Computer vision for dummies » Linear algebra https://www.visiondummy.com A blog about intelligent algorithms, machine learning, computer vision, datamining and more. Tue, 04 May 2021 14:17:31 +0000 en-US hourly 1 https://wordpress.org/?v=3.8.39 A geometric interpretation of the covariance matrix https://www.visiondummy.com/2014/04/geometric-interpretation-covariance-matrix/ https://www.visiondummy.com/2014/04/geometric-interpretation-covariance-matrix/#comments Thu, 24 Apr 2014 11:09:38 +0000 http://www.visiondummy.com/?p=440 In this article, we provide an intuitive, geometric interpretation of the covariance matrix, by exploring the relation between linear transformations and the resulting data covariance. Most textbooks explain the shape of data based on the concept of covariance matrices. Instead, we take a backwards approach and explain the concept of covariance matrices based on the [...]

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]]> Introduction

In this article, we provide an intuitive, geometric interpretation of the covariance matrix, by exploring the relation between linear transformations and the resulting data covariance. Most textbooks explain the shape of data based on the concept of covariance matrices. Instead, we take a backwards approach and explain the concept of covariance matrices based on the shape of data.


In a previous article, we discussed the concept of variance, and provided a derivation and proof of the well known formula to estimate the sample variance. Figure 1 was used in this article to show that the standard deviation, as the square root of the variance, provides a measure of how much the data is spread across the feature space.

Normal distribution

Figure 1. Gaussian density function. For normally distributed data, 68% of the samples fall within the interval defined by the mean plus and minus the standard deviation.

We showed that an unbiased estimator of the sample variance can be obtained by:

(1)   \begin{align*} \sigma_x^2 &= \frac{1}{N-1} \sum_{i=1}^N (x_i - \mu)^2\\ &= \mathbb{E}[ (x - \mathbb{E}(x)) (x - \mathbb{E}(x))]\\ &= \sigma(x,x) \end{align*}

However, variance can only be used to explain the spread of the data in the directions parallel to the axes of the feature space. Consider the 2D feature space shown by figure 2:

Data with a positive covariance

Figure 2. The diagnoal spread of the data is captured by the covariance.

For this data, we could calculate the variance \sigma(x,x) in the x-direction and the variance \sigma(y,y) in the y-direction. However, the horizontal spread and the vertical spread of the data does not explain the clear diagonal correlation. Figure 2 clearly shows that on average, if the x-value of a data point increases, then also the y-value increases, resulting in a positive correlation. This correlation can be captured by extending the notion of variance to what is called the ‘covariance’ of the data:

(2)   \begin{equation*} \sigma(x,y) = \mathbb{E}[ (x - \mathbb{E}(x)) (y - \mathbb{E}(y))] \end{equation*}

For 2D data, we thus obtain \sigma(x,x), \sigma(y,y), \sigma(x,y) and \sigma(y,x). These four values can be summarized in a matrix, called the covariance matrix:

(3)   \begin{equation*} \Sigma = \begin{bmatrix} \sigma(x,x) & \sigma(x,y) \\[0.3em] \sigma(y,x) & \sigma(y,y) \\[0.3em] \end{bmatrix} \end{equation*}

If x is positively correlated with y, y is also positively correlated with x. In other words, we can state that \sigma(x,y) = \sigma(y,x). Therefore, the covariance matrix is always a symmetric matrix with the variances on its diagonal and the covariances off-diagonal. Two-dimensional normally distributed data is explained completely by its mean and its 2\times 2 covariance matrix. Similarly, a 3 \times 3 covariance matrix is used to capture the spread of three-dimensional data, and a N \times N covariance matrix captures the spread of N-dimensional data.

Figure 3 illustrates how the overall shape of the data defines the covariance matrix:

The spread of the data is defined by its covariance matrix

Figure 3. The covariance matrix defines the shape of the data. Diagonal spread is captured by the covariance, while axis-aligned spread is captured by the variance.

Eigendecomposition of a covariance matrix

In the next section, we will discuss how the covariance matrix can be interpreted as a linear operator that transforms white data into the data we observed. However, before diving into the technical details, it is important to gain an intuitive understanding of how eigenvectors and eigenvalues uniquely define the covariance matrix, and therefore the shape of our data.

