasilva1966]

thank you so much, was about to post asking about this.

[@吴瀚宇]

I stuck on this for like 10 mins, until I saw the comments here...

[@maitreyverma2996]

Perfect. I was about to write the same.

[@Tutkumsdream]

Thanks to him! I passed Linear Algebra.. I watched his videos for 4 days before final exam and I got 74 from final.. If I couldnt watch Dr.Strang's lectures, I would probably fail...

[@snoefnone9647]

For some reason i thought you were saying Dr. Strange's lecture!

[@apocalypse2004]

I think Strang leaves out a key point in the difference equation example, which is that the n unique eigenvectors form a basis for R^n, which is why u0 can be expressed as a linear combination of the eigenvectors.

[@alessapiolin]

thanks!

[@wontbenice]

I was totally confused until you chimed in. Thx!

[@seanmcqueen8498]

Thank you for this comment!

[@arsenron]

in my opinion it is so obvious that it is not worth stopping on it

[@dexterod]

I think Strang assumed that A has n independent eigenvectors since most matrices do not have repeated eigenvalues.

[@albertacristie99]

This is magnificiant!! I have no words to express how thankful I am towards the exposure of this video

[@eye2eyeerigavo777]

Math surprises you everytime...🤔 Never thought that connections between rate of growth in system dynamics, fibonacci Series and diagnalization of an INDEPENDENT vectors will finally boil down INTO GOLDEN RATIO OF EIGENVALUES at END! 😳

[@BirnieMac1]

You know you’re in for some shenanigans when they pull out the “little trick” Professor Gilbert is an incredible teacher; I struggled with Eigenvalues and vectors in a previous course and this series of lectures has really helped understand it better Love your work Professor Gilbert

[@christoskettenis880]

The explanations of this professor of all those abstract theorems and blind methodologies are simply briliant

[@ranabhatashim]

There is no mistake at 32:00. The thing about c1x1 is c1 is a number while x1 is a vector, keep this in mind. we have u0 = c1x1....cnxn. Multiply by A both sides, Au0 = Ac1x1..Acnxn. Bring c1 to the front since its a number, c1Ax1 + c2Ax2..., now since x1...xn are eigenvectors Ax1 = lambda x1, so Au0 = c1*lambda*x1. We can now factor out the lamba as a eigenmatrix so Au0 = lambdamatrix(c1x1 + c2x2 .. cnxn). Remeber than x is a vector and c is a number? Therefore we can do c1x1+....cnxn as Xc, here X is eigenvector matrix which is S and c is a vector of (c1 c2...). Therefore Au0 = lambda * S * c.

[@keremdirlik]

Nope, be careful. It's S *lambda *c

[@georgesadler7830]

From this latest lecture , I am learning more about eigenvalues and eigenvectors in relation to diagonalization of a matrix. DR. Strang continues to increase my knowledge of linear algebra with these amazing lectures.

[@syedsheheryarbokhari2780]

There is a small writing mistake at 32:30 by Prof Strang. He writes (eigenvalue matrix)^100 multiplying (eigenvector matrix) multiplying c's (constants). It ought to be (eigenvector matrix) multiplying (eigenvalue matrix)^100 multiplying c's. At the end of the lecture Professor Strang does narrate the correct formula but it is easier to miss.

[@clutterbrainx]

Yeah I was confused for a very long time there

[@go_all_in_777]

At 28:07, uk = (A^k)uo, can also be written as uk = S*(Lambda)^k*(S^-1)*uo. Also, we can write uo = S*c as explained at 30:00. therefore, uk = S*(Lambda)^k*(S^-1)*S*c=S*(Lambda)^k*c

[@jeanpierre-st7rl]

Hi @ 29:46 Uo = C1X1 + C2X2 +C3X3... Is U0 a vector? If so, How can split this U0 in to a combination of eigen vectors? What is Ci ? If you have any info pleases let me know. Thanks.

[@SimmySimmy]

through single matrix transformation, the whole subspace will expand or shrink with the rate of eigenvalues in the direction of its eigenvectors, suppose you can decompose a vector in this subspace into the linear combination of its eigenvectors, so after many times of the same transformation, the random vector will ultimately land on one of its eigenvectors with the largest eigenvalue.

