Finally, v3 is the vector that is perpendicular to both v1 and v2 and gives the greatest length of Ax with these constraints. && x_2^T - \mu^T && \\ If we can find the orthogonal basis and the stretching magnitude, can we characterize the data ? These three steps correspond to the three matrices U, D, and V. Now lets check if the three transformations given by the SVD are equivalent to the transformation done with the original matrix. Understanding the output of SVD when used for PCA, Interpreting matrices of SVD in practical applications. When we deal with a matrix (as a tool of collecting data formed by rows and columns) of high dimensions, is there a way to make it easier to understand the data information and find a lower dimensional representative of it ? The V matrix is returned in a transposed form, e.g. By focusing on directions of larger singular values, one might ensure that the data, any resulting models, and analyses are about the dominant patterns in the data. relationship between svd and eigendecomposition; relationship between svd and eigendecomposition. The matrix X^(T)X is called the Covariance Matrix when we centre the data around 0. when some of a1, a2, .., an are not zero. \newcommand{\vtau}{\vec{\tau}} Graphs models the rich relationships between different entities, so it is crucial to learn the representations of the graphs. Is a PhD visitor considered as a visiting scholar? We also have a noisy column (column #12) which should belong to the second category, but its first and last elements do not have the right values. \newcommand{\fillinblank}{\text{ }\underline{\text{ ? Math Statistics and Probability CSE 6740. given VV = I, we can get XV = U and let: Z1 is so called the first component of X corresponding to the largest 1 since 1 2 p 0. && x_1^T - \mu^T && \\ That is, the SVD expresses A as a nonnegative linear combination of minfm;ng rank-1 matrices, with the singular values providing the multipliers and the outer products of the left and right singular vectors providing the rank-1 matrices. https://hadrienj.github.io/posts/Deep-Learning-Book-Series-2.8-Singular-Value-Decomposition/, https://hadrienj.github.io/posts/Deep-Learning-Book-Series-2.12-Example-Principal-Components-Analysis/, https://brilliant.org/wiki/principal-component-analysis/#from-approximate-equality-to-minimizing-function, https://hadrienj.github.io/posts/Deep-Learning-Book-Series-2.7-Eigendecomposition/, http://infolab.stanford.edu/pub/cstr/reports/na/m/86/36/NA-M-86-36.pdf. Move on to other advanced topics in mathematics or machine learning. The main shape of the scatter plot, which is shown by the ellipse line (red) clearly seen. If A is an mp matrix and B is a pn matrix, the matrix product C=AB (which is an mn matrix) is defined as: For example, the rotation matrix in a 2-d space can be defined as: This matrix rotates a vector about the origin by the angle (with counterclockwise rotation for a positive ). It is important to note that if you do the multiplications on the right side of the above equation, you will not get A exactly. October 20, 2021. \newcommand{\permutation}[2]{{}_{#1} \mathrm{ P }_{#2}} A similar analysis leads to the result that the columns of \( \mU \) are the eigenvectors of \( \mA \mA^T \). We can measure this distance using the L Norm. The span of a set of vectors is the set of all the points obtainable by linear combination of the original vectors. Do roots of these polynomials approach the negative of the Euler-Mascheroni constant? So their multiplication still gives an nn matrix which is the same approximation of A. We know that ui is an eigenvector and it is normalized, so its length and its inner product with itself are both equal to 1. The matrix manifold M is dictated by the known physics of the system at hand. This result indicates that the first SVD mode captures the most important relationship between the CGT and SEALLH SSR in winter. I downoaded articles from libgen (didn't know was illegal) and it seems that advisor used them to publish his work. All the Code Listings in this article are available for download as a Jupyter notebook from GitHub at: https://github.com/reza-bagheri/SVD_article. If we use all the 3 singular values, we get back the original noisy column. X = \left( All the entries along the main diagonal are 1, while all the other entries are zero. Follow the above links to first get acquainted with the corresponding concepts. So if vi is the eigenvector of A^T A (ordered based on its corresponding singular value), and assuming that ||x||=1, then Avi is showing a direction of stretching for Ax, and the corresponding singular value i gives the length of Avi. Is there a proper earth ground point in this switch box? (27) 4 Trace, Determinant, etc. The matrix product of matrices A and B is a third matrix C. In order for this product to be