At the 8'30" mark I was puzzled by the rapid way (N^t*N)^-1 became 1/3 given that it's fundamental component of Least Squares via matrices. So I checked: it's correct. Thinking about it I realised that here N is a 1-D column vector making N^t*N a dot product with a scalar result. Very neat. Another advantage of projecting onto a 1-D space compared to a higher dimension space. TFTV
Matrix vector multiplication is distributive: en.wikipedia.org/wiki/Matrix_multiplication#Distributivity So, it if follows from I*b = P*b + P_n*b that I*b= (P+P_n)*b Therefore, I is the same transformation as (P+P_n) and they are equal. Hopefully that is what you were asking about.
@@theblinkingbrownie4654 If I turn x + y - z = 0 into matrix A = [1 1 -1], (-1, 1, 0) and (1, 0, 1) will be the basis of the null space N(A), right? But why basis of null space become the basis of the plane? I'm a bit confused here.
@@jeffabc1997 Because the dot product of (-1,1,0) and the plane is 0 (since the dot product of A and (x, y, z) is 0), since the vector A is perpendicular to the plane, A's nullspace is the plane
@@theblinkingbrownie4654😮 hi sir, I know I am late.. seems like you are an expert in this area. Would you kindly share some online material to learn about finding basis from an equation of plane
@@empty8537empty Haha, I am just a beginner in linear algebra and I've only gone through Gilbert Strang's 18.06 course (was it fall 2007 or 2011?), so if you want to use that, just search "Gilbert Strang Ax = 0" and watch videos from there to get the concepts you need. I only recommended that regardless of what you use, do some exercises for it to stick. (Where you get the exercises doesn't matter as long as it's relevant.)
from the coefficient of the equation, we get [x y z]^T * [1 1 -1] = 0, hence the basis of the null space for [x y z]^T(i.e. the original plane) is [1 1 -1]^T, by the definition of "a vector is the normal vector for the plane if it perpendicular to all the vectors in the plane", the [1 1 -1] is the normal vector
When he calculates (N_transpose N)^-1, how is he able to directly write 1/3. Though it is correct, but how can one get an intuition to that. I thought it would be 1x1 matrix with only entry of 1/3, why am I wrong? I am wrong because if we take it as [1/3], then we wouldn't be able to multiply it with row vector [1 1 -1].
You are right, it is a 1x1 matrix. We can multiply it to a row vector because the dimensions allow so -> (1x1) x (1x3). What you could not do would be post multiply it by a column vector -> (1x1) x (3x1) => not possible! But again if you pre mulitply (which is happening here) by a column vector then -> (3x1) x (1x1) => possible!
@@anikets4148 Thanks for pointing out this. I should have tried myself proceeding with [3]_1x1 matrix. And, the tutor should not have omitted those square brackets [3] because he just treated as a scalar though the answer in both cases would be same.
@@shizahkhalid9893 find a basis form this equation,e.g. let x=1,z=0 then from the equation , y = -1; and let x=1,y=0,then z must be 1.So we get the basis (1 -1 0)and(1 0 1).And you can let x,y,z be something else,but they do fit the equation.
You're even using some techniques that prof. Strang haven't taught us yet. This is a bad recitation. By watching this video, we're being more confused instead of being more familiar to the lecture.
Sometimes you need to use your brain to come up with new, and smarter results. I believe this is a common skill for all MIT students. Gosh, dude! Stop complaining, he's helping us with it. It's not like you're lying around naked and afraid. Hahaha
"In a humanely fast fashion" That's the best
In vector analysis, projection onto the plane is just v - (v • û), where û is the unit normal.
He is good! Very good recitation! Thank you for that alternate explanation!
3:27 1 over WHAT!! Prof. would be damned lol
What’s the issue? It’s 1 over the determinant
Determinant. What's the problem? It's a number like any other.
At the 8'30" mark I was puzzled by the rapid way (N^t*N)^-1 became 1/3 given that it's fundamental component of Least Squares via matrices. So I checked: it's correct. Thinking about it I realised that here N is a 1-D column vector making N^t*N a dot product with a scalar result. Very neat. Another advantage of projecting onto a 1-D space compared to a higher dimension space. TFTV
7:08 I don't understand why division on both sides is available. This is not a property that can be used in matrix calculation I think.
Matrix vector multiplication is distributive:
en.wikipedia.org/wiki/Matrix_multiplication#Distributivity
So, it if follows from I*b = P*b + P_n*b that I*b= (P+P_n)*b
Therefore, I is the same transformation as (P+P_n) and they are equal.
Hopefully that is what you were asking about.
Isn't (1,-1,0) and (1,0,1) the basis for Null Space?
And not the Col Space.
It's just the vectors which are satisfying the plane I think and may span the plane
cant we just compute projection of b then subtract it from b to find the projection onto normal vector?
I think camera men is his girlfriend
Fr lmao
How he take basis vector for A please tell me....
