Not only did I learn how to find Eigenvalues, I also learned how to factor a cubic function! Thank you so much for all the good videos Sal, they help me out a lot!
I'm so glad this video is online, I'm currently taking a class on Quantum Mechanics (and I'm only 17), and the matrix algebra is incredibly complex, so this made it easier to understand...
First of all, thank you for the great tutorial. Though, I prefer (A- lamda l) but offcourse both work. I suggest everyone to use horner's method to reduce the equation. Way faster and in my opinion easier ;).
Sal, can you PLEASE make a multi variable calc playlist? And thank you so much for everything! I think it's hard to comprehend the help you are giving us all!
Ive spent a good 3 hours on a question trying to calculate eigenvalues and then I found this video and worked it out straight away! Thumbs up and thank you
@TerminatorSe7en You can do it either way; given that you derive the equation from the definition of an eigenvalue, where A(v)=lambda(v), lambda(v) - A(v) = 0, but also, A(v) - lambda(v) = 0. So, you can use either det(A-lambda In), or det(lambda In - A), they're effectively the same thing.
+Prince K3V Either works. -det(lambda-A)=det(A-lambda). You just factor out a -1. In both cases you are setting it equal to zero. So you could also do 2det(A-lambda)=0 which would be equivalent to det(2A-2lambda)=0
I felt it a lot easier to row-reduce before finding the determinant. Taking Row3 - Row2, we're left with a zero in the bottom-left element (3, 1). Factor out (3 - L) from that row, and expanding across the bottom row, it isn't as much of a mission as the one in the video, with an L^3, finding roots, etc. But hey, that's just me! Many thanks, Sal!
Shawn Ryan Yep row reducing the matrix before the A-(Lamda)i or (Lambda)i-A step will change its eigen values. But Amish means row reducing after doing A-(Lamda)i or (Lambda)i-A to make finding the determinant easier, as row reducing does not change a matrix determinant, and thus you will get the same eigenvaues, hope this makes sense
WOW! I missed it too. Everyone ignore this comment like the original commentor said to. We all thought it said minus lambda^2 but it actually is "+ lambda^2"
If you use the row echelon for of Ix-A to make the second row and the first column zero, then you can easily factorize x-3. The way in this video may not work if the solutions are not integers (so we cant guess the roots).
@Crilleakaroffe Unfortunately, an incorrect correction. Rewind the video to before he circles the coefficients and multiply it out; he has got it right but accidentally circled over the + sign to make it look like a - sign later on in the vid (when we all--in our heads-- go back and double check our calculations for errors like this)
What's the rule he used to find the determinant? I'm not familiar with it, does it work for all matrices or is this a special case due to size or because A = Transpose(A) ?
The roots are suppose to be both positive or negative right? But awesome video saved my ass on so many occassions lol. Wish me luck on my finals in two days =/
First of all, I continue to be amazed and thankful for all the videos. Just brilliant and so calmly and precisely explained. I do have a question about the solving of the characteristic equation which was cubic in lambda. I don't disagree with his method and it's simple and commonsensical so my question in based solely on curiosity. When I first saw the equation, I though immediately it has to be (lambda - 3) squared times (lambda + 3). Why? Explanation: When we have a polynomial in standard form, where the first coefficient is 1, we think through the structure as the first term having a power of n, the next a power of n-1, until we get to the last term where there is no independent variable and hence the power is n-n or zero. However, this view only takes the independent variable, in this case lambda, into account. We could simultaneously think of the power of scalars being multiplied by each other and these are in reverse order when compared to the independent variable. In other words, the first term is scalars to the 0th power, hence the standard coefficient of 1, the next would be to the power of 1, the next to 2, and so on. In the case of a cubic polynomial, the final scalar power, i.e., in the last term, would be 3. (Note that by "power" I mean the number of times scalars are being multiplied together. I'm providing all this context because when I saw the characteristic equation, my immediate thought was simply, what number would multiply out to +27, and if two of them, to -9, and if one of them, to -3? It seemed clear from the signs that there has to be at least one negative term, and for the final term to be +27, there must be two negative terms that therefore cancel, and the only three numbers that can would be -3, -3, and +3 since no other combination of or subset of combinations (subsets due to the lower power of other terms in the characteristic equation) would lead to -9 or -3. Hence, the answer upon inspection and without real calculations must be lambda minus three twice (hence squared) and lambda plus three. My question is, was this the right instinctive approach to get such an answer instantly or was it just dumb luck? For example, using my method, if the characteristic equation would have been lambda cubed - 4 lambda squared - 16 lambda + 64 lambda without really thinking much at all you could see it must be two factors of lambda - 4 and one factor of lambda + 4 for the same reason as outlined above. Interesting trick or out to lunch? I have thick skin so I welcome any corrections...