dened, A must have the same number of columns as B has rows. We know that the eigenvalues of A are orthogonal which means each pair of them are perpendicular. They both split up A into the same r matrices u iivT of rank one: column times row. Please note that by convection, a vector is written as a column vector. Results: We develop a new technique for using the marginal relationship between gene ex-pression measurements and patient survival outcomes to identify a small subset of genes which appear highly relevant for predicting survival, produce a low-dimensional embedding based on . Now that we know that eigendecomposition is different from SVD, time to understand the individual components of the SVD. (You can of course put the sign term with the left singular vectors as well. It also has some important applications in data science. That is because the element in row m and column n of each matrix. the variance. \newcommand{\vs}{\vec{s}} Also called Euclidean norm (also used for vector L. \newcommand{\powerset}[1]{\mathcal{P}(#1)} In other terms, you want that the transformed dataset has a diagonal covariance matrix: the covariance between each pair of principal components is equal to zero. Figure 1 shows the output of the code. \newcommand{\mK}{\mat{K}} However, computing the "covariance" matrix AA squares the condition number, i.e. \renewcommand{\smallosymbol}[1]{\mathcal{o}} and each i is the corresponding eigenvalue of vi. A matrix whose columns are an orthonormal set is called an orthogonal matrix, and V is an orthogonal matrix. For example in Figure 26, we have the image of the national monument of Scotland which has 6 pillars (in the image), and the matrix corresponding to the first singular value can capture the number of pillars in the original image. 2. For each label k, all the elements are zero except the k-th element. Truncated SVD: how do I go from [Uk, Sk, Vk'] to low-dimension matrix? So the singular values of A are the length of vectors Avi. Another important property of symmetric matrices is that they are orthogonally diagonalizable. Figure 18 shows two plots of A^T Ax from different angles. When you have a non-symmetric matrix you do not have such a combination. where $v_i$ is the $i$-th Principal Component, or PC, and $\lambda_i$ is the $i$-th eigenvalue of $S$ and is also equal to the variance of the data along the $i$-th PC. The right field is the winter mean SSR over the SEALLH. The singular values can also determine the rank of A. Consider the following vector(v): Lets plot this vector and it looks like the following: Now lets take the dot product of A and v and plot the result, it looks like the following: Here, the blue vector is the original vector(v) and the orange is the vector obtained by the dot product between v and A. What is the molecular structure of the coating on cast iron cookware known as seasoning? Check out the post "Relationship between SVD and PCA. Replacing broken pins/legs on a DIP IC package. Its diagonal is the variance of the corresponding dimensions and other cells are the Covariance between the two corresponding dimensions, which tells us the amount of redundancy. Remember that we write the multiplication of a matrix and a vector as: So unlike the vectors in x which need two coordinates, Fx only needs one coordinate and exists in a 1-d space. In an n-dimensional space, to find the coordinate of ui, we need to draw a hyper-plane passing from x and parallel to all other eigenvectors except ui and see where it intersects the ui axis. The length of each label vector ik is one and these label vectors form a standard basis for a 400-dimensional space. But before explaining how the length can be calculated, we need to get familiar with the transpose of a matrix and the dot product. \newcommand{\setsymmdiff}{\oplus} So the objective is to lose as little as precision as possible. \newcommand{\set}[1]{\lbrace #1 \rbrace} Figure 22 shows the result. I hope that you enjoyed reading this article. (You can of course put the sign term with the left singular vectors as well. We will find the encoding function from the decoding function. \DeclareMathOperator*{\asterisk}{\ast} You can find more about this topic with some examples in python in my Github repo, click here. relationship between svd and eigendecomposition. %PDF-1.5 To understand how the image information is stored in each of these matrices, we can study a much simpler image. Full video list and slides: https://www.kamperh.com/data414/ That is because LA.eig() returns the normalized eigenvector. In this case, because all the singular values . PCA and Correspondence analysis in their relation to Biplot, Making sense of principal component analysis, eigenvectors & eigenvalues, davidvandebunte.gitlab.io/executable-notes/notes/se/, the relationship between PCA and SVD in this longer article, We've