Eigenvectors and eigenvalues

  Рет қаралды 27,641

Brant Carlson

Brant Carlson

Күн бұрын

Пікірлер: 22
@BrodyPianist
@BrodyPianist 10 жыл бұрын
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.
@robertfraser2269
@robertfraser2269 9 жыл бұрын
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.
@the-fantabulous-g
@the-fantabulous-g 4 жыл бұрын
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!
@cebrailcoskun3235
@cebrailcoskun3235 4 жыл бұрын
i got some slightly different results : for lambda 1= 0 ==> e-vect 1=( 1 -i) , for lambda=2 ==> e-vect2 =(1 i)
@elizabethbatts410
@elizabethbatts410 4 жыл бұрын
I got the same thing as G.
@erbazkhan266
@erbazkhan266 3 жыл бұрын
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
@yanwang248
@yanwang248 3 жыл бұрын
@@elizabethbatts410 Use matlab to find them. Cebrail is right
@albertliu2599
@albertliu2599 5 ай бұрын
@@cebrailcoskun3235 I second this answer.
@betadistribution6534
@betadistribution6534 4 жыл бұрын
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?
@MiguelGarcia-zx1qj
@MiguelGarcia-zx1qj 3 жыл бұрын
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 ...
@shawzhang4498
@shawzhang4498 8 жыл бұрын
19:49 I think v1, v2 are column vector
@betadistribution6534
@betadistribution6534 4 жыл бұрын
+1
@turboleggy
@turboleggy 3 жыл бұрын
Me too.
@albertliu2599
@albertliu2599 5 ай бұрын
I second this.
@MrWoopydalan
@MrWoopydalan 9 жыл бұрын
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
@gaidid716
@gaidid716 9 жыл бұрын
it should be -1+(lambda^2)-1=0 so that lambda^2=2, which gives you lambda=+-sqrt(2)
@kontiimanalatit8987
@kontiimanalatit8987 Жыл бұрын
​@@gaidid716yes
@gummybears6125
@gummybears6125 4 жыл бұрын
at 10:46 you made a mistake, instead of -1 - lambda^2 - 1 = 0, it should read -1 + lambda^2 - 1 = 0
@Dekoherence-ii8pw
@Dekoherence-ii8pw 11 ай бұрын
3:00 Eigenvectors means "characteristic vectors" in this context.
@manishsingh-vk8if
@manishsingh-vk8if 5 жыл бұрын
Is it a convention to use 1 for x in determining eigenvector ?
@cebrailcoskun3235
@cebrailcoskun3235 4 жыл бұрын
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)
@ankitmishra2723
@ankitmishra2723 3 жыл бұрын
Thanks sir
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