[@MsAlarman]

Mindbending

[@dwijdixit7810]

33:40 Correction: Eigenvalue matrix be multiplied to S from the right. That has been made in the book. Probably, it slipped off Prof. Strang in the flow.

[@eroicawu]

It's getting more and more interesting when differential equations are involved!

[@sathviktummala5480]

44:00 well that's an outstanding move

[@neoneo1503]

A*S=S*Lambda (Using the linear combination view (Ax1=b1 column part) of Matrix Multiplication), That is Brilliant and Clear! Thanks!

[@neoneo1503]

Also expressing the state u_0 to u_k as linear combination of eigenvectors (at 30:00 and 50:00)

[@Wabbelpaddel]

Well, it's - if you interpret it that way - just a basis transformation from the standard base (up to isomorphism, then just additionally multiply the transforms of the alternate basis) onto the eigenvector basis. Provided of course, that either the characteristic polynomial factors distinctly, or that geometric and algebraic multiplicity match (because then the eigenspaces distinctly span the vector space up to isomorphism; if they weren't, you'd just have a subspace as a generating system). For anyone who wanted one more run-through.

[@neoneo1503]

@@Wabbelpaddel Thanks! =)

[@ozzyfromspace]

For the curious: F_100 = (a^99 - b^99) * b/sqrt(5) + a^99 , where a = (1 + sqrt(5))/2 and b = (1 - sqrt(5))/2 are the two eigenvalues of our system of difference equations. Numerically, F_100 = ~3.542248482 * 10^20 ... it's a very large number that grows like ~1.618^k 😲 Overall, great lecture Professor Strang! Thank you for posting, MIT OCW ☺️

[@starriet]

Notes for future ref.) (7:16) there are _some_ matrices that do _NOT_ have n-independent eigenvectors, but _most_ of the matrices we deal with do have n-independent eigenvectors. (17:14) If all evalues are different, there _must_ be n-indep evectors. But if there are same evalues, it's possible _no_ n-indep evectors. (Identity matrix is an example of having the same evalues but still having n-indep evectors) * Also, the position of Lambda and S should be changed(32:36). You'll see why by just thinking matrix multiplication, and it can also be viewed by knowing A^100=S*Lambda^100*S^-1 and u_0=S*c. Thus, it should be S*Lambda^100*c, and this also can be thought of as 'transformation' between the two different bases - one of the two is the set of the egenvectors of A. * Also, (43:34) How prof. Strang could calculate that?? Actually that number _1.618033988749894..._ is called the 'golden ratio'. * (8:15) Note that A and Lambda are 'similar'. (And, S and S_-1(S inverse) transforms the coordinates.. you know what I mean.. both A and Lambda can be though of as some "transformation" based on different basis.. and S(or S_-1) transforms the coord between those two world.)

[@shadowByte99]

I spent a few hours on the second point before figuring it out :(

[@bastudil94]

There is a MISTAKE on the formula of the minute 32:31. It must be S(Λ^100)c in order to work as it is supposed. However it is an excellent lecture, thanks a lot. :)

[@YaguangLi]

Yes, I am also confused by this mistake.

[@sammao8478]

Yaguang Li agree with you.

[@AdrianVrabie]

+Bryan Astudillo Carpio why not S(Λ^100)S^{-1}c ???

[@apocalypse2004]

u0 is Sc, so S inverse cancels out with the S

[@daiz9109]

You're right... it confused me too...

[@cuinuc]

I love professor Strang's great lectures. Just one small correction at 32:30: It should have been S * LAMBDA^100 * c instead of LAMBDA^100 * S * c.

[@starriet]

Nice catch!