He probably has a writing tablet that gives the same inputs as a mouse. I have one. You simply hover over the tablet to move your cursor, and touch-down to click. There's a side-button on the stylus to right-click.
why are you not my linear algebra professor. My idiot professor during lecture just told us "the characteristic polynomials of a 3x3 will be given because it's just too complicated." Lo and behold, on the test, it was not given. Even the book he uses said that this should be given, but no examples were given. Then he showed us his solution, which skips this whole step because he just copied it out of some textbook that didn't explain it.
Hey, I have a question. When I use the 'usual' method for solving the determinant, I got : (labda + 1)(labda^2- 4labda + 4 - 1) < I miss the minus 1 in the method you are using... Can someone explain to me where it went?
isnt -4x^2+-1x^2 = -5x^2. he put -3x^2 when solving for the characteristic equation (p(x) = x^3 - 3x^2 -9x + 27). I just wanted to point that out if you are in fact trying to learn from this
If you only have two eigenvalues, you technically still have three, because you have a repeated Eigenvalue. If a cubic has a root on the x-axis that coincides with a turning point, it technically counts as two roots, or a repeated root.
U can do both, but i prefer using A - (Lambda)i = 0 as you only need to minus lambda from the diagonals of A as opposed to minusing every value of A from (Lambda)i
I think there is a mistake in the video. the matrix is symmetric and non triangular, so there shouldn't be any repeated lambdas. my det was (L for lambda) L^3-3L^2-L+3=0 from here L1=3,L2=1, L3=-1. and wolfram says the same.
Great explaination. i would be careful to include the negative integers in the factors of the 27 that you used since some of solutions only contain negative values.
Not only did I learn how to find Eigenvalues, I also learned how to factor a cubic function! Thank you so much for all the good videos Sal, they help me out a lot!
thats great but for people that have exams in 8 hours it would be nice if he simplified it
@@AriKariG did you pass? haha
@@admirald.rifter1819 5 years later lmao
@@AriKariG i want to know too. Did you pass? Have you graduated?
since this video was created in 2009, indeed the explanation is beautiful, easy to be recalled even after 11 years.
I'm so glad this video is online, I'm currently taking a class on Quantum Mechanics (and I'm only 17), and the matrix algebra is incredibly complex, so this made it easier to understand...
12 years later 👀
I've always learned as "A - lambda times I", so all you do is substract a lambda in each of the matrix diagonal.
That's what that means
yeah, same thing, but you get slightly uglier polynomials. Lambda - A is cleaner usually
@@kaya-sem Dude, it's been 12 years. I don't even know what a matrix is anymore.
@@samus88 bruhhh
@@abhiramkrishnan7202 For real tho, I haven't been in a math class in over a decade and I legit don't even know what this video is about xD
First of all, thank you for the great tutorial. Though, I prefer (A- lamda l) but offcourse both work. I suggest everyone to use horner's method to reduce the equation. Way faster and in my opinion easier ;).
