When I was younger, I watched Gilbert Strang's MIT lectures, felt like I'd never understood Linear Algebra until then. Seeing those last 3 rules showed me that I still have much to learn - thank you!
Very cool explanation! Overall very high quality and interesting subject matter 😊 My only tip would be to go a little slower on the equation explanations. For me at least, it can be pretty hard to follow the equations changing so fast on screen. I am also less knowledgeable about this than maybe your desired audience is, so keep that in mind. Keep it up!
Great video but: 1) you skipped the intuition for trace part as you gave for the determinant one... 2) also why is fact 3 and 5 right ? Can you please give an intuitive answer...
Thank you! The intuition for the trace can be though of similarly as the determinant as when the matrix is reduced to echelon form, it will still have the same trace, and it's diagonal matrix will have the eignevalues in its diagonal. Let's talk about 3, The sum of rows being the same makes it an eignevalue. For this, consider matrix A where the sum of each row is some constant k. Let's multiply it with a vector with all 1s: v = (1 . . . 1). Av = (k . . . k) by matrix multiplication, which can be factored out as k(1 . . . 1) = kv, so k is a scaled version of (1. . .1) therefore an eigenvalue. To sum it up, the (1 . . . 1) vector is scaled my the constant k because of how matrix multiplication works. For 5, use the same logic as 3 for the first part of getting the first eignevalue = sum of #'s in a row or the one mentioned in the video. For the second part consider the matrix with 4s and 2s like in the video (5:51). If we subtracted from it a 4 x 4 matrix containing all 2's then we can get an all 0 matrix with 2s in the diagonals. Here, we can use this logic (6:00) and these 2s will be the other eigenvalues as obtained by computing diagonal # - other # I hope this helps!
Ya it did ❤ and 3 is beautiful !...but why in 5 you subtracted by a 4*4 matrix with all 2's and is it allowed will it not change the matrix and thus the eigen value ? Also can you tell why echolean form preserves the trace and det ?
That is great! Let M be the original matrix with 4s and 2s, and A be the matrix with all 2s. For the eigenvector v, the property (M-A)v=Mv-Av=Mv needs to be true true so we would just need to find the vector that satisfies the condition. This is possible with vectors (-1 1 0 0) , (-1 0 1 0), (-1 0 0 1) in this case. So if you use this property you will notice that the eigenvectors that correspond to the eigenvalues obtained by subtracting the M - A having eigen vectors that take the diagonal number and subtract the other number. And when you have the matrix (M-A) which equals only the diagonal numbers, these vectors are essentially just subtracting my 0 so the number on the diagonal are the eigenvalues. I hope that makes sense! And for the echelon form as long as you are only subtracting rows with a scalar multiple of other rows and not scaling the rows themselves, the diagonal numbers retain the same value because in reducing to echelon form, the pivots remain the same, so determinants are preserved. The idea for the trace is a bit different and requires a more rigorous proof, the echelon form way is just a way to remember it. But if you would like to explore that, you can find a way to connect the determinant to the trace and use the determinant rule to prove it.
Wait but how did you get the vectors (-1,1,0,0) ; (-1,0,1,0);(-1,0,0,1) ? because I got only (0,0,0,0) as answer because if Mv-Av=Mv is equal to Av=0 => v=0 can you pls explain the 1st part because I still didn't understood it...😢
Hi! Let me clarify what I meant, (M-A)v = Mv was the property that needs to be try to show that they both indeed have the same eigens . So if you try to solve this, you will end up solving for (M-A-M)v which = Av. for this you will just end up solving a system of linear equations with all 2s, so the result will be that each entry of the vector is the negative sum of all the other entries. Therefore, the vectors mentioned in the previous comment (which are the eigenvectors of M) satisfy this, showing that it is okay to subtract with the matrix A. I hope that clarifies things!
When I was younger, I watched Gilbert Strang's MIT lectures, felt like I'd never understood Linear Algebra until then. Seeing those last 3 rules showed me that I still have much to learn - thank you!
Thank you for the video! I know all of these facts, we studied it in linear algebra and numerical methodes
this was... so lovely??? like, i really enjoyed this. you have my sub. very calm, intuitive, enlightening, and wonderful. thank you :D
@@Zachariah-Abueg thank you so much! :D
The poles of the Green’s function are the eigenvalues of the operator.
