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EECS 16B Lectures - Shared screen with speaker view
Ayush Sharma
27:42
JAKE! :D
Vanshaj Singhania
27:46
jake!
Jake Whinnery
27:58
Lol sorry I’m late
Subham Dikhit
28:39
is that Bnew, I?
Anton Zabreyko
29:01
So if the diagonalized system is unstable, then the system is unstable as well?
Anton Zabreyko
29:47
Yes!
Gaurav Rohit Ghosal
30:38
What about continuous time systems?
Seth SANDERS
30:52
Hang on!
Stephen Wang
31:24
does n have anything to do with dimensions of A?
Seth SANDERS
31:43
yes, A is n x n
Sarina Sabouri
34:10
Is A the same as the diagonal Eigen value matrix?
Sarina Sabouri
34:22
Or is A the original matrix?
Anton Zabreyko
34:28
A is the original
Benjamin Tait
34:49
Since we’re talking now about matrices that can’t be made into the diagonal eigenvalue form
Ryan Zhao
36:54
is true that for all upper diagonal matrices, you can read the eigenvalues off the diagonal?
Ayush Sharma
37:34
Don't we also have to take into account the input for stability conditions here?
Oscar Chan
38:49
@Ryan, for an upper diagonal matrices, yes, you can just read the numbers off the diagonal as the Eigenvalues (because the other non lambda terms 0s out)
Carter Fogelman
40:57
@ Ayush No, we just need |lambda|<1 for stability. As long as that's true, z(t->infty) is bounded.
Oscar Chan
41:20
@Ryan Correction: the terms that are not along the diagonal will 0 out due to a zero existing at least once from the bottom left part of the matrix
Benjamin Tait
41:21
@Ryan the definition of upper diagonal is a matrix such that the eigenvalues are on the diagonal, and just also has other values above the diagonal
Balaji Veeramani
41:27
The upper triangular matrix is written with n lambdas across the diagonal. Are we assuming that there are n distinct eigenvalues?
Carter Fogelman
41:47
@Balaji no
Balaji Veeramani
42:03
Do the lambdas just refer to the values across the diagonal then?
Bryan Ngo
43:04
yea
Ayush Sharma
43:12
@Carter Oh huh, really? My thinking is that even if the input is held constant at 1 for all timesteps and |lambda| < 1, then every step we'll be increasing by at least 1 because of the input, right?
Balaji Veeramani
43:22
Okay, thanks
Oscar Chan
44:17
The det(Lambda * I - A) equation can be read off the diagonal of an upper diagonal matrix. As a result, you can get the eigenvalues. Any properties will then have to be determined as a result of seeing the eigenvalues has to be determined afterwards (by experience, you can find properties by inspection, but you gotta know the patterns first).
Hetal Shah
45:33
What about when one of the eigenvalues is equal to 1?
Carter Fogelman
45:57
@Hetal Then bounded input can cause unbounded output.
Hetal Shah
46:15
Thank you
Ayush Sharma
47:36
Aha, OK, I see what you were saying now @Carter. Thank you!
Ryan Zhao
48:01
@Oscar, could you clarify how you do this? "The det(Lambda * I - A) equation can be read off the diagonal of an upper diagonal matrix"
Carter Fogelman
48:38
@Ayush My pleasure!
Anton Zabreyko
48:47
Couldn't we also put it into a lower diagonal form? If so, why do we go with upper diagonal instead?
Gilbert Feng
49:16
How do we get a matrix into upper diagonal form? Why is it guarranteed that matrices (even non-diagonalizable ones) have that upper diagonal form?
Oscar Chan
49:17
@Ryan Remember how we get the Eigenvalues as a result of det(Lambda * I - A)?
Oscar Chan
49:27
We get an equation then solve for 0
Benjamin Tait
49:34
I think they would be equivalent mathematically and you could still reach the same conclusion about how the eigenvalues must be less than 1
Balaji Veeramani
49:45
How do we know there are n eigenvalues?
Carter Fogelman
50:04
We aren't assuming they're distinct
Balaji Veeramani
50:09
But we're assuming there are n?
