Eigenvectors and eigenvalues
ฝัง
- เผยแพร่เมื่อ 20 ก.ค. 2024
- A description of eigenvalues and eigenvectors, what they mean, how to find eigenvalues, how to find eigenvectors, diagonalization, and hermitian transformations.
Note: there is an error in the discussion for slide 7. I say that the matrix S that diagonalizes A has the eigenvectors for rows, when it should be that S is the matrix with eigenvectors for columns.
(This lecture is part of a series for a course based on Griffiths' Introduction to Quantum Mechanics. The Full playlist is at th-cam.com/users/playlist?list=...)
great, I love your description of the physical interpretation of what eigenvectors and values actually are at the start. I find i need these geometric visualizations to grasp such concepts.
I can't follow why the matrix S in the diagonalization section has the eigenvectors in its rows rather than its columns. I suspect this was a mistake since the given vector multiplication only works with column eigenvectors, but if not I would love to hear why.
19:49 I think v1, v2 are column vector
+1
Me too.
I second this.
I'm a bit confused around 25:32, the expression [A'a]b doesn't type check (column * column). Perhaps it ought to be [A'a]'b?
From 24:49, Brant got carried away in an infundibulum of confusing matrix algebra. My problem is that I, knowing a lot about linear algebra, I'm unable to detect a deeper meaning that might have escaped me.
I vote for a missing dagger as the culprit ...
Check your understanding
I got lambda1=0, e-vect1 = (1 i); and lambda2 = 2, e-vect2 = (1 -i). Let me know if anyone has differing answers or gets the same answer!
i got some slightly different results : for lambda 1= 0 ==> e-vect 1=( 1 -i) , for lambda=2 ==> e-vect2 =(1 i)
I got the same thing as G.
I got the same eigenvalues, but the eigenvectors are swapped.
For lambda1 = 0; e-vec = [1 -i]
For lambda2 = 2; e-vec = [1 i]
Same as cebrail coskun
@@elizabethbatts410 Use matlab to find them. Cebrail is right
@@cebrailcoskun3235 I second this answer.
For your eigenvalue example (@10:44), I believe the eigen value should be plus/minus sqrt(2)i, you forgot to bring the negative sign over when you added the two to the RHS
it should be -1+(lambda^2)-1=0 so that lambda^2=2, which gives you lambda=+-sqrt(2)
@@gaidid716yes
at 10:46 you made a mistake, instead of -1 - lambda^2 - 1 = 0, it should read -1 + lambda^2 - 1 = 0
Is it a convention to use 1 for x in determining eigenvector ?
i found one of eigen values = 0. So does that mean, this eigen value changes any vector on corresponding eigen vector to become zero vector? ( since by definition: eigen value is a coefficient of eigen vector)
3:00 Eigenvectors means "characteristic vectors" in this context.
Thanks sir