[@jeffery777]

haha I think so

[@eyuptarkengin816]

yeah, i though of the same thing and scrolled down the comments for a approval. Thanks mate :D

[@ranabhatashim]

There is no mistake. The thing about c1x1 is c1 is a number while x1 is a vector, keep this in mind. we have u0 = c1x1....cnxn. Multiply by A both sides, Au0 = Ac1x1..Acnxn. Bring c1 to the front since its a number, c1Ax1 + c2Ax2..., now since x1...xn are eigenvectors Ax1 = lambda x1, so Au0 = c1*lambda*x1. We can now factor out the lamba as a eigenmatrix so Au0 = lambdamatrix(c1x1 + c2x2 .. cnxn). Remeber than x is a vector and c is a number? Therefore we can do c1x1+....cnxn as Xc, here X is eigenvector matrix which is S and c is a vector of (c1 c2...). Therefore Au0 = lambda * S * c.

[@Zumerjud]

This is so beautiful!

[@ozzyfromspace]

Did we ever prove that if the set of eigenvalues are distinct, the set of eigenvectors are linearly independent? I ask because at ~ 32:00 taking u_o = c1*x1 + c2*x2 + ... + cn*xn requires the eigenvectors to form a basis for an n-dimensional vector space (i.e. span the column space of an invertible matrix). It feels right but I have no solid background for how to think about it

[@roshinis9986]

The idea is easy for 2d. If you have two distinct eigenvalues and their corresponding eigenvectors, you don't just have one eigenvector per eigenvalue, the whole span of that vector (its multiples forming a line) are also the eigenvectors associated with that eigenvalue. If the original eigenvectors were to be dependent, they would lie in the same line making it impossible for them to scale by a factor of two distinct eigenvalues simultaneously. I haven't yet been able to extend this intuition to 3 or higher dimensions though as now dependence need not mean lying in the same line.

[@jeanpierre-st7rl]

@@roshinis9986 Hi @ 29:46 Uo = C1X1 + C2X2 +C3X3... Is U0 a vector? If so, How can split this U0 in to a combination of eigen vectors? What is Ci ? If you have any info pleases let me know. Thanks.

[@kanikabagree1084]

This teacher made fall in love with linear algebra thankyou ❤️

[@maoqiutong]

32:41 There is a slight error here. The result Λ^100 * S * C may be wrong. I think it should be S * Λ^100 * C.

[@meetghelani5222]

Thank you for existing MITOCW and Prof. Gilbert Strang.

[@coreconceptclasses7494]

I got 70 out of 75 in my final linear algebra exam thanks MIT...

[@muyuanliu3175]

32:42, should be S lambda^100 c, great lecture, 3rd time I learn this

[@Huayuan-p4z]

I have learned about the Fibonacci sequence in my high school, and it is so good to have a new perspective on the magical sequence.I think the significane of learning lies in the collection of new perspectives.😀

[@nguyenbaodung1603]

I read something on SVD without even knowing about eigenvalues and eigenvectors, then watch a youtube video, explaining that V is actually the eigenvector decomposition of A^TA. Which is extremely insane when I got to see this video oh my godness. Now even haven't watched your SVD lecture, I can even tell the precise concept of it. Oh my godness Math is so perfect!!

[@rolandheinze7182]

Hard lecture to get through personally but does illustrate some of the cool machinery for applying eigenvectors

[@florianwicher]

Really happy this is online! Thank you Professor :)

[@dalisabe62]

The golden ratio arose from the Fibonacci sequence and has nothing to do with eigenvectors or eigenvalues. The beauty of using the eigenvectors and eigenvalue of a matrix though is limiting the effect of the transformation to the change in magnitude only, which reduces dynamics systems such as population growth that is a function of several variables to be encoded in a matrix computation without worrying about the effect of direction or rotation typically associated with matrix transformation. Since eigenvectors and eigenvalues change the magnitude of the parameter vector only, the idea of employing the Eigen transformation concept is quite genius. The same technique could be used in any dynamic system that could be modeled as a matrix transformation but one that produces a change in magnitude only.

[@Arycke]

Hence the title of his *example* as "Fibonacci Example." Nowhere was it stated explicitly sthat the golden ratio didn't arise from the Fibonacci sequence, so I don't see where you got that from. The example has a lot to do with eigenvalues and eigenvectors by design, and is using a simple recurrence relation to show a use case. The Fibonacci sequence isn't unique anyway.

[@ccamii__]

Absolutely amazing! This lecture really helped me to understand better the ideas about Linear Algebra I've already had.

