Principal Component Analysis fully explained

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Principal Component Analysis fully explained - Quiz 1

1 / 4

Amongst the principal components from PCA, which one is the most important?

2 / 4

What are the usages of PCA? Choose all that apply.

3 / 4

What best describes the transformation given by using PCA?

4 / 4

Which of the following is additive for independent variables?

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Although being used intensively in many applications, from feature engineering to data compression, from tabular to image data, the Principal Component Analysis technique (a.k.a PCA) seems to remain vague for many people. Most of the tutorials about this topic are either high-level or contain extensive maths with lots of abbreviations assuming readers’ extensive knowledge base.

This article, thus, was born with the attempt to make PCA crystal clear to anyone who wishes to understand it thoroughly, step-by-step, in both high and low-level concepts.

An overview

The aim of PCA is to summarize the data.

When a dataset is collected, there may be many features that are similar to each other. For example, in the well-known Ames Housing dataset, which collects information of houses to predict their prices, the 2 features GarageArea (size of the garage in square feet) and GarageCars (size of the garage in car capacity) are highly correlated.

This overlapping information makes it inefficient for processing: data occupies more disk and RAM space, the computational time increases, the predictive models are confused (due to multicollinearity). To facilitate this problem, there are 2 methods in general:

  • Choose and remove some redundant features. For instance, we may want to remove the entire column about GarageCars.
  • Combine the correlated features into one. In this example, an option is to standardize GarageArea and GarageCars, then combine them by taking the average of these 2 values.

PCA follows the latter option. However, it is a bit more complex than just taking the average, instead, it transforms the entire problem space into a smaller-dimensional space. Suppose, originally, our dataset has 30 features, which means the problem space has 30 dimensions, we may use PCA to project it onto a smaller, say, 10 dimensional-space. Each of the 10 dimensions is a distinct linear combination of the original 30.

If you feel confused about the above definition, don’t worry. Let’s make it easier using a simpler example. Suppose we have a 2 dimensional-space and we want to reduce it to 1. Look at this figure:

data points on 2 dimensional-space
data points on 2 dimensional-space.

We have 9 samples and 2 features. As you can see, these 2 features (x_1 and x_2) are highly correlated, when one goes up, the other tends to increase as well.

For the purpose of illustration, let me highlight several random points. Let’s call them P1 and P2, where P1 is in location (3, 2) and P2 is in (7, 9):

highlight 2 random points, call them P1 and P2
highlight 2 random points, call them P1 and P2.

Now, as we want to project these 9 points onto a 1 dimensional-space, we should first define what the new space is. This new space may be x_1 alone (we just remove x_2 from the dataset), or x_2 alone (remove x_1), or a combination of x_1 and x_2. What does a combination mean? Well, it is something like g = 0.6x_1 + 0.5x_2 + 0.3. In other words, it is a line in the original space. Let’s make a line g:

draw a line g, which acts as the new basis
draw a line g, which acts as a new basis.

We say: g is our new space, we don’t need x_1 and x_2 anymore, every point is now being measured by the g-dimension only. We discard the 2 old dimensions and project the points on the new dimension:

project the points on the new basis
project the points on g – a new basis.

Yes! We just compressed the data, from 2 features into 1.

At first, our data looks like this:

Data PointX_1X_2

Now, it becomes:

Data Pointg

Are we done with the PCA? Of course not. We just know that PCA helps us with compressing the features, from d features into k ones, with k \leq d (in the above, it is from 2 features to 1 feature). We have not learned about how PCA actually does that and proof of its correctness. This will be elaborated in the next sections.

Another example of compressing data is with television. We know we are living in a 3-dimensional world, however, the TV is flat, it can only show 2-dimensional movies. In fact, there are infinite numbers of ways to choose a 2-dimensional basis from a 3-dimensional one (i.e. you can place your eyes in an infinite number of positions). The director (or the cameraman) tries to select the best place to put the camera on, this is analogous to us trying to find the best sub-dimensional space to project the data points so that the most information is retained.

At this point, it is the right time to mention that: everything comes with a cost, as we get the advantage of reducing the dimensionality, we have to trade-off by losing some amount of information from the data, and we try our best to make this loss as small as possible.

Let’s go back to the example above when we have 30 features in our dataset. Suppose we try to apply PCA here, then at the end of the day, PCA gives something like this:

Information retained by each new feature
Information retained by each new feature, from new-feature-1 to new-feature-30.

This bar-plot shows how much information we can retain using each new feature. For example, if we compress the original set of features into only 1 feature, 45% of the information is kept; if we compress it into 2 features, 45% + 28% = 73% is kept; if we compress it into 3 features, 86% is kept; for 4 features, the number is 92%, etc.

