To know for doing these projection problems. In your medium-term memory because it's a good thing It's OK if you don't memorize it, but temporarily, put this This in your everyday life five or ten years from now, so So then you have A transposeĪ, and you take the inverse of it. The middle, you have these two guys switched around. Gonna have something in the middle, and then you have A Symmetry or a way of- you could say it's A times, you're Gives you a headache, but there's a certain pattern or Of x onto V would be equal to- and this is kind of hard. So A is a matrix whose columnsĪre the basis for our subspace, then the projection If A is a matrix who's columnsĪre the basis for the subspace, so let's say A isĮqual to 1 0 0 1, 0 1 0 1. Up with a general way to figure this out. Of R4, and I want to figure out a transformation So let's say that X- we'reĭealing in R4 here, right? Let's say that x is a member Projection of any arbitrary vector onto this subspace. If we can find out the transformation matrix for the To a 1 there, so they're linearly dependent. There's no way you can take someĬombination of this guy to somehow get a 1 there. Span a subspace are a basis for that subspace. Two vectors that are linear- orĪny set of vectors that are linearly independent and that And you can see that theseĪre going to be a basis. Is 1 0 0 1, and the second vector is 0 1 0 1. Subspaces, and it's equal to the span of two vectors in R4. V, which tends to be our favorite letter for but some components of the output vector might be zero making them vectors of a smaller subspace. To conclude: yes the transformation matrix is symmetric as it is a general matrix that transforms R4 vectors to R4. So if you project something onto V, you are bound to lose the 3rd component (9which is what we find from the matrix which has 3rd row full of zeroes only). Why is the 3rd row all zeroes? note that all the basis of V have zeroes in the 3rd position, So this subspace can span vectors that have components in 1s, 2nd and 4th dimension only. So the projection matrix takes a vector in R4 and returns a vector in R4 whose 3rd component is 0 (so it is kind of like in R3). But the interesting thing here is that the 3rd row is zero. In general the projection will be a vector in R4 so the matrix is 4x4. Since in this case we are dealing with R4, we expect a vector of R4 as input so the final transformation matrix has 4 columns. Keep in mind that even if A itself is not invertible, A^T A is invertible since A consists of linearly independent columns. However, If A is not invertible, then apparently there are some elements of Rn that are not in the column space of A, and so it makes since to speak of the projection of arbitrary vectors into C(A), which can be computed using the projection matrix A (A^T A) ^(-1) A^T. This implies that the columns of A are a basis for Rn(since they are linearly independent and they span Rn) and that therefore any projection of an arbitrary vector x onto the subspace spanned by the columns of A is simply x, since x is already in the columns space of A. In fact, this makes since because when A is invertible, the system Ax=b has a unique solution for every b in Rn. So the projection of x onto the column space is simply x. In such a case, the simplification A (A^T A) ^(-1) A^T =A A^(-1) A^T^(-1) A^T=I would be valid. Their product A^T A is defined because the number of rows in A^T is equal to the number of columns in A. If A is invertible, then it follows that A^T is also invertible. The property (AB)^-1=(B)^-1*(A)^-1 is valid only when both A and B are invertible and when matrix multiplication between them is defined.
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