[@RolfBazuin]

Who would have guessed, when this guy explains it, it almost sounds easy! You, dear dr. Strang, are a master at what you do...

[@cecilimiao]

@cuinuc I think they are actually the same, because LAMBDA is a diagonal matrix, you can have a try.

[@aattoommmmable]

the lecture and the teacher of my life!

[@serden8804]

bro yasiyor musun

[@Afnimation]

well i got impressed at the begining, but when he stated the second eigenvalue i realized it is just the golden ratio... That does not demerits him, he's great!

[@mospehraict]

@PhilOrzechowski he does it to make first order difference equations system out of second order

[@uzferry5524]

bruh the fibonacci example just blew my mind. crazy how linear algebra just works like that!!

[@mike-yj5mm]

I don't understand 11:25 why A square can be written in the way on the blackboard. I think A^2 should be (S Lambda S^-1)^T (S Lambda S^-1), the result differs from the one on the blackboard. Could someone explain this?

[@mike-yj5mm]

Okay, I figured it out. The S is an orthogonal matrix under the n independent eigenvector assumption, the inverse of which equals to its transpose.

[@APaleDot]

@@mike-yj5mm No, it doesn't require S to be an orthogonal matrix. n independent eigenvectors ≠ n orthogonal eigenvectors of unit length, which would be required to make S an orthogonal matrix. At this point in the lecture we've already proven that A = S ∧ S^-1 and therefore it follows immediately that A^2 = AA = (S ∧ S^-1)(S ∧ S^-1). All the matrices are square, so there is no conflict in their dimensions.

[@zyctc000]

If any one ever asks you about why the Fibonacci and the golden ratio phi is connected , point him/her to this video. Thank you Dr. Strang

[@shadownik2327]

Now I get it, so its like breaking the thing ( vector or matrix or system really) we want to transform into little parts and then transforming them individually cz thats easier as the parts get transformed in the same direction and then adding up all those pieces. E vectors tell us how to make the pieces and e values how to make the transformation with the given matrix or system. Wow thanks ! It’s like something fit in in my mind and became very simple. Basically this is like finding the easiest way to transform. Thanks to @MIT and Professor Strang for making this available online for free.

[@Mohamed1992able]

a big thanks tothis prof for his efforts to give us cours about linear algebra

[@alexspiers6229]

This is one of the best in the series

[@jojowasamanwho]

19:21 I would sure like to see the proof that if there are no repeated eigenvalues, then there are certain to be n linearly independent eigenvectors

[@eren96lmn]

43:36 that moment when your professor's computational abilities goes far beyond standart human capabilities

[@BalerionFyre]

Yeah wtf? How did he do that in his head?? lol

[@BalerionFyre]

Wait a minute! He didn't do anything special. 1.618... is the golden ratio! He just knew the first 4 digits. Damn that's a little anticlimactic. Bummer.

[@AdrianVrabie]

+Stephen Lovejoy Damn! :D Wow! AWESOME! I have no words! Nice spot! I actually checked it in Octave and I was amazed the prof could do it in his head. But I guess he knew the Fibonacci is related to the golden ratio.

[@IJOHN84]

All students should know the solution to that golden quadratic by heart.

[@ozzyfromspace]

Fun fact since we're all talking about the golden ratio. The Fibonacci sequence isn't that special. Any sequence F_(k+2) = F_(k+1) + F_k for any seeds F_0 = a and F_1 = b != -a generate a sequence that grows at the rate (1+sqrt(5))/2 .. your golden ratio. Another fun way to check this: take the limit of the ratio of numbers in your arbitrary sequence with your preferred software :) edit: that's a great excuse to write a bit of code lol

[@gomasaanjanna2897]

Iam from india I love your teaching

[@abdulghanialmasri5550]

The best math teacher ever.

[@LAnonHubbard]

I've only just learnt about eigenvalues and eigenvectors from KhanAcademy and Strang's Lecture 21 so a lot of this went whoooosh over my head, but managed to find the first 20 minutes useful. Hope to come back to this when I've looked at differential equations (which AFAIK are very daunting), etc and understand more of it.