That said, with using just 4 new features, we manage to keep 92% of the information from the original set of 30 features, which is pretty impressive: the number of features reduced by 86.7% while the loss information is just 8%.

You might have been wondering all along about how information is measured. Information is a vague definition and it is nearly impossible to be measured quantitatively like what we did (45%, 73%, etc.). You are right. What we actually measured here is not really information, but an estimation of information, which is the variance of the data.

Why do we use variance to estimate information? There are several reasons:

  • Intuitively, the variability of the data makes it informative. Imagine an extreme case when all samples have the same value of a feature, does that feature have any impact on the analysis/predictive process? Seems not.
    Regarding the Ames Housing dataset we mentioned above, suppose there is a feature named Location, and all the data points have their Location value: On Earth. It is obvious that we cannot use this feature, which has zero variance, for our analysis.
  • The variance is additive for independent random variables. In the above example, compress the 30 features into 2 new features retains 45% + 28% = 73% of the variance (we use the term variance here, not information anymore).
    If instead of variance, we use standard deviation (std), this addition does not make sense since std is not additive. Thus, by using the variance, we have a clearer view of how much information is retained for each number of new features. (Amoeba gave a great answer about this on StackExchange.)
  • For mathematical convenience. Well, the formulas turn out to be very beautiful, as we will see in the next sections.

Hey, so what are the independent random variables? By independent here, we mean the set of new features has its elements perpendicular to each other (or say, orthogonal). In every sense, they should be orthogonal. Let us elaborate:

  • They, together, form a new set of bases for the data points to be projected on. We can think of this as a coordinate system, it is very familiar that the axes (i.e. Ox, Oy, Oz) are orthogonal to each other, isn’t it?
  • By being orthogonal, they are most efficient in summarizing data. Suppose you try to summarize a book (e.g. 300 pages, or 3000 sentences) in 10 sentences in 1 page. It is probable that each of the 10 sentences is completely different from the other 9, right? If that is not true, e.g. 2 sentences contain some overlapping content, that would be a waste and may make you run out of writing space.
    Well, we admit that for literature, a little bit of overlapping is acceptable to connect the sentences together, however, in maths, we are not that generous. We make it as efficient as possible, thus the new features generated by PCA are orthogonal, i.e. independent of each other.


Here we zoom a little bit more into the variance of the data, which is our estimation of the amount of information, about how it is computed.

Look at the example that we have 9 points on a plane and try to compress (or project) the data onto a single line (axis). Projecting these points on the g-axis as follow:

Project points on the g-axis
Project points on the g-axis.

So the values we get are [0.9, 1.2, 1.5, 2.9, 3.0, etc.]. Notice the bold dot at position 4.0, it is the mean (\overline{x}) of the 9 points. Then, we compute the variance. The sample variance, as you probably know, is computed by:

\begin{aligned}\sigma^2 = \frac{\sum_{1 \leq i \leq n} (x_i - \overline{x})^2}{n-1} \end{aligned}

In this example, \sigma^2 \approx 7.51. That means by using this g-axis, we keep about 7.51 units of the variance of the data.

Instead of g, we may choose to use many other axes (in fact, we may choose from an infinite number of axes), for example, g1, g2, and g3 as depicted below.

example of other possible axes
example of other possible axes

Each of these axes, when the data points are projected on, gives a different variance. And as we concurred before, we try to find the axis that makes the largest variance, as this indicates the most information from the original data is retained. Just with our eyes, we can say that g2 seems to be a bad choice since the points when being projected on it are condensed in a small range, which causes the variance to be small.

Now again, we come back to the g-axis. If you have ever seen a Linear regression or an OLS, you might tell that the g-axis is kind of similar to a linear regressor that uses x_1 to predict x_2 (the best-fitted line). That is true, however, they are quite similar but not the same. In PCA, we want to maximize the variance, that is, to maximize the sum (or average) of the squared distances from the mean to all the points:

PCA's goal is to maximize the sum of the square of the brown lengths.
PCA’s goal is to maximize the sum of the square of the brown lengths.

Interestingly, because the projection is perpendicular, each point forms a right triangle with the mean, and because of the distance between a point and the mean is constant, according to the Pythagorean theorem, maximizing the squared distance between the project-point and the mean is equivalent to minimizing the squared distance between the original point and the project-point:

 maximizing the squared distance between the project-point and the mean is equivalent to minimizing the squared distance between the original point and the project-point
As c is constant, maximizing a^2 is equivalent to minimizing b^2. (Pythagorean theorem: a^2 + b^2 = c^2.)

So what does that mean? That means we can say the objective of PCA is to minimize the sum (or average) of the distance between the points and the new axis.

