First time user, sorry if this is the wrong forum.
I am looking for a way to express the following in pure linear algebra vector notation, i.e. remove the element-wise operations.
I am hoping that would make deriving the gradient and Hessian easier.
In MATLAB:
sum((W'*p - r).^2 .* m)
where W is a matrix, p, r and m are vectors.
In R:
sum(t(W) %*% p - r)^2 * m)
Thanks
The element-wise squaring makes the vector notation a bit more obscure in terms of everyday linear algebra operators.
The Hadamard product is the mathematical term for element-wise multiplication and will be denoted as ○.
Using the Hadamard product, we can write the expression as
In brackets, we have the element-wise multiplication, and then we left-multiply with m's transpose to take the sum (vector form of the dot-product).
We could also use the trace (denoted here are "tr") and diag operations to get an equivalent expression:
Here, we use diag to create a square matrix, square the square matrix (which is a valid operation), perform a matrix-vector multiplication, diag the resulting vector, and take its trace. This one looks like code, but I think that's just because code is trying to look like it ;).
Related
I have a k*k squared matrix with diagonal elements x>0 and all other elements y>0. The values of k, x, y are all subject to change.
Now I need the determinant of this matrix. I know there won't be a closed-form formula for it, but is there a way to calculate it faster than the commonly used LU-decomposition which takes O(K^3) time complexity (considering its special structure)?
(I am using R as my coding language, and the built-in det() function in R uses the LU-decomposition.)
I am trying to calculate P^100 where P is my transition matrix. I want to do this by diagonalizing P so that way we have P = Q*D*Q^-1.
Of course, if I can get P to be of this form, then I can easily calculate P^100 = Q*D^100*Q^-1 (where * denotes matrix multiplication).
I discovered that if you just do P^5 that all you'll get in return is a matrix where each of your entries of P were raised to the 5th power, rather than the fifth power of the matrix (P*P*P*P*P).
I found a question on here that asks how to check if a matrix is diagonalizable but not how to explicitly construct the diagonalization of a matrix. In MATLAB it's super easy but well, I'm using R and not MATLAB.
The eigen() function will compute eigenvalues and eigenvectors for you (the matrix of eigenvectors is Q in your expression, diag() of the eigenvalues is D).
You could also use the %^% operator in the expm package, or functions from other packages described in the answers to this question.
The advantages of using someone else's code are that it's already been tested and debugged, and may use faster or more robust algorithms (e.g., it's often more efficient to compute the matrix power by composing powers of two of the matrix rather than doing the eigenvector computations). The advantage of writing your own method is that you'll understand it better.
So I solve the eigenvectors for a matrix in Maxima.
a:matrix([10,10],[-4,-3]);
\\outputs matrix
vec:eigenvectors(a);
[[[5,2],[1,1]],[[[1,-1/2]],[[1,-4/5]]]]
I've hand calculated the eigenvalues, and vectors as (1x2) 5: [-2,1]. 2:[-5,4], which are correct. What is Maxima outputting?
Eigenvectors are only determined up to a multiplicative constant. That is, if x is an eigenvector, then so is a*x where a is a scalar. I think if you look at your result and Maxima's result, you'll see that they are equivalent in that sense.
There are different normalization schemes. Looks like Maxima makes the first element 1. Another common scheme is to make the norm of the eigenvector equal to 1. Or one can just leave them unnormalized.
I need to take the inner product of every pair of columns in a particular matrix, which I achieve by calculating
t(M) %*% M
However this naturally produces a symmetrical result, doing just over twice the necessary work (I don't need the diagonal either). Obviously I could break the multiply down into individual inner product operations, but is there a better way to calculate just the upper triangular part of the product?
From the description in help("crossprod"):
Given matrices x and y as arguments, return a matrix cross-product.
This is formally equivalent to (but usually slightly faster than) the
call t(x) %*% y (crossprod) or x %*% t(y) (tcrossprod).
Thus, use crossprod(M).
I wonder if it is possible (and if it is then how) to re-present an arbitrary M3 matrix transformation as a sequence of simpler transformations (such as translate, scale, skew, rotate)
In other words: how to calculate MTranslate, MScale, MRotate, MSkew matrices from the MComplex so that the following equation would be true:
MComplex = MTranslate * MScale * MRotate * MSkew (or in an other order)
Singular Value Decomposition (see also this blog and this PDF). It turns an arbitrary matrix into a composition of 3 matrices: orthogonal + diagonal + orthogonal. The orthogonal matrices are rotation matrices; the diagonal matrix represents skewing along the primary axes = scaling.
The translation throws a monkey wrench into the game, but what you should do is take out the translation part of the matrix so you have a 3x3 matrix, run SVD on that to give you the rotation+skewing, then add the translation part back in. That way you'll have a rotation + scale + rotation + translate composition of 4 matrices. It's probably possible to do this in 3 matrices (rotation + scaling along some set of axes + translation) but I'm not sure exactly how... maybe a QR decomposition (Q = orthogonal = rotation, but I'm not sure if the R is skew-only or has a rotational part.)
Yes, but the solution will not be unique. Also you should rather put translation at the end (the order of the rest doesn't matter)
For any given square matrix A there exists infinitely many matrices B and C so that A = B*C. Choose any invertible matrix B (which means that B^-1 exists or det(B) != 0) and now C = B^-1*A.
So for your solution first decompose MC into MT and MS*MR*MSk*I, choosing MT to be some invertible transposition matrix. Then decompose the rest into MS and MR*MSk*I so that MS is arbitrary scaling matrix. And so on...
Now if at the end of the fun I is an identity matrix (with 1 on diagonal, 0 elsewhere) you're good. If it is not, start over, but choose different matrices ;-)
In fact, using the method above symbolically you can create set of equations that will yield you a parametrized formulas for all of these matrices.
How useful these decompositions would be for you, well - that's another story.
If you type this into Mathematica or Maxima they'll compute this for you in no time.