After 20 years, it turns out that I was fooled by my Math teachers for not sharing these beautiful facts in linear algebra. Thank you Dr. Peyam, you are the best.
I guess your math teachers didn't know any better, haha. I, too, loved this video and had never seen this before. It was quite exciting. It's got me even more interested in block matrices now and possibly even embedding operators inside the theoretical continuous matrix representations of operators. I'm going to try to put some thought into that now. You never know. What I found interesting about this video is that it shows that a new decomposition of A of the same type was created by just jumbling the numbers (almost, anyway). For those who don't know how to do obtain the decomposition of A Dr. Peyam presented, all you need to do is take the real and imaginary parts of a Av = (lambda)v for one of the lambdas, treat them as two separate equations, and then join the two equations together into a matrix equation.
Muchas Gracias Dr. Peyam. Como Estadístico aplicado, normalmente uso matrices simétricas y, claro, los eigenvalores son reales. Pero...disfruté mucho este video, muy interesante.
i was once shown the wonderful 'complete the square' technique, and i've ditched the dull quadratic formula ever since. this together with vieta's formulas will make anyone the master of solving quadratic equations in no time! ;)
sidenote: i've recently found out that the roots of the polynomial Q(x) = a_0 x^n + a_1 x^{n-1} ... + a_{n-1} x + a_n are the reciprocals of the nonzero roots of the polynomial P(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0, i.e. if α ≠ 0 and P(α) = 0, then Q(1/α) = 0. so for example: since the roots of P(x) = x^2 - 5x + 6 are α=2 and α=3, the roots of Q(x) = 6x^2 - 5x + 1 are α=1/2 and α=1/3. how cool is that?! :)
Remembering that the eigenvalues can be complex, and what is a rotational matrix, are very useful insights when dealing with N-D Complex & Circular numbers. For example, the matrix representation for a 3-D complex number (j³ = - 1) is: [ a; - c; - b ] [ b; a; - c ] [ c; b; a ] While for a "circular" number (j³ = 1) is: [ a; c; b ] [ b; a; c ] [ c; b; a ] And, to easily calculate logs, exps, powers and roots, the diagonalization is a must-have in hand, where the eigenvalues are complex numbers. IIRC one matrix P for the 3-D complex numbers is: [ + 1; w²; - w ] [ - 1; w; w² ] / sqrt(3) [ + 1; 1; 1 ] where w := (- 1)^(1 / 3) = exp{i pi / 3}
Hello Sir, Thank you for the session. Can you please suggest a way to find a 2x2 matrix whose eigen values are complex numbers and eigen vectors are not known. Thank you
Does "null space of [ 1+3i , 5 2 , -1+3i ]" mean "the Kernel of the space generated by the vectors ( (1+3i,2) , (5, -1+3i) )" ? I'm learning in another country with different notations, so I want to be sure
I wonder... since complex numbers are isomorphic to the 2x2 matrix SR, what do quaternions look like? I expect a 4x4 matrix, but does it truly map correctly?
@@thedoublehelix5661 well in the time since i'd already learned about quaternions in matrix form, and they are indeed 4x4 matrices, equivalent to the 2x2 complex matrix (a+bi, c+di, -c+di, a-bi)
@@MrRyanroberson1 okay thats sort of rude to suggest that I didn't do my research because I did. According to Wikipedia the set of unit quaternions forms a double covering of SO(3, R)
After 20 years, it turns out that I was fooled by my Math teachers for not sharing these beautiful facts in linear algebra. Thank you Dr. Peyam, you are the best.
I guess your math teachers didn't know any better, haha. I, too, loved this video and had never seen this before. It was quite exciting. It's got me even more interested in block matrices now and possibly even embedding operators inside the theoretical continuous matrix representations of operators. I'm going to try to put some thought into that now. You never know. What I found interesting about this video is that it shows that a new decomposition of A of the same type was created by just jumbling the numbers (almost, anyway). For those who don't know how to do obtain the decomposition of A Dr. Peyam presented, all you need to do is take the real and imaginary parts of a Av = (lambda)v for one of the lambdas, treat them as two separate equations, and then join the two equations together into a matrix equation.
Muchas Gracias Dr. Peyam. Como Estadístico aplicado, normalmente uso matrices simétricas y, claro, los eigenvalores son reales. Pero...disfruté mucho este video, muy interesante.