[@rolandheinze7182]

Don't think you need diff EQ at all to understand the algebra. Maybe the applications

[@jasonhe6947]

absolutely a brilliant example for how to apply eigenvalues to real world problem

[@gianlucacococcia2384]

Can I ask you why A1 x x1 is lambda x1?

[@gianlucacococcia2384]

Zhixun He

[@wendywang4232]

something wrong with this lecture, 32:39, A^{100}u_0=SM^100c. Here I use M to substitute the eigenvalue diagonal matrix. The professor said A^{100}u_0=M^100Sc which is not correct.

[@PaulHobbs23]

@lolololort 1/2(1 + sqrt(5)) is also the golden ratio! Math is amazing =] I'm sure the professor knew the answer and didn't calculate it in his head on the spot.

[@kunleolutomilayo4018]

Thank you, Prof. Thank you, MIT.

[@ItsKhabib]

true masterpiece!

[@suzukikenta1079]

Could someone explain why vector, lambda and 1 is a solution to the nullspace at 48:43? First component, (1-lambda)*lambda+1 is not 0 or different from lambda^2 - lambda -1.

[@antoniolewis1016]

Not sure what you mean, but the professor was saying the vector [lambda ] [ 1 ] is in the nullspace of the matrix A-lambda* I, and not the nullspace of A. And since this vector is in that nullspace, it must be an eigenvector of A attached to eigenvalue lamda.

[@jasarinvorawathanabuncha6620]

(1-lambda)*lambda+1 is indeed 0 because of the determinant next to it: lambda^2 - lambda -1= 0

[@roronoa_d_law1075]

that makes lambda^2 - lambda + 1 and not -1 I'm confused

[@roronoa_d_law1075]

nevermind

[@rolandheinze7182]

Do the matrix multiplication and see fr yourself, it evaluates to 0 fr the eigenvalues. Also these eigenvectors give you the same equation as characteristic equation (flipped to other side of 0=...) So must be a solution

[@stumbling]

7:33 Surprise horn?

[@mayurkulkarni755]

wtf was that xD

[@canmumcu9804]

probably grinding of a chair xd

[@dexterod]

I'd say if you play this video at speed 1.5, it's even more awesome!

[@abdulbasithashraf5480]

Trueee

[@ozzyfromspace]

1x all the way. I savor the learning ☺️

[@pelemanov]

@LAnonHubbard You don't really need to know about differential equations to understand this lecture. Just watch lessons 1 to 20 as well ;-). Takes you only 15h :-D.

[@eugenek951]

He is my linear algebra super hero!🙂

[@benzhang7261]

Master Yoda passed on what he has learnt by fibonacci and 1.618.

[@dennisyangji]

A great lecture showing us the wonderful secret behind linear algebra

[@tomodren]

Thank you for posting this. These videos will allow me to pass my class!

[@khanhdovanit]

15:02 interested information inside matrix - eigenvalues

[@iebalazs]

At 32:32 the expression is actually S*Lamdba^100*c, and not Lambda^100*S*c .

[@thomassun3046]

Here comes a question, How U0 is equal to c1x1+c2x2...+cnxn. at 29:50, confused, could anyone explain it to me?

[@Tman1000-be7op]

He is just writing that the initial condition is equal to a combination of n independent eigen vectors.

[@putrijulianarahayu7975]

39:03 where the A (1,1,1,0) comes from? help.....

[@finalfantasy1112]

It comes from the first order system on the left.

[@NisargJain]

You can read my comment where I explained it.

[@praduk]

Fibonacci numbers being solved for as an algebraic equation with linear algebra was pretty cool.

[@technoshrink]

U0 == "you know it" First time I've heard his boston accent c:

[@phononify]

very nice discussion about Fibonacci ... great !

[@laurencerousseau23]

Someone know why the x1 in he nullspace is (1 0) and not (0 1) as he said 26:36.

[@giulianoguaragna9962]

(0,1) don't have sense ,because 0 *((0,0)t) + 1*((1,0)t) is not equal to ((0,0)t). Think in the null space of (A -2I)

[@mohammedal-haddad2652]

The first equation is x2=0 so x1 is arbitrary and he chose x1=1 so this gives (1 0)'.

[@faizanmohsin3685]

Because 0 1will not give 00

[@amyzeng7130]

What a brilliant lecture !!!