Anton Zabreyko
50:14
No
Ryan Zhao
50:16
no
Gilbert Feng
50:17
No
Oscar Chan
50:20
Never heard of cofactor
Subham Dikhit
50:21
no
Ryan Zhao
50:22
we only did determinant for 2D matrix
Subham Dikhit
50:44
read??? whats that
Ryan Zhao
50:46
that would bbe great!
Benjamin Tait
50:47
We did learn about the process of det(A - Lambda*I) = 0
Kunaal Sundara
50:53
could also jusut look at it in terms of invertibility
Balaji Veeramani
50:57
What do we do in the case when the A has fewer than n eigenvalues?
Felix Yu
51:09
:d
Alex Yang
51:15
O.o
Seth SANDERS
52:00
Every n x n matrix has n eigenvalues
Kunaal Sundara
52:14
A - lambdaI is not invertible if any of the diagonal entries are zero -> lambda must be one of the eigenvalues
Benjamin Tait
52:40
Yes
Oscar Chan
52:44
We can have fewer than n eigenvalues, especially if there are duplicates
Kunaal Sundara
53:27
this would not be the case if we restrict to the reals though right?
Balaji Veeramani
55:27
I thought over complex numbers we can only guarantee at least one eigenvalue?
Oscar Chan
56:17
Ooh, I just remembered what was cofactor. We only used Jupyter Notebooks in EECS 16A with Numpy to solve for matrices at and more than 3 x 3, so we never talked about it.
Gilbert Feng
56:47
what is the process to upper diagonalize a matrix?
Oscar Chan
58:01
Matrices created from Eigenvalues are automatically upper diagonalizable (like Sigma and Lambda, usually what we see)
Oscar Chan
58:47
*Matrix Big Lambda that represents a matrix form of all Eigenvalues
Stephen Wang
59:10
think we used tau
Bryan Ngo
01:00:00
same conditions as discrete case
Ashwin Rammohan
01:00:04
positive lambda
Vainavi Viswanath
01:00:09
Negative lambdas will converge
Connie Shi
01:00:11
lambda < 0
Zhiping Gu
01:00:11
Lambda < 0 stable
Ryan Zhao
01:00:13
negative remains bounded
Anton Zabreyko
01:00:13
Lambda < 1
Anton Zabreyko
01:00:31
Haha wait < 0
Vainavi Viswanath
01:05:29
If the real part is 0, is it stable or unstable?
Connie Shi
01:05:48
marginally stable (basically stable only if no inputs)
Stephen Wang
01:05:59
love the writing ASMR
Stephen Wang
01:16:26
don't we usually do A - lambda*I?
Benjamin Tait
01:16:41
Its an equivalent form
Ayush Sharma
01:16:51
Yup, the extra negatives would cancel.
Sarina Sabouri
01:16:55
How do we know term in square root will be less than k/m?
dylanbrater
01:16:55
Works out the same
Carter Fogelman
01:17:11
Because we're subtracting something positive
Carter Fogelman
01:17:28
Or nonnegative, I suppose
Benjamin Tait
01:25:31
@Sarina because sqrt of (K/m)^2 normally is k/m, so subtracting from (K/m)^2 within then taking the sqrt would result in a value less than k/m
Stephen Wang
01:28:23
cos?
Ryan Zhao
01:35:45
very pretty diagram :)
Ayush Sharma
01:35:53
^^^
Subham Dikhit
01:44:50
GOOD LUCK ON THE MT EVERYONE!!!
dylanbrater
01:44:57
Thanks
Jennifer Zhou
01:45:01
YALL GOT THIS
Vanshaj Singhania
01:45:10
gooooood luck!
Ayush Sharma
01:45:11
Thank y'all!
Jake Whinnery
01:45:11
Good luck tonight y’all!!
Stephen Wang
01:45:11
thanks professors
Kunaal Sundara
01:45:14
thanks!
Eugenia Chien
01:45:16
thank you!
Subham Dikhit
01:45:16
thanks professors
Vanshaj Singhania
01:45:17
thank you guys!
Abhik
01:45:18
thank you!
Anton Zabreyko
01:45:18
Thanks everyon!
Gilbert Feng
01:45:19
thank you!