EECS 16B Lectures - Shared screen with speaker view
Lol sorry I’m late
is that Bnew, I?
So if the diagonalized system is unstable, then the system is unstable as well?
Gaurav Rohit Ghosal
What about continuous time systems?
does n have anything to do with dimensions of A?
yes, A is n x n
Is A the same as the diagonal Eigen value matrix?
Or is A the original matrix?
A is the original
Since we’re talking now about matrices that can’t be made into the diagonal eigenvalue form
is true that for all upper diagonal matrices, you can read the eigenvalues off the diagonal?
Don't we also have to take into account the input for stability conditions here?
@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)
@ Ayush No, we just need |lambda|<1 for stability. As long as that's true, z(t->infty) is bounded.
@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
@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
The upper triangular matrix is written with n lambdas across the diagonal. Are we assuming that there are n distinct eigenvalues?
Do the lambdas just refer to the values across the diagonal then?
@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?
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).
What about when one of the eigenvalues is equal to 1?
@Hetal Then bounded input can cause unbounded output.
Aha, OK, I see what you were saying now @Carter. Thank you!
@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"
@Ayush My pleasure!
Couldn't we also put it into a lower diagonal form? If so, why do we go with upper diagonal instead?
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?
@Ryan Remember how we get the Eigenvalues as a result of det(Lambda * I - A)?
We get an equation then solve for 0
I think they would be equivalent mathematically and you could still reach the same conclusion about how the eigenvalues must be less than 1
How do we know there are n eigenvalues?
We aren't assuming they're distinct
But we're assuming there are n?
Never heard of cofactor
we only did determinant for 2D matrix
read??? whats that
that would bbe great!
We did learn about the process of det(A - Lambda*I) = 0
could also jusut look at it in terms of invertibility
What do we do in the case when the A has fewer than n eigenvalues?
Every n x n matrix has n eigenvalues
A - lambdaI is not invertible if any of the diagonal entries are zero -> lambda must be one of the eigenvalues
We can have fewer than n eigenvalues, especially if there are duplicates
this would not be the case if we restrict to the reals though right?
I thought over complex numbers we can only guarantee at least one eigenvalue?
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.
what is the process to upper diagonalize a matrix?
Matrices created from Eigenvalues are automatically upper diagonalizable (like Sigma and Lambda, usually what we see)
*Matrix Big Lambda that represents a matrix form of all Eigenvalues
think we used tau
same conditions as discrete case
Negative lambdas will converge
lambda < 0
Lambda < 0 stable
negative remains bounded
Lambda < 1
Haha wait < 0
If the real part is 0, is it stable or unstable?
marginally stable (basically stable only if no inputs)
love the writing ASMR
don't we usually do A - lambda*I?
Its an equivalent form
Yup, the extra negatives would cancel.
How do we know term in square root will be less than k/m?
Works out the same
Because we're subtracting something positive
Or nonnegative, I suppose
@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
very pretty diagram :)
GOOD LUCK ON THE MT EVERYONE!!!
YALL GOT THIS
Good luck tonight y’all!!
thank you guys!