[@iDiAnZhu]

At around 32:45, Prof. Strang writes Lambda^100*S*c. Notation wise, shouldn't this be S*Lambda^100*c?

[@sharmabu]

absolutely beautiful

[@kebabsallad]

@PhilOrzechowski , he says that he just adds it to create a system of equations.

[@Hindusandaczech]

Bravo!!! Very much the best and premium stuff.

[@niraj_ds]

@ 44:00 why summation of both eigen values are 1?? have i missed any concept behind this?? : (

[@ElectricTeaCup]

Yes, "The sum of the n eigenvalues equals the sum of the n diagonal entries". The sum of the diagonal entries is 1.

[@胯下蜈蚣長老]

Excuse me, why doesn't the calculation of F_100 at "46:08" time the eigenvector X_1? Am i missing something?

[@9888565407]

can you elaborate your concern ?

[@胯下蜈蚣長老]

Hello, the formula at "33:12" shows that A^100*u_0= c_1*lambda_1^100*x_1+c2*....., so i think perhaps F_100 equals to c1*lambda_1^100*x_1? Thanks for replying~

[@santiagotheone]

@@胯下蜈蚣長老 Nope. F_100 is a scalar, but x_1 is a vector (in R^2 in this case). I guess the explicit thought in your question is actually A^100 * u_0 = A^100 [F_1 ; F_0] = c_1 * lambda_1^100 * x_1 + c_2 * lambda_2^100 * x_2 = [F_101 ; F_100] Focus on the term F_100 in second row. By using x_1 and x_2 (in 49:07), we know F_100 = c_1 * lambda_1^100 * 1 (element in second row of x_1) + c_2 * lambda_2^100 * 1 (element in second row of x_2) = c_1 * {[1 + sqrt(5)] / 2}^100 + c_2 * {[1 - sqrt(5)] / 2}^100. c_2 * {[1 - sqrt(5)] / 2}^100 can be omitted since {[1 - sqrt(5)] / 2}^100 is too small compared to {[1 + sqrt(5)] / 2}^100. Then we get the professor's approximation in 46:08.

[@shayanghanbari6838]

@@santiagotheone thanks for your clarifying

[@theshreyansjain]

Is there an error at 32:30? Shouldn't S be multiplied before (lamda matrix)^100?

[@zionen01]

Great stuff. I was able to do my homework with this lecture. I will definitely be getting Strang's book.

[@MaproXiZ]

I dont undertand why the eigenvectors are [lamba_1 1] and [lamba_2 1] at 49:19... since it is NOT true that ((1 - lamba) * (lamba)) + 1 is lamba^2 - lamba - 1 ... or it is? or what is happening?

[@youcefyahiaoui1465]

He's just using the original definition of the eigenvalues. We already have the characteristic equation lamd^2-lamda-1=0 as the polynomial equation the solution of which is both eigenvalues. Then he recognized that by writing the A-lamda.I by the vector [lamda 1] will generate this same characteristic equation. Hence, the eigenvectors are just [lamda 1]

[@rohitsaxena22]

Why u_0 can be written as a linear combination of eigenvector of A? 29:55

[@Samurai_Jack__]

4 years for the question but I'll still answer it if the eigenvectors are independent it means they span the whole n-dimensional space so any vector of size n can be written as a combination of this vectors. they can form a basis for the space.

[@MeridianLights]

In the Fib example, it seems impossible to find a c1 and c2 s.t. c1*x1 + c2*x2 = [1 0]

[@jasarinvorawathanabuncha6620]

You can! Try using elimination to easily see that c1 = 1/sqrt5 c2 = -1/sqrt5

[@ozzyfromspace]

@@jasarinvorawathanabuncha6620 not true, c1 = (a-1)*(1+(a-1)/(b-a)) and c2 = -(a-1)*(b-1)/(b-a) where a is the positive eigenvector and b is the negative eigenvector of our problem. Also worth noting, the eigenvectors have the form x = [1/(lambda - 1), 1], not x = [-lambda, 1] as the professor wrote :) There were a few mistakes in the way to the solution so whatever answer we arrived at was simply not correct Lol I just watched the video now but obviously this is really late for you, hopefully someone else finds this useful. Best wishes friend

[@tanjiaqing1294]

@@ozzyfromspace [1/(lambda - 1), 1] = [-lambda, 1] in this case, u can plug in lambda = (1+sqrt(5))/2 and (1-sqrt(5))/2 to check. So the c1 and c2 got by @Jassarin should be correct.