The two basis vectors are two solutions of x y z to the equation of the plane A
They are vectors that lie in the plane
How'd he get the basis for the plane??
x+y-z=0 if y = 1 and z = 0 then x = -1, if y = 0 and z = 1 then x = 1 so (-1, 1, 0) and (1, 0 -1) are special solutions.
@@theblinkingbrownie4654 If I turn x + y - z = 0 into matrix A = [1 1 -1], (-1, 1, 0) and (1, 0, 1) will be the basis of the null space N(A), right? But why basis of null space become the basis of the plane? I'm a bit confused here.
@@jeffabc1997 Because the dot product of (-1,1,0) and the plane is 0 (since the dot product of A and (x, y, z) is 0), since the vector A is perpendicular to the plane, A's nullspace is the plane
@@theblinkingbrownie4654😮 hi sir, I know I am late.. seems like you are an expert in this area. Would you kindly share some online material to learn about finding basis from an equation of plane
@@empty8537empty Haha, I am just a beginner in linear algebra and I've only gone through Gilbert Strang's 18.06 course (was it fall 2007 or 2011?), so if you want to use that, just search "Gilbert Strang Ax = 0" and watch videos from there to get the concepts you need.
I only recommended that regardless of what you use, do some exercises for it to stick. (Where you get the exercises doesn't matter as long as it's relevant.)
isn't the basis for the plane is (-1,1,0) and (1,0,1) if we put the equation into parametric form?
u right
Both are right. You need two vectors in the plane which are not linearly dependent
one of the best teacher
How to get normal vector N(1 1 -1)?
from the coefficient of the equation, we get [x y z]^T * [1 1 -1] = 0, hence the basis of the null space for [x y z]^T(i.e. the original plane) is [1 1 -1]^T, by the definition of "a vector is the normal vector for the plane if it perpendicular to all the vectors in the plane", the [1 1 -1] is the normal vector
The plane is defined by that normal vector, it is a given quantity in the problem.
@@alileo1578 true but its wholesome that you can multiply the normal with the coefficient and get zero
Formula for a plane is Ax+By+Cz+D=0. A,B,and C are the components of the normal vector. D=0. So you get a normal vector of [1 1 -1]
When he calculates (N_transpose N)^-1, how is he able to directly write 1/3. Though it is correct, but how can one get an intuition to that. I thought it would be 1x1 matrix with only entry of 1/3, why am I wrong? I am wrong because if we take it as [1/3], then we wouldn't be able to multiply it with row vector [1 1 -1].
You are right, it is a 1x1 matrix.
We can multiply it to a row vector because the dimensions allow so -> (1x1) x (1x3).
What you could not do would be post multiply it by a column vector -> (1x1) x (3x1) => not possible!
But again if you pre mulitply (which is happening here) by a column vector then -> (3x1) x (1x1) => possible!
@@anikets4148 Thanks for pointing out this. I should have tried myself proceeding with [3]_1x1 matrix. And, the tutor should not have omitted those square brackets [3] because he just treated as a scalar though the answer in both cases would be same.
He just applied the general formula for the inverse of a 2x2 matrix:1/(ad-bc) * [[d, -b], [-c, -a]].
One can directly see, that 1/(2*2-1*1) = 1/3
what reasoning did he use to get the basis columns for the plane at the beginining
never mind he explained it later on
@@thedailyepochs338 i dont get how he obtained those basis columns. could you explain?
@@shizahkhalid9893 find a basis form this equation,e.g.
let x=1,z=0 then from the equation , y = -1; and let x=1,y=0,then z must be 1.So we get the basis (1 -1 0)and(1 0 1).And you can let x,y,z be something else,but they do fit the equation.
is (i-Pn) really equals to P here??
it is
@@Tom-ef6kw haha true i see it now
i felt so stupid once i knew that i was ignoring the fraction.
here you solved only projection matrix, but it is not orthogonal ?
how he found a1 and a2?
Any vector which satisfies the given plane equation can be a1 and a2
any set of 2 INDEPENDANT vectors
It need not be orthogonal. But it must be independent
a orthonormal needs to be square and invertible, so its imposible find a orthonormal matrix for a plane in R3
both solution are not matching !!! something is wrong
They are matching, Redo you math.
great
bruh, a1 and a2 are not even independent, how can they be the basis for the plane -_-
(no hatred, the other parts of the recitation was pretty good)
They are independent, there is no non-zero constant c for which ca2 = a1 since the third element of a1 is zero and the third element of a2 is non-zero
You're even using some techniques that prof. Strang haven't taught us yet. This is a bad recitation. By watching this video, we're being more confused instead of being more familiar to the lecture.
Sometimes you need to use your brain to come up with new, and smarter results. I believe this is a common skill for all MIT students. Gosh, dude! Stop complaining, he's helping us with it. It's not like you're lying around naked and afraid. Hahaha