added a "Necessary cookies only" option to the cookie consent popup. An important reason to find a basis for a vector space is to have a coordinate system on that. As a result, we already have enough vi vectors to form U. \newcommand{\vsigma}{\vec{\sigma}} Each matrix iui vi ^T has a rank of 1 and has the same number of rows and columns as the original matrix. Specifically, section VI: A More General Solution Using SVD. \newcommand{\dataset}{\mathbb{D}} \newcommand{\mQ}{\mat{Q}} So the eigendecomposition mathematically explains an important property of the symmetric matrices that we saw in the plots before. Hence, $A = U \Sigma V^T = W \Lambda W^T$, and $$A^2 = U \Sigma^2 U^T = V \Sigma^2 V^T = W \Lambda^2 W^T$$. If in the original matrix A, the other (n-k) eigenvalues that we leave out are very small and close to zero, then the approximated matrix is very similar to the original matrix, and we have a good approximation. Given the close relationship between SVD, aging, and geriatric syndrome, geriatricians and health professionals who work with the elderly are very likely to encounter those with covert SVD in clinical or research settings. So A^T A is equal to its transpose, and it is a symmetric matrix. Similar to the eigendecomposition method, we can approximate our original matrix A by summing the terms which have the highest singular values. So we first make an r r diagonal matrix with diagonal entries of 1, 2, , r. Another example is the stretching matrix B in a 2-d space which is defined as: This matrix stretches a vector along the x-axis by a constant factor k but does not affect it in the y-direction. Each pixel represents the color or the intensity of light in a specific location in the image. To learn more about the application of eigendecomposition and SVD in PCA, you can read these articles: https://reza-bagheri79.medium.com/understanding-principal-component-analysis-and-its-application-in-data-science-part-1-54481cd0ad01, https://reza-bagheri79.medium.com/understanding-principal-component-analysis-and-its-application-in-data-science-part-2-e16b1b225620. A tutorial on Principal Component Analysis by Jonathon Shlens is a good tutorial on PCA and its relation to SVD. \newcommand{\mV}{\mat{V}} The left singular vectors $u_i$ are $w_i$ and the right singular vectors $v_i$ are $\text{sign}(\lambda_i) w_i$. Thanks for sharing. So I did not use cmap='gray' and did not display them as grayscale images. Then we only keep the first j number of significant largest principle components that describe the majority of the variance (corresponding the first j largest stretching magnitudes) hence the dimensional reduction. Initially, we have a circle that contains all the vectors that are one unit away from the origin. Remember that in the eigendecomposition equation, each ui ui^T was a projection matrix that would give the orthogonal projection of x onto ui. So label k will be represented by the vector: Now we store each image in a column vector. In NumPy you can use the transpose() method to calculate the transpose. The 4 circles are roughly captured as four rectangles in the first 2 matrices in Figure 24, and more details on them are added in the last 4 matrices. For those significantly smaller than previous , we can ignore them all. The following are some of the properties of Dot Product: Identity Matrix: An identity matrix is a matrix that does not change any vector when we multiply that vector by that matrix. We call it to read the data and stores the images in the imgs array. The SVD is, in a sense, the eigendecomposition of a rectangular matrix. (SVD) of M = U(M) (M)V(M)>and de ne M . SingularValueDecomposition(SVD) Introduction Wehaveseenthatsymmetricmatricesarealways(orthogonally)diagonalizable. Figure 10 shows an interesting example in which the 22 matrix A1 is multiplied by a 2-d vector x, but the transformed vector Ax is a straight line. So a grayscale image with mn pixels can be stored in an mn matrix or NumPy array. (a) Compare the U and V matrices to the eigenvectors from part (c). Using the SVD we can represent the same data using only 153+253+3 = 123 15 3 + 25 3 + 3 = 123 units of storage (corresponding to the truncated U, V, and D in the example above). The Threshold can be found using the following: A is a Non-square Matrix (mn) where m and n are dimensions of the matrix and is not known, in this case the threshold is calculated as: is the aspect ratio of the data matrix =m/n, and: and we wish to apply a lossy compression to these points so that we can store these points in a lesser memory but may lose some precision. relationship between svd and eigendecomposition. Now the eigendecomposition equation becomes: Each of the eigenvectors ui is normalized, so they are unit vectors. The close connection between the SVD and the well known theory of diagonalization for symmetric matrices makes the topic immediately accessible to linear algebra teachers, and indeed, a natural extension of what these teachers already know. Remember that if vi is an eigenvector for an eigenvalue, then (-1)vi is also an eigenvector for the same eigenvalue, and its length is also the same. In addition, though the direction of the reconstructed n is almost correct, its magnitude is smaller compared to the vectors in the first category. Why is this sentence from The Great Gatsby grammatical? This transformed vector is a scaled version (scaled by the value ) of the initial vector v. If v is an eigenvector of A, then so is any rescaled vector sv for s R, s!