[@Zoro3120]

In the computation of the Eigen values for A², he used A = SʌSˉ¹ to derive that ʌ² represents its Eigen value matrix. However this can be true only if S is invertible for A², which need not be always true. For example, for the matrix below (say A), the Eigen values are 1, -1(refer previous lecture). This would imply that A² has only one Eigen value of 1. This would imply that S has 2 columns which are same (if it has only one column then it is no longer square and hence inverse doesn't apply) and hence non invertible. This implies that this proof cannot be used for all the cases of the matrix A. _ _ │ 0 1 │ │ 1 0 │ ¯ ¯ Is there something I'm missing here?

[@hinmatth]

Please check 17:32

[@sviswesh3555]

Which null space does Prof mean, at @24:36 ?

[@atefehpeimani5488]

the null space of A - 2I which gives the answer for X in: (A - 2I)X = O

[@sviswesh3555]

@@atefehpeimani5488 got it, thanks! :)

[@safatkhan6839]

49:35 Why is F1 equal to 1? Isn't the first number in the sequence 0?

[@Jason-ke4jf]

Yeah, it starts from zero. F0 = 0, F1 = 1 and so on.

[@ashutoshtiwari4398]

First number in the sequence is F0 = 0 and second number is F1 = 1.

[@coffle1]

Can anyone direct me to a good proof for 19:36?

[@cartmansuperstar]

allegedly 26:00 , but i didn´t get it either...

[@АлександрСницаренко-р4д]

MIT, thanks you!

[@shamsularefinsajib7778]

Gilbert strang a great math teacher............

[@abhi220]

I have a doubt in Difference equations part. He writes u_0 as a combination of eigen-vectors of A. Why should this be true?

[@olfchandan]

eigen vectors span the entire space (Remeber - S is square invertible matrix). So, U0 will be a linear combination of eigen vectors.

[@suziiemusic]

A set of n independent eigenvectors, each one with n components, is a basis for Rn, and therefore any vector in Rn (including u0) can be written as a linear combination of these n eigenvectors. We could choose any other set of n independent vectors as a basis and do the same thing. The "standard" basis would be the columns of the identity matrix, which in 3 dimensions correspond to the x,y and z axes.

[@thedailyepochs338]

for anyone wondering how he turned the fibonaci sequence to a matrix @ 37:00 you are not alone , check this video out kzbin.info/www/bejne/n4exoHytjpWIjJo

[@aditiprasad5549]

Got stuck on the same thing ..Thanks for this!

[@anonymous.youtuber]

Just wondering...what keeps us from calling the eigenvector matrix E instead of S ? Is E already used for something else ?

[@Itsmeaboud]

Yes, it is used for elimination matrix.

[@rambohrynyk8897]

It always shits me how quickly the students clammer to get out of the class….how are you not absolutely dumbfounded by the profundity of what this great man is laying down!!!!

[@ashutoshtiwari4398]

Why the skew-symmetric matrix have zero or imaginary eigenvalue?

[@sidaliu8989]

By the definition of the skew-symmetric matrix (A^T=-A), all entries in the diagonal of the matrix must be 0. So when we come up with the characteristic equation, it will be lambda^n+b^2=0 (since the trace is zero and the determinant is some square), and this will give us pure imaginary solutions if b^2>0.

[@ricardocesargomes7274]

Thanks for uploading.!

[@utxeee]

So, the second component of u(k+1) is useless, right? The actual value is given by the first component.

[@thovinh5386]

Yep, you can use u(k) = u(k) and it still works.

[@jamesmcpherson3924]

I had to pause to figure out how he got the eigenvectors at the end. Plugging in Phi works but it wasn’t until I watched again that I noticed he was pointing to the lambda^2-lambda-1=0 relationship to reveal the vector.