If I have 3 different matrices, one for rotation (R), one for translation(T), and one for scaling (S), how to achieve the reverse effect of these matrices by manipulating the one caused it?
What I gathered so far is if I transposed the rotation matrix, I will achieve what I desire (is this correct?). And what about the other two?
And if there is a common way, are there any special cases where these ways will not suffice?
The inverse of the rotation matrix R is indeed its transpose RT.
The inverse of the scaling matrix S is easy as it contains only diagonal elements (the first three rows, as the last row is always equals to (0 0 0 1))
So you just replace each diagonal si with 1/si.
Finally the translation matrix T is an identity matrix with the translation vector on the last column. The inverse is achieved by replacing those elements with their negative.
Also the inverse of the product of 3 matrices, is the reverse product
(S T R)-1 = R-1 T-1 S-1
Related
I am trying to find conditions then a certain matrix is invertible or not (which is problematic as the matrix is random). The matrix results from the following:
$A=\Tilde{A}+diag(n)$.
Furthermore $\Tilde{A}$ results from the pointwise multiplication of a random symmetric matrix (consisting of 0 and 1, but necessarily 0 on the diagonal) with a random vector which constsis of $\alpha$ and $\beta$ entries.
Does anyone have any ideas how to deduce some criterions for the invertibility of matrix $A$?
Thank you so much!
I already tried thinking about LU decomoposition, but could not deduce any criterion. Obviosly, it fully depends on how the random matrices look and linear dependence between the rows are less likely if the dimension is higher...
I'm working of matrices having rank >1. It is possible to reduce the rank of a matrix to rank=1 substituing some values to zeros?
Rank in a matrix refers to how many of the column vectors are independent and non-zero (Or row vectors, but I was taught to always use column vectors). So, if you're willing to lose a lot of the information about the transformation your matrix is defining, you could create a matrix that's just the first non-zero column of your matrix, and everything else set to zero. Guaranteed to be rank 1.
However, that loses a whole lot of information about the transformation. Perhaps a more useful thing to do would be project your matrix onto a space of size 1x1. There are ways to do this in such a way that can create an injection from your matrix to the new space, guaranteeing that no two matrices produce an equivalent result. The first one that comes to mind is:
Let A be an n x m matrix
Let {P_i} be the ith prime number.
Let F(A) = {sum from i to (n * m)} {P_i} ^ (A_(i div n),(i mod m))
While this generates a single number, you can think of a single number as a 1 x 1 matrix, which, if non-zero, has rank 1.
All that being said, rank 1 matrices are kinda boring and you can do cooler stuff with matrices if you keep it at rank != 1. In particular, if you have an n x n matrix with rank n, a whole world of possibility opens up. It really depends on what you want to use these matrices for.
You might want to look at the singular value decomposition, which can be used to write your matrix as a sum of weighted outer products (see here). Choosing only the highest-weighted component of this sum will give you the closest rank-1 approximation to the decomposed matrix.
Most common linear algebra libraries (Eigen, OpenCV, NumPy) have an SVD implementation.
I have two sparse matrices "Matrix1" and "Matrix2" of the same size p x n.
By sparse matrix I mean that it contains a lot of exactly zero elements.
I want to show the two matrices under the same colormap and a unique colorbar. Doing this in MATLAB is straightforward:
bottom = min(min(min(Matrix1)),min(min(Matrix2)));
top = max(max(max(Matrix1)),max(max(Matrix2)));
subplot(1,2,1)
imagesc(Matrix1)
colormap(gray)
caxis manual
caxis([bottom top]);
subplot(1,2,2)
imagesc(Matrix2)
colormap(gray)
caxis manual
caxis([bottom top]);
colorbar;
My problem:
In fact, when I show the matrix using imagesc(Matrix), it can ignore the noises (or backgrounds) that always appear with using imagesc(10*log10(Matrix)).
That is why, I want to show the 10*log10 of the matrices. But in this case, the minimum value will be -Inf since the matrices are sparse. In this case caxis will give an error because bottom is equal to -Inf.
What do you suggest me? How can I modify the above code?
Any help will be very appreciated!