Thank You
Wow you are amazing and you reply all questions soon you'll become next 3BLUE1BROWN ❤
Thank you! You are so kind!
Very cool explanation! Overall very high quality and interesting subject matter 😊 My only tip would be to go a little slower on the equation explanations. For me at least, it can be pretty hard to follow the equations changing so fast on screen. I am also less knowledgeable about this than maybe your desired audience is, so keep that in mind. Keep it up!
Thank you!
How do you make your videos, Can you make some tutorial?
@@nazishahmad1337 Maybe in the future :)
However, this video was made using blender and Keynote
Got something special....
Great video but:
1) you skipped the intuition for trace part as you gave for the determinant one...
2) also why is fact 3 and 5 right ? Can you please give an intuitive answer...
Thank you!
The intuition for the trace can be though of similarly as the determinant as when the matrix is reduced to echelon form, it will still have the same trace, and it's diagonal matrix will have the eignevalues in its diagonal.
Let's talk about 3, The sum of rows being the same makes it an eignevalue. For this, consider matrix A where the sum of each row is some constant k. Let's multiply it with a vector with all 1s: v = (1 . . . 1). Av = (k . . . k) by matrix multiplication, which can be factored out as k(1 . . . 1) = kv, so k is a scaled version of (1. . .1) therefore an eigenvalue. To sum it up, the (1 . . . 1) vector is scaled my the constant k because of how matrix multiplication works.
For 5, use the same logic as 3 for the first part of getting the first eignevalue = sum of #'s in a row or the one mentioned in the video. For the second part consider the matrix with 4s and 2s like in the video (5:51). If we subtracted from it a 4 x 4 matrix containing all 2's then we can get an all 0 matrix with 2s in the diagonals. Here, we can use this logic (6:00) and these 2s will be the other eigenvalues as obtained by computing diagonal # - other #
I hope this helps!
Ya it did ❤ and 3 is beautiful !...but why in 5 you subtracted by a 4*4 matrix with all 2's and is it allowed will it not change the matrix and thus the eigen value ? Also can you tell why echolean form preserves the trace and det ?
That is great!
Let M be the original matrix with 4s and 2s, and A be the matrix with all 2s. For the eigenvector v, the property (M-A)v=Mv-Av=Mv needs to be true true so we would just need to find the vector that satisfies the condition. This is possible with vectors (-1 1 0 0) , (-1 0 1 0), (-1 0 0 1) in this case. So if you use this property you will notice that the eigenvectors that correspond to the eigenvalues obtained by subtracting the M - A having eigen vectors that take the diagonal number and subtract the other number. And when you have the matrix (M-A) which equals only the diagonal numbers, these vectors are essentially just subtracting my 0 so the number on the diagonal are the eigenvalues. I hope that makes sense!
And for the echelon form as long as you are only subtracting rows with a scalar multiple of other rows and not scaling the rows themselves, the diagonal numbers retain the same value because in reducing to echelon form, the pivots remain the same, so determinants are preserved. The idea for the trace is a bit different and requires a more rigorous proof, the echelon form way is just a way to remember it. But if you would like to explore that, you can find a way to connect the determinant to the trace and use the determinant rule to prove it.
Wait but how did you get the vectors (-1,1,0,0) ; (-1,0,1,0);(-1,0,0,1) ? because I got only (0,0,0,0) as answer because if Mv-Av=Mv is equal to Av=0 => v=0 can you pls explain the 1st part because I still didn't understood it...😢
Hi!
Let me clarify what I meant, (M-A)v = Mv was the property that needs to be try to show that they both indeed have the same eigens . So if you try to solve this, you will end up solving for (M-A-M)v which = Av. for this you will just end up solving a system of linear equations with all 2s, so the result will be that each entry of the vector is the negative sum of all the other entries. Therefore, the vectors mentioned in the previous comment (which are the eigenvectors of M) satisfy this, showing that it is okay to subtract with the matrix A.
I hope that clarifies things!
this works only for 2x2 matrix.
@@apoorvvyas52 these should work for any n x n matrix unless stated otherwise
You have forgotten rules for multidiagonal and cyclic matrices
Please feel free to share any additional tricks or methods that you know!
too fast. no time to parse what you are saying
you can slow it down
git gud