Sal, can you PLEASE make a multi variable calc playlist? And thank you so much for everything! I think it's hard to comprehend the help you are giving us all!
Ive spent a good 3 hours on a question trying to calculate eigenvalues and then I found this video and worked it out straight away! Thumbs up and thank you
@TerminatorSe7en You can do it either way; given that you derive the equation from the definition of an eigenvalue, where A(v)=lambda(v), lambda(v) - A(v) = 0, but also, A(v) - lambda(v) = 0. So, you can use either det(A-lambda In), or det(lambda In - A), they're effectively the same thing.
Thank you
so.. much... writing... but still clearer than my prof. thanks man!
12:07 I just came over from professor Dave's video on this. Really helpful
This guy is good. He proves that Math is not a secret society .
Why couldn't Patrick jmt make a video on this :(
he has man, he did it on a 2x2 matrix it is the same process just the determinant will be a bit more tedious to calculate
@@azzzzzz640 not the same process at all
@@awesome7732 elaborate
Even in the year 2021, Sal Khan is still a lifesaver.
Save me God from this on exam. :|
nah it was easy
realfan91 Hope i can say the same tomorrow :D
good luck mate
thx so much
And passed it :D
You are my hero!
Also:
12:39
That's what she said!
I come from 2020. Welcome.
No he is right, I thought the same thing but when he circled the +lambda^2 term it looked like it was negative but it wasnt
yeah i also think its det(A-lambda), all my uni text books say that for finding eigenvalues and also later on when diagonlising matrices
thnks for your helps it makes me feel good and iam happy now after i have watched this vedio . please make it more for benefit
Shouldn't it be A-(lamda).(I)?
+Prince K3V It is equal to zero anyway
+Prince K3V Either works. -det(lambda-A)=det(A-lambda). You just factor out a -1. In both cases you are setting it equal to zero. So you could also do 2det(A-lambda)=0 which would be equivalent to det(2A-2lambda)=0
My teacher just taught us (a-lambaI) and now I'm confused.
This has made my work easier! Thanks
@rb44 if you times both sides by -1 then it will be. Either notation is right.
I felt it a lot easier to row-reduce before finding the determinant. Taking Row3 - Row2, we're left with a zero in the bottom-left element (3, 1). Factor out (3 - L) from that row, and expanding across the bottom row, it isn't as much of a mission as the one in the video, with an L^3, finding roots, etc.
But hey, that's just me! Many thanks, Sal!
Shawn Ryan
Yep row reducing the matrix before the A-(Lamda)i or (Lambda)i-A step will change its eigen values. But Amish means row reducing after doing A-(Lamda)i or (Lambda)i-A to make finding the determinant easier, as row reducing does not change a matrix determinant, and thus you will get the same eigenvaues, hope this makes sense
WOW! I missed it too. Everyone ignore this comment like the original commentor said to. We all thought it said minus lambda^2 but it actually is "+ lambda^2"
great 🧎
@kramaster3
Yea its actually correct either way as they're both derived from Ax=λIx, i.e
Ax=λIx
=> 0=λIx-Ax
=> 0=(λI-A)x
out of all videos this is the most useful
go back awhile ago, and check before he encircled the +lambda^2
thank you , that was really clear to understand. Much better than my professor.
@rb44 I think (lamda I - A) is correct because i've seen it some of the other examples, but i guess it doesn't matter which way you put it :)
If you use the row echelon for of Ix-A to make the second row and the first column zero, then you can easily factorize x-3. The way in this video may not work if the solutions are not integers (so we cant guess the roots).
Meant to be a +ve lambda in the row before so it should be
-4lambda^2 + lambda^2 instead of -4lambda^2 - lambda^2
@Crilleakaroffe Unfortunately, an incorrect correction. Rewind the video to before he circles the coefficients and multiply it out; he has got it right but accidentally circled over the + sign to make it look like a - sign later on in the vid (when we all--in our heads-- go back and double check our calculations for errors like this)
This just blew my mind! Thank you!