Woah (cue Dr Peyam waoh clip) never would've guessed that you could do something like that with matrices
HAHAHAHAHAHAHA I know what that "woah clip" is!!!
I’ve become a living meme 😂
The best living meme
"I think it actually works in infinite dimensional spaces and such." Yup! Useful with Hilbert spaces in quantum mechanics.
Example of this happening in hilbert spaces in quantum mechanics?
i was once shown the wonderful 'complete the square' technique, and i've ditched the dull quadratic formula ever since. this together with vieta's formulas will make anyone the master of solving quadratic equations in no time! ;)
sidenote: i've recently found out that the roots of the polynomial Q(x) = a_0 x^n + a_1 x^{n-1} ... + a_{n-1} x + a_n are the reciprocals of the nonzero roots of the polynomial P(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0, i.e. if α ≠ 0 and P(α) = 0, then Q(1/α) = 0.
so for example: since the roots of P(x) = x^2 - 5x + 6 are α=2 and α=3, the roots of Q(x) = 6x^2 - 5x + 1 are α=1/2 and α=1/3.
how cool is that?! :)
Wow! That's awesome to learn!
I'm sure some day in the near future it'll be very useful for me!
The quality of video is excellent. (1080p60)
Remembering that the eigenvalues can be complex, and what is a rotational matrix, are very useful insights when dealing with N-D Complex & Circular numbers.
For example, the matrix representation for a 3-D complex number (j³ = - 1) is:
[ a; - c; - b ]
[ b; a; - c ]
[ c; b; a ]
While for a "circular" number (j³ = 1) is:
[ a; c; b ]
[ b; a; c ]
[ c; b; a ]
And, to easily calculate logs, exps, powers and roots, the diagonalization is a must-have in hand, where the eigenvalues are complex numbers.
IIRC one matrix P for the 3-D complex numbers is:
[ + 1; w²; - w ]
[ - 1; w; w² ] / sqrt(3)
[ + 1; 1; 1 ]
where w := (- 1)^(1 / 3) = exp{i pi / 3}
This series related to matrices are awesome!!
Sir I just become your fan.
ooohh Thanks Dr. Tigre Peyam, you are the best!!!
You have a weird accent but it doesn't matter. Your teaching is brilliant.
The explanation is clear. Maybe I will embed some of your linear algebra videos in my courses.
Thank you! 🙂
Thank you so much! It is very clear!
Thanks a lot D peyam السلام عليكم
Hello Sir,
Thank you for the session.
Can you please suggest a way to find a 2x2 matrix whose eigen values are complex numbers and eigen vectors are not known.
Thank you
Does "null space of
[ 1+3i , 5
2 , -1+3i ]"
mean "the Kernel of the space generated by the vectors ( (1+3i,2) , (5, -1+3i) )"
?
I'm learning in another country with different notations, so I want to be sure
Correct
The star is the same thing as "dagger" notation right? If you're talking about conjugate transposes
Sir would this trick work for generic n*n case? Or is it limited only to 2*2?
The trick only for 2x2, but the theorem at the end for nxn
This should not be "surprising", because you can identify the complex numbers with R^2
Prisma d, Karalho
Please example for w=ui_vi
Parabéns felicidades somatorias de Sucessos MDC MAX MMC MMX a aula tá dentro está-r dentro... Kells
I wonder... since complex numbers are isomorphic to the 2x2 matrix SR, what do quaternions look like? I expect a 4x4 matrix, but does it truly map correctly?
3x3 SR matrices right?
@@thedoublehelix5661 well in the time since i'd already learned about quaternions in matrix form, and they are indeed 4x4 matrices, equivalent to the 2x2 complex matrix (a+bi, c+di, -c+di, a-bi)
@@MrRyanroberson1 I thought they're meant to encapsulate 3d rotation so 3x3 matrices seemed like the way to go
@@thedoublehelix5661 just look up quaternions it's not that hard to use wikipedia or google
@@MrRyanroberson1 okay thats sort of rude to suggest that I didn't do my research because I did. According to Wikipedia the set of unit quaternions forms a double covering of SO(3, R)
Ah yes, real Jordan Normalform joins the fight
Implicação d, Karalho
Apogeu d, Karalho
First, yeaah
13:30 wtf