Expressing a projection on to a line as a matrix vector prod | Linear Algebra | Khan Academy

Expressing a Projection on to a line as a Matrix Vector prod
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[@Vulkura]

You explained this better than my professor and teaching assistant. My hat goes off to you this day, sir.

[@fashionvella730]

Khan academy please can you tell me where you find out all this stuff in details because in books there is always something missing. But in your videos, you give the whole proof of everything.

[@atiramgine]

Khaaaaan. I love you Khan. I can almost feel my Lin Alg grade rising...

[@doordie7560]

i had never seen tutorials like this

[@stonemysterioserusss]

do or die until about a week ago

[@ratnakarbachu2954]

thanks sir.god bless u.

[@LAnonHubbard]

@EvgenijM86 Sal is applying the projection to the first column of the identity matrix (the "x" basis vector [1 0]). The projection formula was already defined to be (vector_to_project "dot" unit_vector) * unit_vector. In this case vector_to_project is the "x" basis vector.

[@scientist1492]

Hello everyone, Could someone explain how the solutions to (x.u)u = Ax could be taken as a span of [1 0] and [0 1] as explained in the video at 9:47 point? Any links to explaning the process could be helpful. Thank You

[@mingtianni]

The answer is given in lecture 56. Here is a summary: T(x) = T(I*X) = T(x1*e1 + x2*e2) = x1*T(e1) + x2*T(e2) = [T(e1), T(e2)]*X, where I is identity matrix and e1 and e2 are basis vectors. The idea has been that any vector is a is a linear combination of basis vectors. Applying a linear transformation to that vector is equivalent to applying the same transformation to basis vectors first and then combining afterwards.

[@zhenyishen5102]

Thank you, I have better understanding about the projection and It has some strong conncetion with the matrix norm.

[@alaaabdullahmohammedalsaad7447]

you expressed very well the meaning of projection , thank you so much

[@97021840]

excellent

[@s0m0c]

Gracias.

[@mt2yo]

Bravo, bravo, bravo! Very elegant explanation thank you so much!

[@gekorio]

I think I understand your thinking, but the dot product give us a scalar, so it can't be a projection, right?

[@alkalait]

I've come to think of the dot product also as the projection of a onto b, as if b was a unit vector, but with that projection scaled to times the length of b

[@martinhvezda6535]

i love you

[@alkalait]

yes, apologies. the projection is a vector, my mistake. replace the word "projection" with "length of projection", to what I said before.

[@suyashneelambugg]

You don't explain the reason why you choose identity matrix for finding A and what if we use any other matrix instead of identity.

[@mt2yo]

What happens if you choose another matrix?

[@saakethchintha5385]

He chose an identity matrix because it's also the standard matrix in R2..meaning whatever cud be done to the standard matrix cud be done to any other matrix in R2

[@Judobrinez]

... ?

[@Judobrinez]

***** nice ;D