As we saw in figure 3, the covariance matrix defines both the spread (variance), and the orientation (covariance) of our data. So, if we would like to represent the covariance matrix with a vector and its magnitude, we should simply try to find the vector that points into the direction of the largest spread of the data, and whose magnitude equals the spread (variance) in this direction.

If we define this vector as \vec{v}, then the projection of our data D onto this vector is obtained as \vec{v}^{\intercal} D, and the variance of the projected data is \vec{v}^{\intercal} \Sigma \vec{v}. Since we are looking for the vector \vec{v} that points into the direction of the largest variance, we should choose its components such that the covariance matrix \vec{v}^{\intercal} \Sigma \vec{v} of the projected data is as large as possible. Maximizing any function of the form \vec{v}^{\intercal} \Sigma \vec{v} with respect to \vec{v}, where \vec{v} is a normalized unit vector, can be formulated as a so called Rayleigh Quotient. The maximum of such a Rayleigh Quotient is obtained by setting \vec{v} equal to the largest eigenvector of matrix \Sigma.

In other words, the largest eigenvector of the covariance matrix always points into the direction of the largest variance of the data, and the magnitude of this vector equals the corresponding eigenvalue. The second largest eigenvector is always orthogonal to the largest eigenvector, and points into the direction of the second largest spread of the data.

Now let’s have a look at some examples. In an earlier article we saw that a linear transformation matrix T is completely defined by its eigenvectors and eigenvalues. Applied to the covariance matrix, this means that:

(4)   \begin{equation*} \Sigma \vec{v} = \lambda \vec{v} \end{equation*}

where \vec{v} is an eigenvector of \Sigma, and \lambda is the corresponding eigenvalue.

If the covariance matrix of our data is a diagonal matrix, such that the covariances are zero, then this means that the variances must be equal to the eigenvalues \lambda. This is illustrated by figure 4, where the eigenvectors are shown in green and magenta, and where the eigenvalues clearly equal the variance components of the covariance matrix.

Eigenvectors of a covariance matrix

Figure 4. Eigenvectors of a covariance matrix

However, if the covariance matrix is not diagonal, such that the covariances are not zero, then the situation is a little more complicated. The eigenvalues still represent the variance magnitude in the direction of the largest spread of the data, and the variance components of the covariance matrix still represent the variance magnitude in the direction of the x-axis and y-axis. But since the data is not axis aligned, these values are not the same anymore as shown by figure 5.

Eigenvectors with covariance

Figure 5. Eigenvalues versus variance

By comparing figure 5 with figure 4, it becomes clear that the eigenvalues represent the variance of the data along the eigenvector directions, whereas the variance components of the covariance matrix represent the spread along the axes. If there are no covariances, then both values are equal.

Covariance matrix as a linear transformation

Now let’s forget about covariance matrices for a moment. Each of the examples in figure 3 can simply be considered to be a linearly transformed instance of figure 6:

White data

Figure 6. Data with unit covariance matrix is called white data.

Let the data shown by figure 6 be D, then each of the examples shown by figure 3 can be obtained by linearly transforming D:

(5)   \begin{equation*} D' = T \, D \end{equation*}

where T is a transformation matrix consisting of a rotation matrix R and a scaling matrix S:

(6)   \begin{equation*} T = R \, S. \end{equation*}

These matrices are defined as:

(7)   \begin{equation*} R = \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\[0.3em] \sin(\theta) & \cos(\theta) \end{bmatrix} \end{equation*}

where \theta is the rotation angle, and:

(8)   \begin{equation*} S = \begin{bmatrix} s_x & 0 \\[0.3em] 0 & s_y \end{bmatrix} \end{equation*}

where s_x and s_y are the scaling factors in the x direction and the y direction respectively.

In the following paragraphs, we will discuss the relation between the covariance matrix \Sigma, and the linear transformation matrix T = R\, S.

Let’s start with unscaled (scale equals 1) and unrotated data. In statistics this is often refered to as ‘white data’ because its samples are drawn from a standard normal distribution and therefore correspond to white (uncorrelated) noise:

Whitened data

Figure 7. White data is data with a unit covariance matrix.