For OLS, assume we use x_1 to predict x_2, its objective is to find a linear regressor that minimizes the sum (or average) of the residuals. Look at the visual below, the squares of the grey lines are what PCA and OLS try to minimize.

PCA vs OLS in terms of the objective function.
PCA vs OLS in terms of the objective function.

Interesting, isn’t it?

Ok, we have done with the OLS. Before moving on, let me mention some characteristics of the projections:

  • When we change the axis from g to a g’ and g’ is parallel to g, the position of the project-points change, yet their variance remains the same. (If you are not clear about this, draw them on and you will get it.) Thus, to make it simple, we assume the best axis we are finding goes through a specific point (in the example above, we assume it goes through the mean point, that is why the distance between each point and the mean is a constant). Normally, we assume it goes through the origin of the original axes.
  • Call g an axis helps us understand the concept, however, to work on the math, let’s call g a vector. Furthermore, as the length of a vector does not affect how points are projected on it, we assume g has length 1.
Test your understanding

Principal Component Analysis fully explained - Quiz 2

1 / 6

When finding each Principal component, PCA tries to ... (choose all that apply)

2 / 6

Is the compression made by PCA lossy or lossless?

3 / 6

Let's say we project a set of points P onto a vector V to get a set of points Pv. We also project P onto a vector U to get a set of points Pu. If U and V are parallel, ...

4 / 6

Someone says that: applying PCA is just changing the coordinate system of the data space. Is it true?

5 / 6

Which quantitative value does PCA use to estimate the amount of useful information?

6 / 6

With PCA, what do we assume about the Principal component vectors? Choose all that apply.

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Full explanation

In this section, we will have a thorough view, at a lower level, of PCA and how to conduct it. At first, let us summarize the overview we have had, as well as refreshing some common knowledge we would need in the subsequent proofs.

Summary of the overview

  • PCA is a technique usually used for compressing or summarizing the data.
  • PCA assumes the variance of the data is the estimation of the information it contains. Why?
  • PCA projects the data points, which are lying on their original space, onto a new space, with each axis of this new space is a linear combination of the axes from the original space. Illustration.
  • The new axes of this new space are orthogonal to each other. Why? We may even assume that those new axes go through the Origin and each of them has a length of 1. Why?
  • Essentially, if the original dimensionality is d, the new space also has d dimensions, however, these new d dimensions are sorted in order from the most important (the axis that expresses the most variance) to the least important (the axis that expresses the least variance). The unit in use is often the percentage (i.e. what percentage of the original variance that an axis inherits). Illustration.
  • To reduce the dimensionality of the feature space (or say, to summarize the data, to compress the features), we remove a number of the least important dimensions. In this example, we keep only 4 dimensions while removing the other 26, this summarization cost us only 8% of the overall information (variance).

Other prerequisite knowledge

Covariance (matrix)

We already know that the variance is a measure of the variability of a set (or sequence) of values. It depends on the distance from each point to the mean of the set. The formula for variance is:

\begin{aligned}\text{var(x)} = \sigma^2 = \frac{\sum_{1 \leq i \leq n} (x_i - \overline{x})(x_i - \overline{x})}{n-1}\end{aligned}

The covariance, on the other hand, involves 2 sequences of values; its aim is to measure how each of the 2 sequences varies from its mean but with regard to the other sequence.

\begin{aligned} \text{cov(x, y)} = \frac{\sum_{1 \leq i \leq n} (x_i - \overline{x})(y_i - \overline{y})}{n-1} \end{aligned}

It is also worth to note that cov(x, y) = cov(y, x), obviously.

People seem to be more familiar with correlation than covariance. These 2 terms are very similar, however, the correlation is more suitable for general usage and comparison since it is unitless. In fact, correlation can be thought of as a standardized covariance, as for any given input, the corresponding correlation is always in the range [-1, 1].

As a side note, the correlation is computed by:

\begin{aligned} \text{corr(x, y)} = \frac{\text{cov}(x, y)}{\sigma_x \sigma_y} \end{aligned}

Suppose we have d sequences of values, the covariance matrix for these sequences is a d x d matrix whose element at position [i, j] is the covariance of the i-th and j-th sequence, i.e. cov(sequence_i, sequence_j). Notice that by this definition, the element at position [i, i] is the variance of the i-th sequence.