Thank you ! really helpful.
thank you, possibly just saved my degree :)
Thank you, thank you and thank you...
Really, you are a hero, sir.
All my respect and god bless you!!!
Thanks Very Easy TO Understand
What is the "rule of Saris" at 5:02? I know how the rule works, but I've never heard it called that. Did I spell Saris correctly, and who is Saris?
I believe it's spelled Saurus
P(λ)= (A+(−λ*determinant))
This is only for symmetric matrices, way to be extremely specific and not give a general solution..
Great video, thank you very much!!!!!!
This guy is a HERO
What's the rule he used to find the determinant? I'm not familiar with it, does it work for all matrices or is this a special case due to size or because A = Transpose(A) ?
if you do X-lambda you do not need to reverse the signs of the other values, correct?
I liked the video before watching it coz I knew I am gonna love it :)
The roots are suppose to be both positive or negative right? But awesome video saved my ass on so many occassions lol. Wish me luck on my finals in two days =/
I like it .am finalizing this course this semester
i wish u did a video related to DIAGONALIZATION......
very useful for my semester
Bruh. This video. Yes.
The video was very helpful
amazing content. !!! Thank you
First of all, I continue to be amazed and thankful for all the videos. Just brilliant and so calmly and precisely explained. I do have a question about the solving of the characteristic equation which was cubic in lambda. I don't disagree with his method and it's simple and commonsensical so my question in based solely on curiosity. When I first saw the equation, I though immediately it has to be (lambda - 3) squared times (lambda + 3). Why? Explanation: When we have a polynomial in standard form, where the first coefficient is 1, we think through the structure as the first term having a power of n, the next a power of n-1, until we get to the last term where there is no independent variable and hence the power is n-n or zero. However, this view only takes the independent variable, in this case lambda, into account. We could simultaneously think of the power of scalars being multiplied by each other and these are in reverse order when compared to the independent variable. In other words, the first term is scalars to the 0th power, hence the standard coefficient of 1, the next would be to the power of 1, the next to 2, and so on. In the case of a cubic polynomial, the final scalar power, i.e., in the last term, would be 3. (Note that by "power" I mean the number of times scalars are being multiplied together. I'm providing all this context because when I saw the characteristic equation, my immediate thought was simply, what number would multiply out to +27, and if two of them, to -9, and if one of them, to -3? It seemed clear from the signs that there has to be at least one negative term, and for the final term to be +27, there must be two negative terms that therefore cancel, and the only three numbers that can would be -3, -3, and +3 since no other combination of or subset of combinations (subsets due to the lower power of other terms in the characteristic equation) would lead to -9 or -3. Hence, the answer upon inspection and without real calculations must be lambda minus three twice (hence squared) and lambda plus three. My question is, was this the right instinctive approach to get such an answer instantly or was it just dumb luck? For example, using my method, if the characteristic equation would have been lambda cubed - 4 lambda squared - 16 lambda + 64 lambda without really thinking much at all you could see it must be two factors of lambda - 4 and one factor of lambda + 4 for the same reason as outlined above. Interesting trick or out to lunch? I have thick skin so I welcome any corrections...
nice tutorial
but does method of multiplication and subtraction to get the characteristic polynomial works with 4x4 matrix??
@Crilleakaroffe why is it -5lamda^2?
how is his handwriting so good with a mouse?!
He probably has a writing tablet that gives the same inputs as a mouse. I have one. You simply hover over the tablet to move your cursor, and touch-down to click. There's a side-button on the stylus to right-click.
thank you sal. you are always the best.
OH MY GOD, THANK YOU FOR YOUR EXISTENCE.
We were taught by using A-λI, but why does this show λI-A?
Jeremy Tan After reading the comment's, I got my answer =)
😂
Thank you very much :)
You saved my life
no mate the yellow lambda^2 is positive not negative, so it gives a sum of 3lambda^2
arent the values on A matrix substracted by the lambda matrix and not the other way around?