= 0. In fact, x2 and t2 have the same direction. Now consider some eigen-decomposition of $A$, $$A^2 = W\Lambda W^T W\Lambda W^T = W\Lambda^2 W^T$$. This time the eigenvectors have an interesting property. If you center this data (subtract the mean data point $\mu$ from each data vector $x_i$) you can stack the data to make a matrix, $$ Now that we are familiar with the transpose and dot product, we can define the length (also called the 2-norm) of the vector u as: To normalize a vector u, we simply divide it by its length to have the normalized vector n: The normalized vector n is still in the same direction of u, but its length is 1. \newcommand{\dox}[1]{\doh{#1}{x}} So. in the eigendecomposition equation is a symmetric nn matrix with n eigenvectors. S = \frac{1}{n-1} \sum_{i=1}^n (x_i-\mu)(x_i-\mu)^T = \frac{1}{n-1} X^T X To calculate the inverse of a matrix, the function np.linalg.inv() can be used. Matrix A only stretches x2 in the same direction and gives the vector t2 which has a bigger magnitude. So it is not possible to write. Is it correct to use "the" before "materials used in making buildings are"? For rectangular matrices, we turn to singular value decomposition (SVD). In this section, we have merely defined the various matrix types. If the set of vectors B ={v1, v2, v3 , vn} form a basis for a vector space, then every vector x in that space can be uniquely specified using those basis vectors : Now the coordinate of x relative to this basis B is: In fact, when we are writing a vector in R, we are already expressing its coordinate relative to the standard basis. So we get: and since the ui vectors are the eigenvectors of A, we finally get: which is the eigendecomposition equation. In fact u1= -u2. \newcommand{\entropy}[1]{\mathcal{H}\left[#1\right]} How to use Slater Type Orbitals as a basis functions in matrix method correctly? The SVD gives optimal low-rank approximations for other norms. Relationship between eigendecomposition and singular value decomposition. Let me try this matrix: The eigenvectors and corresponding eigenvalues are: Now if we plot the transformed vectors we get: As you see now we have stretching along u1 and shrinking along u2. Interested in Machine Learning and Deep Learning. Categories . You can now easily see that A was not symmetric. This is not true for all the vectors in x. We can store an image in a matrix. \newcommand{\vr}{\vec{r}} kat stratford pants; jeffrey paley son of william paley. Now we decompose this matrix using SVD. We can also add a scalar to a matrix or multiply a matrix by a scalar, just by performing that operation on each element of a matrix: We can also do the addition of a matrix and a vector, yielding another matrix: A matrix whose eigenvalues are all positive is called. The images show the face of 40 distinct subjects. We start by picking a random 2-d vector x1 from all the vectors that have a length of 1 in x (Figure 171). If LPG gas burners can reach temperatures above 1700 C, then how do HCA and PAH not develop in extreme amounts during cooking? The result is shown in Figure 4. u2-coordinate can be found similarly as shown in Figure 8. \newcommand{\ndatasmall}{d} Now if we use ui as a basis, we can decompose n and find its orthogonal projection onto ui. If we multiply both sides of the SVD equation by x we get: We know that the set {u1, u2, , ur} is an orthonormal basis for Ax. And therein lies the importance of SVD. What video game is Charlie playing in Poker Face S01E07? Since \( \mU \) and \( \mV \) are strictly orthogonal matrices and only perform rotation or reflection, any stretching or shrinkage has to come from the diagonal matrix \( \mD \). Thanks for your anser Andre. In addition, the eigenvectors are exactly the same eigenvectors of A. The vectors fk will be the columns of matrix M: This matrix has 4096 rows and 400 columns. Now we can summarize an important result which forms the backbone of the SVD method. Before going into these topics, I will start by discussing some basic Linear Algebra and then will go into these topics in detail. In linear algebra, the Singular Value Decomposition (SVD) of a matrix is a factorization of that matrix into three matrices. Let $A \in \mathbb{R}^{n\times n}$ be a real symmetric matrix. Now we can write the singular value decomposition of A as: where V is an nn matrix that its columns are vi. