Eigenvectors and eigenvalues

[@BrodyPianist]

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]

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.

[@betadistribution6534]

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]

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]

19:49 I think v1, v2 are column vector

[@betadistribution6534]

+1

[@turboleggy]

Me too.

[@albertliu2599]

I second this.

[@MrWoopydalan]

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]

it should be -1+(lambda^2)-1=0 so that lambda^2=2, which gives you lambda=+-sqrt(2)

[@kontiimanalatit8987]

​@@gaidid716yes

[@the-fantabulous-g]

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]

i got some slightly different results : for lambda 1= 0 ==> e-vect 1=( 1 -i) , for lambda=2 ==> e-vect2 =(1 i)

[@elizabethbatts410]

I got the same thing as G.

[@erbazkhan266]

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]

@@elizabethbatts410 Use matlab to find them. Cebrail is right

[@albertliu2599]

@@cebrailcoskun3235 I second this answer.

[@gummybears6125]

at 10:46 you made a mistake, instead of -1 - lambda^2 - 1 = 0, it should read -1 + lambda^2 - 1 = 0

[@manishsingh-vk8if]

Is it a convention to use 1 for x in determining eigenvector ?

[@cebrailcoskun3235]

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)

[@Dekoherence-ii8pw]

3:00 Eigenvectors means "characteristic vectors" in this context.

[@ankitmishra2723]

Thanks sir