Obviously not an Engineer are you.
Use it ALL. THE. TIME.
why are you not my linear algebra professor. My idiot professor during lecture just told us "the characteristic polynomials of a 3x3 will be given because it's just too complicated." Lo and behold, on the test, it was not given. Even the book he uses said that this should be given, but no examples were given. Then he showed us his solution, which skips this whole step because he just copied it out of some textbook that didn't explain it.
Hey, I have a question. When I use the 'usual' method for solving the determinant, I got : (labda + 1)(labda^2- 4labda + 4 - 1) < I miss the minus 1 in the method you are using...
Can someone explain to me where it went?
isnt -4x^2+-1x^2 = -5x^2. he put -3x^2 when solving for the characteristic equation (p(x) = x^3 - 3x^2 -9x + 27). I just wanted to point that out if you are in fact trying to learn from this
just little silly mistakes haha, but yeah i do see it, its at @8:50 if anyone is wondering
he had the wrong sign before it is actually correct
is the way he solved for a factor of the 3rd degree polynomial always supposed to work if the polynomial can be factored?
If it can be factored, yes. If it can't be factored, you likely need to use a numeric method, or the monster cubic formula.
@rb44
it doesn't matter if it's the opposite way
I forgot the division method. Is there another method to find the eigenvalues?
Maheen Hashmi Yeah, you could factor by grouping!
Why do you only have two eigen values for a 3x3 matrix?
If you only have two eigenvalues, you technically still have three, because you have a repeated Eigenvalue. If a cubic has a root on the x-axis that coincides with a turning point, it technically counts as two roots, or a repeated root.
Thanks. It's actually all just number crunching if you understand the equation that difines eigenvalues.
is it (A-lambaI)=0 or is it (lambaI-A)=0 ? normally we need to take the absolute value..but you didnt? whyy?
U can do both, but i prefer using A - (Lambda)i = 0 as you only need to minus lambda from the diagonals of A as opposed to minusing every value of A from (Lambda)i
Oh my god... you are always keeping up with my lectures all the time!! we just started this unit today!!!
u still alive???
@@fahadmohamed346 haha yes. I am now 29
@@NotmyYTchannel lol replying after all these years, what a legend
p(lamdba) = lamdba^3 - 5lamdba^2 - 9lamdba + 27 = 0
Correction?
Thank you so much for sharing this it really helped me understand
@rb44 It can be both.
thanks
thnkq for the lemda how to find .
I think there is a mistake in the video. the matrix is symmetric and non triangular, so there shouldn't be any repeated lambdas. my det was (L for lambda) L^3-3L^2-L+3=0 from here L1=3,L2=1, L3=-1.
and wolfram says the same.
+Oshanii Yea I believe the second (llambda + 1) was supposed to be 2(llambda + 1).
This is what I got aswell. -1,1,3
Na he's right. Found my mistake.
Im looking at my notes and its the other way around. I mean that instead of boing (lambda)I - A it's A - (ambda)i
What about eigen vectors
OMGOSH, THE CALCULATIONS :O
practice it, you will get much faster.
this video was good but starting at 11:32 it started to be badly explain im sorry
Thanks alot man :)
for the record i do not think he's a clown, prob much smarter than myself
is it (A-lambaI)=0 or is it (lambaI-A)=0 ?
wont finding the determinant by cofactor be easier?
both, those statements are equivalent
7 months later i realize that the above "correction" is an incorrect correction. IGNORE!
What about the other factors of 27, will those give more eigenvalues?
Great explaination. i would be careful to include the negative integers in the factors of the 27 that you used since some of solutions only contain negative values.
amazing.
is it just me or this guy the guy from ownageprank calls
Depending on your field you might be surprised.
what's the name of the software used?
Synthetic division would have sped things up.
Much easier to do det(A-yI)=0 it avoids all the negatives