The covariance matrix of this ‘white’ data equals the identity matrix, such that the variances and standard deviations equal 1 and the covariance equals zero:

(9)   \begin{equation*} \Sigma = \begin{bmatrix} \sigma_x^2 & 0 \\[0.3em] 0 & \sigma_y^2 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\ \end{bmatrix} \end{equation*}

Now let’s scale the data in the x-direction with a factor 4:

(10)   \begin{equation*} D' = \begin{bmatrix} 4 & 0 \\[0.3em] 0 & 1 \\ \end{bmatrix} \, D \end{equation*}

The data D' now looks as follows:

Data with variance in the x-direction

Figure 8. Variance in the x-direction results in a horizontal scaling.

The covariance matrix \Sigma' of D' is now:

(11)   \begin{equation*} \Sigma' = \begin{bmatrix} \sigma_x^2 & 0 \\[0.3em] 0 & \sigma_y^2 \\ \end{bmatrix} = \begin{bmatrix} 16 & 0 \\[0.3em] 0 & 1 \\ \end{bmatrix} \end{equation*}

Thus, the covariance matrix \Sigma' of the resulting data D' is related to the linear transformation T that is applied to the original data as follows: D' = T \, D, where

(12)   \begin{equation*} T = \sqrt{\Sigma'} = \begin{bmatrix} 4 & 0 \\[0.3em] 0 & 1 \\ \end{bmatrix}. \end{equation*}

However, although equation (12) holds when the data is scaled in the x and y direction, the question rises if it also holds when a rotation is applied. To investigate the relation between the linear transformation matrix T and the covariance matrix \Sigma' in the general case, we will therefore try to decompose the covariance matrix into the product of rotation and scaling matrices.

As we saw earlier, we can represent the covariance matrix by its eigenvectors and eigenvalues:

(13)   \begin{equation*} \Sigma \vec{v} = \lambda \vec{v} \end{equation*}

where \vec{v} is an eigenvector of \Sigma, and \lambda is the corresponding eigenvalue.

Equation (13) holds for each eigenvector-eigenvalue pair of matrix \Sigma. In the 2D case, we obtain two eigenvectors and two eigenvalues. The system of two equations defined by equation (13) can be represented efficiently using matrix notation:

(14)   \begin{equation*} \Sigma \, V = V \, L \end{equation*}

where V is the matrix whose columns are the eigenvectors of \Sigma and L is the diagonal matrix whose non-zero elements are the corresponding eigenvalues.

This means that we can represent the covariance matrix as a function of its eigenvectors and eigenvalues:

(15)   \begin{equation*} \Sigma = V \, L \, V^{-1} \end{equation*}

Equation (15) is called the eigendecomposition of the covariance matrix and can be obtained using a Singular Value Decomposition algorithm. Whereas the eigenvectors represent the directions of the largest variance of the data, the eigenvalues represent the magnitude of this variance in those directions. In other words, V represents a rotation matrix, while \sqrt{L} represents a scaling matrix. The covariance matrix can thus be decomposed further as:

(16)   \begin{equation*} \Sigma = R \, S \, S \, R^{-1} \end{equation*}

where R=V is a rotation matrix and S=\sqrt{L} is a scaling matrix.

In equation (6) we defined a linear transformation T=R \, S. Since S is a diagonal scaling matrix, S = S^{\intercal}. Furthermore, since R is an orthogonal matrix, R^{-1} = R^{\intercal}. Therefore, T^{\intercal} = (R \, S)^{\intercal} = S^{\intercal} \, R^{\intercal} = S \, R^{-1}. The covariance matrix can thus be written as:

(17)   \begin{equation*} \Sigma = R \, S \, S \, R^{-1} = T \, T^{\intercal}, \end{equation*}

In other words, if we apply the linear transformation defined by T=R \, S to the original white data D shown by figure 7, we obtain the rotated and scaled data D' with covariance matrix T \, T^{\intercal} = \Sigma' = R \, S \, S \, R^{-1}. This is illustrated by figure 10:

The covariance matrix represents a linear transformation of the original data

Figure 10. The covariance matrix represents a linear transformation of the original data.

The colored arrows in figure 10 represent the eigenvectors. The largest eigenvector, i.e. the eigenvector with the largest corresponding eigenvalue, always points in the direction of the largest variance of the data and thereby defines its orientation. Subsequent eigenvectors are always orthogonal to the largest eigenvector due to the orthogonality of rotation matrices.