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions. And \( \mD \in \real^{m \times n} \) is a diagonal matrix containing singular values of the matrix \( \mA \). \newcommand{\vo}{\vec{o}} The Eigendecomposition of A is then given by: Decomposing a matrix into its corresponding eigenvalues and eigenvectors help to analyse properties of the matrix and it helps to understand the behaviour of that matrix. \newcommand{\inf}{\text{inf}} for example, the center position of this group of data the mean, (2) how the data are spreading (magnitude) in different directions. But the matrix \( \mQ \) in an eigendecomposition may not be orthogonal. For example, suppose that our basis set B is formed by the vectors: To calculate the coordinate of x in B, first, we form the change-of-coordinate matrix: Now the coordinate of x relative to B is: Listing 6 shows how this can be calculated in NumPy. While they share some similarities, there are also some important differences between them. The following is another geometry of the eigendecomposition for A. So the rank of Ak is k, and by picking the first k singular values, we approximate A with a rank-k matrix. But the eigenvectors of a symmetric matrix are orthogonal too. Eigenvalues are defined as roots of the characteristic equation det (In A) = 0. For rectangular matrices, some interesting relationships hold. It only takes a minute to sign up. Suppose that, However, we dont apply it to just one vector. In the previous example, the rank of F is 1. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Eigenvalue decomposition Singular value decomposition, Relation in PCA and EigenDecomposition $A = W \Lambda W^T$, Singular value decomposition of positive definite matrix, Understanding the singular value decomposition (SVD), Relation between singular values of a data matrix and the eigenvalues of its covariance matrix. So they span Ak x and since they are linearly independent they form a basis for Ak x (or col A). \( \mU \in \real^{m \times m} \) is an orthogonal matrix. In fact, in Listing 10 we calculated vi with a different method and svd() is just reporting (-1)vi which is still correct. We can think of a matrix A as a transformation that acts on a vector x by multiplication to produce a new vector Ax. Study Resources. If we only include the first k eigenvalues and eigenvectors in the original eigendecomposition equation, we get the same result: Now Dk is a kk diagonal matrix comprised of the first k eigenvalues of A, Pk is an nk matrix comprised of the first k eigenvectors of A, and its transpose becomes a kn matrix. Let $A \in \mathbb{R}^{n\times n}$ be a real symmetric matrix. So the set {vi} is an orthonormal set. The general effect of matrix A on the vectors in x is a combination of rotation and stretching. bendigo health intranet. I wrote this FAQ-style question together with my own answer, because it is frequently being asked in various forms, but there is no canonical thread and so closing duplicates is difficult. Now the column vectors have 3 elements. $$, measures to which degree the different coordinates in which your data is given vary together. What molecular features create the sensation of sweetness? Note that \( \mU \) and \( \mV \) are square matrices This is a closed set, so when the vectors are added or multiplied by a scalar, the result still belongs to the set. To maximize the variance and minimize the covariance (in order to de-correlate the dimensions) means that the ideal covariance matrix is a diagonal matrix (non-zero values in the diagonal only).The diagonalization of the covariance matrix will give us the optimal solution. Dimensions with higher singular values are more dominant (stretched) and conversely, those with lower singular values are shrunk. Math Statistics and Probability CSE 6740. First, we load the dataset: The fetch_olivetti_faces() function has been already imported in Listing 1. How long would it take for sucrose to undergo hydrolysis in boiling water? Then we filter the non-zero eigenvalues and take the square root of them to get the non-zero singular values. As a result, the dimension of R is 2. \newcommand{\cdf}[1]{F(#1)} So we can now write the coordinate of x relative to this new basis: and based on the definition of basis, any vector x can be uniquely written as a linear combination of the eigenvectors of A. For rectangular matrices, we turn to singular value decomposition. What is the relationship between SVD and eigendecomposition? In SVD, the roles played by \( \mU, \mD, \mV^T \) are similar to those of \( \mQ, \mLambda, \mQ^{-1} \) in eigendecomposition. However, it can also be performed via singular value