Linear Algebra 14TBD: The Direct Algebraic Definition of the Determinant
Vložit
- čas přidán 8. 09. 2024
- bit.ly/PavelPa...
lem.ma/LA - Linear Algebra on Lemma
bit.ly/ITCYTNew - Dr. Grinfeld's Tensor Calculus textbook
lem.ma/prep - Complete SAT Math Prep
Go to LEM.MA/LA for videos, exercises, and to ask us questions directly.
I hadn't the slightest clue what my professor was explaining when he went over this, and in 11 minutes you explained it in a way I completely understood. Thank you so much.
Math is finally beautiful when you explain it. It's clearer than what the textbook explains
This is a terrible explanation.
@@jak9439 How so?
How didnt I know about this before? This is so much simpler than the laplace expansion
Right? These rules for building determinants seemed to come out of nowhere?! I always wondered how this seemingly random combination of matrix-elements came to be - now I finally found the real consistent definition of it that is applicable to any square matrix of order n.
you have the most beautiful handwriting on a blackboard that i have ever seen, respect for that!
this is super Clear and I love your video! the way you speak makes me feel that I can handle this! good luck to everyone in the comment!
So concise and clear, thank you sir!
So is your comment - thank you!
I dont understand why so many linnear algebra skip this defintion its very simple and rigourous
That is a great video, yet I believe that you could make it even better. The transition from perceiving a permutation as a tuple to perceiving it as a function could have been made explicit. It is still absolutely amazing, though!
Thank you so so much for making such a nice explanation for this mess. Almost give up until I found this video. THANKS A LOT.
That is a very good video. Thank you. I like how organized you talk. Your handwriting is also very neat.
Thanks from Brazil. Nice, amazing video.
This video is brilliant, thank you very much.
loved this!
very clear explanation. thank you
it is helpful. thank you. I don't if you can to permutations similarity.
Beautiful as usual.
Thank you! It's in the name of the channel, so it's like an obligation.
Thank you!
glad I found this
Thank you
great explanation, thanks!
Beautiful video!
You're great!
thank you so much, sir this is very helpful
For the sign, you could also trace diagonals in the matrix, the ones that go descending from left to right are positive and the ascending ones are negative, like being the inverse of the slope you get if it was a cartesian plane.
That's a good way of thinking about it.
@@MathTheBeautiful really? I did spot the clockwise order for the positive arrangements [n=3] (1,2,3), (2,3,1), and (3,1,2) and negative for ccw (1,3,2), (2,1,3), and (3,2,1). Your parity of the permutation could I guess be simply put into a formula. I wonder if that would then look like Mirsky's notation in (1955: 3) An Introduction to Linear Algebra? (which is proving difficult to follow by itself)
Mirsky's notation is now understood [e.g. (1,2,3) = sgn(2-1)(3-2)(3-1) = 1]; and I've the circular notation clarified (see Socratica: Cycle notation of permutations) - including transpositions [(135) = (13) (15)] . Elliot Nicholson was also ok on parity of permutations. Michel van Biezen I'd use for the simple alternative type of formula for the sign. Enjoyed getting to this point but a frustrating search , at times.
Nice explanation! I like it very much!
10:26 you only need 3 transpositions, (15)(13)(12) by index notation
Wow! VERY helpful! Thank you!
Thanks! Please check out lem.ma
I've checked out lemma but I don't see a structured course, I see individual topics. Am I missing something?
Awesome
My man here really cleans his blackboard...
broooooooooooooo very gooooooooood
Near the end, I can't understand how you are calculating five swaps. I can see that the given permutation can be expressed as a 4-cycle and a k-cycle is a composite of k-1 transpositions, so the given permutation is a composite of 3 transpositions, this implies that the permutation is odd and hence it's sign is -1.
The number of swaps does not matter, as long as you know if it is even or odd. An odd permutation can never be written as a composition of an even number of steps, and vice versa.
Hi Professor, based on the theorem that det(A) = det(A-transpose), I'm wondering whether it is also true to express the algebraic definition in such a way that the columns are in order from 1 to N, and the rows are from ɛ(1) to ɛ(N). Would that also work as a definition of the determinant? Thanks!
Yes, absolutely. You will need to use the fact that the permutation and its inverse have the same parity.
I finally understood this shit ! Thanks ! Thanks, thanks !
Great video and great courses. Should the second term on the fourth line of the formula for the 4x4 determinant be a14a21a33a42
nice one
Good!
Why does the labelling of terms begin with 1? Is this standard procedure, or do some people start with 0, going a00, a01, a02 etc? I thought of this because its equivolence with counting in ternary. (Perhaps being able to position an element of a matrix by calling out a base 3 number)
in the 5x5 example wouldnt it be 3swaps? first you swap 1 and 2 then 2 and 3 and then 3 and 5?
Yes!
Unajua mzee ungejua elimu ya huku ilivyo ngumu
please! sir show us .find the square matrix A A^2-2A=[-1 0
. 6 3]
I think there are four solutions. One of them is
1 0
3 3
The key is the eigenvalue decomposition..
FINALMENTE
Better finalmente than never
♥️
It can be proven by induction rather easily that our method of deriving determinant by gaussian elimination conforms to this general formulae for any NxN matrix
but why???
Good question! See other videos in the playlist.
"I recommend thinking back to the Russian way of calculating the determinant"
Why do I hear boss music?
I still can't understand what the fuck does this have to do with the determinant by linear transformation definition. Like it's just an awesome coincidence of mathematics or what? i don't understand :'(
Fundi moja
you look like woody harrielson omg
Hello,
I've tried your procedure of counting column permutations to attribute the sign of each term of a 4x4 matrix and compared them to the terms obtained by reducing the 4x4 determinant to 3x3 determinants times the appropriate value of the first row (Indian method). Here a picture of my work : imgur.com/a/tGzpm .
The result is not satisfying. I obtain opposite signs for some terms (-a12 a23 a34 a41 by permutation (same as you at 8:44) and +a12 a23 a34 a41 by doing the reduction of the determinant).
Could you please take a look at my work and give me a hint of why this is not consistent depending of the procedure used?
Thank you for your help and many thanks for all these great videos.
I just found the reason why it wasn't consistent by looking how you choose the points on your 5x5 matrix. They're not align and so combining the Indian method with the American method doesn't work.
To have the good sign it's necessary to reduce the matrix to a 2x2 matrix and then, the Indian crossing give us the correct sign for each value.
I know you gave us a better way to find any values and the correct sign by counting the permutations but each method should be equivalent in my mind.