Conclusion

In this article we showed that the covariance matrix of observed data is directly related to a linear transformation of white, uncorrelated data. This linear transformation is completely defined by the eigenvectors and eigenvalues of the data. While the eigenvectors represent the rotation matrix, the eigenvalues correspond to the square of the scaling factor in each dimension.

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]]> https://www.visiondummy.com/2014/04/geometric-interpretation-covariance-matrix/feed/ 47 What are eigenvectors and eigenvalues? https://www.visiondummy.com/2014/03/eigenvalues-eigenvectors/ https://www.visiondummy.com/2014/03/eigenvalues-eigenvectors/#comments Wed, 05 Mar 2014 14:44:53 +0000 http://www.visiondummy.com/?p=111 Eigenvectors and eigenvalues have many important applications in computer vision and machine learning in general. Well known examples are PCA (Principal Component Analysis) for dimensionality reduction or EigenFaces for face recognition. An interesting use of eigenvectors and eigenvalues is also illustrated in my post about error ellipses. Furthermore, eigendecomposition forms the base of the geometric [...]

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]]> Introduction

Eigenvectors and eigenvalues have many important applications in computer vision and machine learning in general. Well known examples are PCA (Principal Component Analysis) for dimensionality reduction or EigenFaces for face recognition. An interesting use of eigenvectors and eigenvalues is also illustrated in my post about error ellipses. Furthermore, eigendecomposition forms the base of the geometric interpretation of covariance matrices, discussed in an more recent post. In this article, I will provide a gentle introduction into this mathematical concept, and will show how to manually obtain the eigendecomposition of a 2D square matrix.

An eigenvector is a vector whose direction remains unchanged when a linear transformation is applied to it. Consider the image below in which three vectors are shown. The green square is only drawn to illustrate the linear transformation that is applied to each of these three vectors.

eigenvectors

Eigenvectors (red) do not change direction when a linear transformation (e.g. scaling) is applied to them. Other vectors (yellow) do.

The transformation in this case is a simple scaling with factor 2 in the horizontal direction and factor 0.5 in the vertical direction, such that the transformation matrix A is defined as:

A=\begin{bmatrix} 2 & 0 \\ 0 & 0.5 \end{bmatrix}.

A vector \vec{v}=(x,y) is then scaled by applying this transformation as \vec{v}\prime = A\vec{v}. The above figure shows that the direction of some vectors (shown in red) is not affected by this linear transformation. These vectors are called eigenvectors of the transformation, and uniquely define the square matrix A. This unique, deterministic relation is exactly the reason that those vectors are called ‘eigenvectors’ (Eigen means ‘specific’ in German).

In general, the eigenvector \vec{v} of a matrix A is the vector for which the following holds:

(1)   \begin{equation*} A \vec{v} = \lambda \vec{v} \end{equation*}

where \lambda is a scalar value called the ‘eigenvalue’. This means that the linear transformation A on vector \vec{v} is completely defined by \lambda.

We can rewrite equation (1) as follows:

(2)   \begin{eqnarray*} A \vec{v} - \lambda \vec{v} = 0 \\ \Rightarrow \vec{v} (A - \lambda I) = 0, \end{eqnarray*}

where I is the identity matrix of the same dimensions as A.

However, assuming that \vec{v} is not the null-vector, equation (2) can only be defined if (A - \lambda I) is not invertible. If a square matrix is not invertible, that means that its determinant must equal zero. Therefore, to find the eigenvectors of A, we simply have to solve the following equation:

(3)   \begin{equation*} Det(A - \lambda I) = 0. \end{equation*}

In the following sections we will determine the eigenvectors and eigenvalues of a matrix A, by solving equation (3). Matrix A in this example, is defined by:

(4)   \begin{equation*} A = \begin{bmatrix} 2 & 3 \\ 2 & 1 \end{bmatrix}. \end{equation*}

Calculating the eigenvalues

To determine the eigenvalues for this example, we substitute A in equation (3) by equation (4) and obtain:

(5)   \begin{equation*} Det\begin{pmatrix}2-\lambda&3\\2&1-\lambda\end{pmatrix}=0. \end{equation*}

