15. Projections onto Subspaces

MIT OpenCourseWare · 638K views · 2009 · 48:51

Open on YouTube

University lectures & coursesMath & theory (other)lecture

Why Mathematics Stack Exchange recommends it

Mathematics Stack Exchange users cite it as the source lecture when asking about projection matrices, least squares setup, and the algebra Strang writes on the board.

Recommended by 1 people across 1 different months, 2015 to 2020.

What it is

Lecture 15 of Gilbert Strang's MIT 18.06 Linear Algebra course (Spring 2005), covering projection of a vector onto a line and onto subspaces, the projection matrix P = A(A^T A)^-1 A^T, and its link to least squares. It is a chalkboard lecture recorded in a university classroom.

What people said

Recommends Aravindh Vasu · 2020-01 · Mathematics Stack Exchange · on “Transpose of a matrix and the product $A A^\top$”

I've been following 3Blue1Brown's Essence of Linear Algebra [LINK] , basically the question (1) is what is the geometric meaning of the transpose? I've watched Chapter 9-Dot Products and duality [LINK] . I can see that the transpose of a matrix has something to do with duality and dual spaces, but I'm not able to put my finger on it. There is an answer , 3B1B Transpose [LINK] , in this answer there is a sentence, When you transpose a matrix, you are actually making use of this vector-dual vector identification to change your transformation to act on the dual vectors instead of the original vectors. I've been reading this sentence again and again, but I'm having a hard time understanding the

Cites Pedro · 2015-03 · Mathematics Stack Exchange · on “How to divide a $1 \times n$ rowvector by a $n \times 1$ vector and the result of such a divsion”

Dividing by a vector is something that is undefined. Let me give you some intuition about matrices to know why. Generally a matrix can be seen as an operator. Roughly speaking a matrix is something that acts on vectors and gives back an other vector. You have row vector (aka 1xn matrix) and column vectors (aka nx1 matrix). If you use a row vector then you write $xA = y$, and then we say that matrix A manipulated row vector x to produce row vector y. If you use column vectors you write $Ax = y$, and we say that we manipulated x to produce vector y. Notice that with column vectors, we multiply from the right. In most mathematical and engineering fields people use column vectors and we multiply

Cites Nimitz14 · 2015-11 · Mathematics Stack Exchange · on “Confused regarding the interpretation of A in the least squares formula A^T A = A^T b”

So I'm was watching gilbert strang's lecture to refresh my memory on least squares, and there's something that's confusing me (timestamp included). [this video] In the 2D case he has $A=[1,1;1,2;1,3]$. In the lecture he talks about how A is a subspace in our n dimesional (in this case n=2) space. If you look at the top left point of the chalk board in that time stamp you will see he has written down $A=[a_{1},a_{2}]$. Here he was talking about the 3d case, and the $a$s were vectors that spanned a plane onto which we wanted to project. My problem is, I'm not quite sure how I'm supposed to understand the values for the 2D. Clearly the 2nd column is the x values, but can one say they span a space? T

Cites YohanRoth · 2017-04 · Mathematics Stack Exchange · on “Cannot understand part of the lecture on linear algebra”

I cannot understand how projecting vector $b$ onto vector $a$ gives us a side $e$ of a right triangle which we can express as $e = b - p$? Video is here [this video] [LINK]

Cites Swakshar Deb · 2020-05 · Mathematics Stack Exchange · on “How can $a^T(b-xa) = 0 $ means $xa^{T}a = a^Tb$?”

I was watching gilbert strang lectures on the projection matrix [this video] , at 5.17 he wrote that: $a^T(b-xa) = 0 $ That means: $xa^{T}a = a^Tb$ how can we write this expression? In my opinion isn't it should be: $ a^Txa = a^Tb $ ? Actually I am not good at linear algebra, so I know the question may be silly.

Cites Rotkiv · 2020-12 · Mathematics Stack Exchange · on “how can I show that for projection matrix $P^2 = P$ without applying trigonometry”

I am following MIT 18.06 lectures [this video] on linear algebra and would like to understand how one can show that $P^2 = P$ for a projection matrix. I understand that for 2-d case: $P = \frac{a a^t}{a^t a}$ $P^2 = \frac{a a^t}{a^t a} \frac{a a^t}{a^t a}$ I assume I should easily show it by using some properties of transpose matrices/vectors, but I couldn't quite figure out how to derive it. (I am familiar with the proof that relies on trigonometric formulas)