A very important point is that the minimum value in your matrix will always be 0. Leveraging this, a very simple way to address your problem is to add 1 inside the log operation so that values that map to 0 in the original matrix also map to 0 in the log operation. This avoids the -Inf error that you're encountering. In fact, this is a very common way of visualizing the Fourier Transform if you will. Adding 1 to the logarithm ensures that the transform has no negative values in the output, yet the derivative or its rate of change remains intact as the effect is simply a translation of the curve by 1 unit to the left.
Therefore, simply do imagesc(10*log10(1 + Matrix));, then the minimum is always bounded at 0 while the maximum is unbounded but subject to the largest value that is seen in Matrix.
I would like to ask about Savitzky–Golay filter on 2D-images.
What are the best coefficient and order to choose for finding local details in the image.
Moreover, if someone has an explanation for coefficients and the orders one the 2D-images, it would be perfect.
Thanks in advance
Please check out this website:
https://en.wikipedia.org/wiki/Savitzky%E2%80%93Golay_filter#Two-dimensional_convolution_coefficients
UPDATE: (Thank you for the suggestion, #Rasclatt)
Which has been reproduced here:
Two-dimensional smoothing and differentiation can also be applied to tables of data values, such as intensity values in a photographic image which is composed of a rectangular grid of pixels.[16] [17] The trick is to transform part of the table into a row by a simple ordering of the indices of the pixels. Whereas the one-dimensional filter coefficients are found by fitting a polynomial in the subsidiary variable, z to a set of m data points, the two-dimensional coefficients are found by fitting a polynomial in subsidiary variables v and w to a set of m × m data points. The following example, for a bicubic polynomial and m = 5, illustrates the process, which parallels the process for the one dimensional case, above.[18]
The square of 25 data values, d1 − d25
becomes a vector when the rows are placed one after another.
The Jacobian has 10 columns, one for each of the parameters a00 − a03 and 25 rows, one for each pair of v and w values. Each row has the form
The convolution coefficients are calculated as
The first row of C contains 25 convolution coefficients which can be multiplied with the 25 data values to provide a smoothed value for the central data point (13) of the 25.
check out the below links which use SURE(Stein's unbiased risk estimator) to minimizes the mean squared error between your estimate and the image. This method is useful for denoising and data smoothing.
this link is for optimization of parameters for 1D Savitzky Golay Filter(this will be helpful to understand the 2D part)
https://ieeexplore.ieee.org/abstract/document/6331560/?part=1
this link is for optimization of parameters of 2D Savitzky Golay Filter
https://ieeexplore.ieee.org/document/6738095/
I read now tons of different explanations of the gaussian blur and I am really confused.
I roughly understand how the gaussian blur works.
http://en.wikipedia.org/wiki/Gaussian_blur
I understood that we choose 3*sigma as the maxium size for our mask because the values will get really small.
But my three questions are:
How do I create a gaussian mask with the sigma only?
If I understood it correctly, the mask gives me the weights, then
I place the mask on the top left pixel. I multiply the weights for
each value of the pixels in the mask. Then I move the mask to the
next pixel. I do this for all pixels. Is this correct?
I also know that 1D masks are faster. So I create a mask for x and a
mask for y. Lets say my mask would look like this. (3x3)
1 2 1
2 4 2
1 2 1
How would my x and y mask look like?
1- A solution to create a gaussian mask is to setup an N by N matrix, with N=3*sigma (or less if you want a coarser solution), and fill each entry (i,j) with exp(-((i-N/2)^2 + (j-N/2)^2)/(2*sigma^2)). As a comment mentioned, taking N=3*sigma just means that you truncate your gaussian at a "sufficiently small" threshold.
2- yes - you understood correctly. A small detail is that you'll need to normalize by the sum of your weights (ie., divide the result of what you said by the sum of all the elements of your matrix). The other option is that you can build your matrix already normalized, so that you don't need to perform this normalization at the end (the normalized gaussian formula becomes exp(-((i-N/2)^2 + (j-N/2)^2)/(2*sigma^2))/(2*pi*sigma))
3- In your specific case, the 1D version is [1 2 1] (ie, both your x and y masks) since you can obtain the matrix you gave with the multiplication transpose([1 2 1]) * [1 2 1]. In general, you can directly build these 1D gaussians using the 1D gaussian formula which is similar as the one above : exp(-((i-N/2)^2)/(2*sigma^2)) (or the normalized version exp(-((i-N/2)^2)/(2*sigma^2)) / (sigma*sqrt(2*pi)))