Calculating the determinant gives:

(6)   \begin{align*} &(2-\lambda)(1-\lambda) - 6 = 0\\ \Rightarrow &2 - 2 \lambda - \lambda - \lambda^2 -6 = 0\\ \Rightarrow &{\lambda}^2 - 3 \lambda -4 = 0. \end{align*}

To solve this quadratic equation in \lambda, we find the discriminant:

    \begin{equation*} D = b^2 -4ac = (-3)^2 -4*1*(-4) = 9+16 = 25. \end{equation*}

Since the discriminant is strictly positive, this means that two different values for \lambda exist:

(7)   \begin{align*} \lambda _1 &= \frac{-b - \sqrt{D}}{2a} = \frac{3-5}{2} = -1,\\ \lambda _2 &= \frac{-b + \sqrt{D}}{2a} = \frac{3+5}{2} = 4. \end{align*}

We have now determined the two eigenvalues \lambda_1 and \lambda_2. Note that a square matrix of size N \times N always has exactly N eigenvalues, each with a corresponding eigenvector. The eigenvalue specifies the size of the eigenvector.

Calculating the first eigenvector

We can now determine the eigenvectors by plugging the eigenvalues from equation (7) into equation (1) that originally defined the problem. The eigenvectors are then found by solving this system of equations.

We first do this for eigenvalue \lambda_1, in order to find the corresponding first eigenvector:

    \begin{equation*} \begin{bmatrix}2&3\\2&1\end{bmatrix} \begin{bmatrix}x_{11}\\x_{12}\end{bmatrix} = -1 \begin{bmatrix}x_{11}\\x_{12}\end{bmatrix}. \end{equation*}

Since this is simply the matrix notation for a system of equations, we can write it in its equivalent form:

(8)   \begin{eqnarray*} \left\{ \begin{array}{lr} 2x_{11} + 3x_{12} = -x_{11}\\ 2x_{11} + x_{12} = -x_{12} \end{array} \right. \end{eqnarray*}

and solve the first equation as a function of x_{12}, resulting in:

(9)   \begin{equation*} x_{11} = -x_{12}. \end{equation*}

Since an eigenvector simply represents an orientation (the corresponding eigenvalue represents the magnitude), all scalar multiples of the eigenvector are vectors that are parallel to this eigenvector, and are therefore equivalent (If we would normalize the vectors, they would all be equal). Thus, instead of further solving the above system of equations, we can freely chose a real value for either x_{11} or x_{12}, and determine the other one by using equation (9).

For this example, we arbitrarily choose x_{12} = 1, such that x_{11}=-1. Therefore, the eigenvector that corresponds to eigenvalue \lambda_1 = -1 is

(10)   \begin{equation*} \vec{v}_1 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}. \end{equation*}

Calculating the second eigenvector

Calculations for the second eigenvector are similar to those needed for the first eigenvector;
We now substitute eigenvalue \lambda_2=4 into equation (1), yielding:

(11)   \begin{equation*} \begin{bmatrix}2&3\\2&1\end{bmatrix} \begin{bmatrix}x_{21}\\x_{22}\end{bmatrix} = 4 * \begin{bmatrix}x_{21}\\x_{22}\end{bmatrix}. \end{equation*}

Written as a system of equations, this is equivalent to:

(12)   \begin{eqnarray*} \left\{ \begin{array}{lr} 2x_{21} + 3x_{22} = 4x_{21}\\ 2x_{21} + x_{22} = 4x_{22} \end{array} \right. \end{eqnarray*}

Solving the first equation as a function of x_{21} resuls in:

(13)   \begin{equation*} x_{22} = \frac{3}{2}x_{21} \end{equation*}

We then arbitrarily choose x_{21}=2, and find x_{22}=3. Therefore, the eigenvector that corresponds to eigenvalue \lambda_2 = 4 is

(14)   \begin{equation*} \vec{v}_2 = \begin{bmatrix} 3 \\ 2 \end{bmatrix}. \end{equation*}

Conclusion

In this article we reviewed the theoretical concepts of eigenvectors and eigenvalues. These concepts are of great importance in many techniques used in computer vision and machine learning, such as dimensionality reduction by means of PCA, or face recognition by means of EigenFaces.

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