decomposition (SVD) of the data matrix $\mathbf X$. We also know that the set {Av1, Av2, , Avr} is an orthogonal basis for Col A, and i = ||Avi||. One of them is zero and the other is equal to 1 of the original matrix A. LinkedIn: https://www.linkedin.com/in/reza-bagheri-71882a76/, https://github.com/reza-bagheri/SVD_article, https://www.linkedin.com/in/reza-bagheri-71882a76/. Then it can be shown that rank A which is the number of vectors that form the basis of Ax is r. It can be also shown that the set {Av1, Av2, , Avr} is an orthogonal basis for Ax (the Col A). \newcommand{\max}{\text{max}\;} \newcommand{\mR}{\mat{R}} What about the next one ? \newcommand{\vu}{\vec{u}} We call physics-informed DMD (piDMD) as the optimization integrates underlying knowledge of the system physics into the learning framework. Then we try to calculate Ax1 using the SVD method. The problem is that I see formulas where $\lambda_i = s_i^2$ and try to understand, how to use them? Every real matrix A Rmn A R m n can be factorized as follows A = UDVT A = U D V T Such formulation is known as the Singular value decomposition (SVD). As a result, we need the first 400 vectors of U to reconstruct the matrix completely. Anonymous sites used to attack researchers. \newcommand{\mat}[1]{\mathbf{#1}} How does it work? \DeclareMathOperator*{\argmin}{arg\,min} So to write a row vector, we write it as the transpose of a column vector. As a special case, suppose that x is a column vector. So we can say that that v is an eigenvector of A. eigenvectors are those Vectors(v) when we apply a square matrix A on v, will lie in the same direction as that of v. Suppose that a matrix A has n linearly independent eigenvectors {v1,.,vn} with corresponding eigenvalues {1,.,n}. Why do universities check for plagiarism in student assignments with online content? They correspond to a new set of features (that are a linear combination of the original features) with the first feature explaining most of the variance. So the vector Ax can be written as a linear combination of them. \newcommand{\vec}[1]{\mathbf{#1}} You should notice a few things in the output. What is the relationship between SVD and eigendecomposition? NumPy has a function called svd() which can do the same thing for us. \newcommand{\vk}{\vec{k}} \newcommand{\star}[1]{#1^*} The output is: To construct V, we take the vi vectors corresponding to the r non-zero singular values of A and divide them by their corresponding singular values. So the transpose of P has been written in terms of the transpose of the columns of P. This factorization of A is called the eigendecomposition of A. Please answer ALL parts Part 1: Discuss at least 1 affliction Please answer ALL parts . \newcommand{\sup}{\text{sup}} So for the eigenvectors, the matrix multiplication turns into a simple scalar multiplication. So. Figure 35 shows a plot of these columns in 3-d space. @OrvarKorvar: What n x n matrix are you talking about ? By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. The first direction of stretching can be defined as the direction of the vector which has the greatest length in this oval (Av1 in Figure 15). As you see the 2nd eigenvalue is zero. If we reconstruct a low-rank matrix (ignoring the lower singular values), the noise will be reduced, however, the correct part of the matrix changes too. What can a lawyer do if the client wants him to be acquitted of everything despite serious evidence? So the rank of A is the dimension of Ax. The first element of this tuple is an array that stores the eigenvalues, and the second element is a 2-d array that stores the corresponding eigenvectors. Let me start with PCA. Where does this (supposedly) Gibson quote come from. Here the rotation matrix is calculated for =30 and in the stretching matrix k=3. \newcommand{\ndimsmall}{n} What does this tell you about the relationship between the eigendecomposition and the singular value decomposition? Note that the eigenvalues of $A^2$ are positive. \newcommand{\mB}{\mat{B}} In other words, none of the vi vectors in this set can be expressed in terms of the other vectors. Now we go back to the eigendecomposition equation again. \newcommand{\vc}{\vec{c}} In general, an mn matrix does not necessarily transform an n-dimensional vector into anther m-dimensional vector. is an example. now we can calculate ui: So ui is the eigenvector of A corresponding to i (and i). Since it projects all the vectors on ui, its rank is 1. If we now perform singular value decomposition of $\mathbf X$, we obtain a decomposition $$\mathbf X = \mathbf U \mathbf S \mathbf V^\top,$$ where $\mathbf U$ is a unitary matrix (with columns called left singular vectors), $\mathbf S$ is the diagonal matrix of singular values $s_i$ and $\mathbf V$ columns are called right singular vectors.
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