I'm Dave Friday, and I teach mathematics for Macomb Community College in Macomb County, MI, USA.
The majority of what you can find on my channel resulted from the Covid 19 pandemic. When remote learning rose, I rose to the challenge of creating targeted content to students of mathematics. My playlists are sorted by section. I have completed trigonometry, Calculus 1-3, and I have some decent examples of Linear Algebra.
As we slowly come out of the pandemic, I have been creating less new content. However, if you are interested in seeing a topic not covered here, I'll be happy to take on the request. Email me at the address below with requests.
I also have a GREAT interest in competition math. If you have some competition problems you'd like to see done, hit me up!
Recently found out some of your videos cause I'm going through linear algebra right now! Simply Amazing, Straight to the point and Brilliant! Glad to see there's still new uploads
Thanks very much! Post-pandemic and back to the grind of day to day teaching, it's difficult to find the motivation to keep creating content. It's comments like this that help boost me. I appreciate you
I have a question if you don’t mind. With constrained optimization, we look at the bordered Hessian matrix and we say that if the determinant is > 0 we have a MAX. This rule seems counterintuitive and out of line with my previous understanding of D2 > 0 translating to a minimum. Is there a way for me to reconcile these two differences in methods intuitively?
Sure! I will do my best to answer. The previous understanding of D2>0 translating to a minimum is spot on. In the Hessian determinant, we are looking at the four different partial derivatives that exist and creating products. Every statement that follows will be for a function of two variables. Imagine a local maximum... from the 2D perspective, it would be concave down from every orientation, meaning that both the partial derivatives f_xx and f_yy would be less than 0. In the Hessian determinant, they are multiplied, which produces a positive number, meaning now that D>0. The same is true for a local minimum being concave up from every orientation: both f_xx and f_yy are positive, meaning that their product is positive as well. This is why a positive D value means a critical point could be a local max OR a local min. It's not until we look at the values of the individual second partial derivatives (f_xx and f_yy) to determine which of these critical points we have.
@@davidfriday7498 thank you so much for your quick response! I’m actually studying the Fundamentals of Mathematics for Economics by Alpha C. and I hope to finally wrap my head around the concept with time. Thank you so much once again for your help!!
That's what the above does show. In order to show closure under scalar multiplication, we need to show that if A is symmetric, then cA is symmetric as well. The sequence of equalities you just listed shows this.
@@subratadebnath5436 The playlists are in the sequence associated with the order of the sections in the textbook I use, which is "Elementary Linear Algebra", 8th edition, by Larson.
So... not really sure what you are intending with this comment, but it is definitely inaccurate. Expanding the left side would give -♠^2+2♠+3 rather than what we see on the right side of your equation.
As a pure mathematician, my obligatory answer is "I don't care." However, especially in physics when problems require a rotation of a coordinate system, I have little doubt in my mind that an application exists. Come to think of it, there are probably applications in the field of double and triple integrals with the associated Jacobians of the transformations into non-standard bases. However, those don't necessarily have to be linear.
why did you pull the 8 out? when you can simply just divide by 8? since this is a matrix? why was the 8 outside on the side? I thought you can simply just divide it by 8 and that would still be the same matrix ]
The short answer is "no, it would not be the same matrix." Dividing something by 8 changes its value eightfold; literally the definition of division. I believe you are thinking about operations that can be done to both sides of an equation. If you divide both sides of an equation by 8, then yes, the equation is equivalent to its predecessor. But arbitrarily dividing something, like an expression, a matrix, or a row of a matrix, by 8 does indeed change its value.
@@davidfriday7498 how about row reduce echelon form? i was taught that i can switch rows and take division and multiplication to cancel out also would another way to do this , getting it down to row reduce echelon form and doing a multiplication of lamda cross multiplication? Thank you
An ij cofactor is defined as (-1)^(i+j) multiplied by the ij-minor. In the 1,2 entry, that would be (-1)^(1+2) = (-1)^3 = -1. This causes the negative.
Are 2 by 2 matrices with repeated eigenvalues always defective when they are non-diagonal? I think every diagonal 2 by 2 matrix is diagonalizable with or without repeated eigenvalues.
In general, every diagonal matrix is diagonalizable using the identity matrix. I believe your initial assertion is correct; a 2x2 matrix with repeated eigenvalues necessarily has to be defective f it's not a diagonal matrix of its own eigenvalues. In order to be non-defective, you'd need two rows of 0s/two free variables, which is the zero matrix. This is only attainable with the aforementioned condition.
openstax.org/books/calculus-volume-3/pages/4-6-directional-derivatives-and-the-gradient Start at the section labelled "Gradients and level curves" and read through to Theorem 4.14. This proof extends nicely to three dimensions.
