Linear transformations

[@Gismho]

Very good summary of matrix algebra.

[@the-fantabulous-g]

The Check your understanding: hermitian conjugate of x is equal to y. lemme know if anyone has a different answer!

[@akashagnihotri2469]

Yeah I got the same thing. Just to confirm: x = 0 6+2i 3 y = (0 6-2i 3) Is that correct?

[@puritybundi8204]

@@akashagnihotri2469 got same answers, and y is the hermintian conjugate of x.

[@shawzhang4498]

By definition of basis, they are independent, kind of confused why we need all other x_i to represent x_1

[@Swetlana0]

He is not representing |x_1> but actually T|x_1>

[@shawzhang4498]

Professor, I think you kind of mess up with the subscript of T when you try to expend T|x_1> at the beginning.

[@Swetlana0]

The subscripts are correct. Consider the matrix of the T operator as shown in the video. Now operate this matrix to a basis vector say |x_2>. What is the matrix form of |x_2> in our {|x_i>} basis? Well it's going to a column matrix with entries (0 , x_2 , 0 , 0, ...0) Now do the matrix multiplication of the matrix T with matrix of |x_2> and you'll see subscripts which are written out in the video are the correct subscripts we obtain.

[@jasonyao3753]

It’s also in the textbook

[@Aaaaaaaaaaaaaamin]

Thank you.

[@gaidid716]

3:38 how do you get lx_i> from lx_ j>?

[@hershyfishman2929]

xj picks out a particular basis vector from all those making up the vector |α>. Any one of those basis vectors in turn, after we apply a transformation to it, is itself composed of all the basis vectors (as described earlier 1:08) which we sum up using the i notation.

[@karimkhan1312]

pls make me your research scholar -- I am M.Sc (Physics) from Delhi University -- amarjeet advocate delhi high court -- india

[@eliasportenkirchner2531]

Did your plan work out? haha

[@ifrazali3052]

​@@eliasportenkirchner2531 Lol