For example, suppose we have 3 sequences: x, y, and z. Their covariance matrix is:

\begin{bmatrix}var(x) & cov(x, y) & cov(x, z) \\var(y, x) & var(y) & cov(y, z) \\cov(z, x) & cov(z, y) & var(z)\end{bmatrix}

which is also equal to:

\begin{bmatrix}var(x) & cov(x, y) & cov(x, z) \\var(x, y) & var(y) & cov(y, z) \\cov(x, z) & cov(y, z) & var(z)\end{bmatrix}

Linear transformation (projection)

In Linear algebra, a linear transformation (T) can be represented by a transformation matrix A. While A is a m x n matrix, it maps any points from R^n to R^m (i.e. from a n-dimensional space to a m-dimensional space). The transformation is done by multiplying A to the left of the point-to-be-projected. That is:

T(x) = Ax

The reason why any linear transformations can be done using a matrix is very simple and direct:

  • A linear combination is expressed in the form: a_1x_1 + a_2x_2 + a_3x_3 + ... + a_dx_d, with d is the number of dimensions, x_i represents the i-th dimension, a is a sequence of scalars defining the linear transformation.
  • Each row (amongst m rows) of the transformation matrix A acts as a a.

For example, suppose we want to transform points from a 3-dimensional space to a 2-dimensional space that has the 2 axes being: 3x_1 + 2x_2 -1 and -4x_1 +7x_3. The transformation matrix A should be:

\begin{aligned}A = \begin{bmatrix}3 & 2 & -1 \\-4 & 0 & 7\end{bmatrix}\end{aligned}

If you want to map a point P, e.g. P = [8, 15, -5] to the new space, that would be:

\begin{aligned}T(P) = AP = \begin{bmatrix}3 & 2 & -1 \\-4 & 0 & 7\end{bmatrix}\begin{bmatrix}8 \\15 \\-5\end{bmatrix}= \begin{bmatrix}59 \\-67\end{bmatrix}\end{aligned}

Note that any point P is denoted by the vector that goes from the Origin to P, and a vector is assumed to be in column form (\begin{bmatrix}8 \\ 15 \\ -5 \end{bmatrix} instead of [ 8 15 -5 ]).

Transpose of a matrix

The transpose of a matrix A (size m x n) is denoted by A^T (size n x m), which is the flip of A, i.e. A^T[i, j] = A[j, i]. Thus, we say Transpose is a flip operation.

An example of Transpose
An example of Transpose, taken from wiki.

We will also need the property of Transpose on a product:

(AB)^T = B^TA^T

Eigenvalues and eigenvectors

In linear algebra, we have the definition of eigenvalue and eigenvector.

For a square matrix A, a vector v, and a scalar \lambda, if v is not all-zero and:

Av = \lambdav

then v is an eigenvector of the matrix A and \lambda is the eigenvalue corresponding to that eigenvector v.

For example:

\begin{aligned}  \begin{bmatrix}    5 & 6 \\    3 & -2  \end{bmatrix}  \begin{bmatrix}    -2 \\    3  \end{bmatrix}\end{aligned} =   \begin{bmatrix}    8 \\    -12  \end{bmatrix}= -4  \begin{bmatrix}    -2 \\    3  \end{bmatrix}\end{aligned}

We say: \begin{bmatrix}    -2 \\    3  \end{bmatrix} is a eigenvector of matrix \begin{bmatrix}  5 & 6 \\    3 & -2  \end{bmatrix} and -4 is the corresponding eigenvalue of this eigenvector. Note that if you scale the eigenvector to whatever amount (e.g. \begin{bmatrix}    -4 \\   6  \end{bmatrix}, \begin{bmatrix}    200 \\    -300  \end{bmatrix}), you still get an eigenvector with the same eigenvalue.

Only square matrices have eigenvectors and eigenvalues. Not all square matrices have, however, if a d x d matrix has, it has exactly d pairs of eigenvectors and eigenvalues (except the eigenvectors that are scaled version of some other). These eigenvalues may be distinct or some of them are the same.

For convenience, we also assume all eigenvectors have length 1.

In general, the set of eigenvectors of a matrix are not necessarily orthogonal. Yet, for a symmetric matrix (A = A^T), the eigenvectors are always orthogonal to each other. Prove:

For any real matrix A and 2 vectors x and y, we have:

\langle Ax, y \rangle = \langle x, A^Ty \rangle

Suppose x and y are 2 eigenvectors of a symmetric matrix A with 2 corresponding eigenvalues \lambda_x and \lambda_y, respectively. We have:

\begin{aligned}\lambda_x \langle x, y \rangle &= \langle \lambda_x x, y \rangle \\&= \langle Ax, y \rangle \\&= \langle x, A^Ty \rangle \\&= \langle x, Ay \rangle \\&= \langle x, \lambda_y y \rangle \\&= \lambda_y \langle x, y \rangle \\\end{aligned}

Thus (\lambda_x - \lambda_y) \langle x, y \rangle = 0. As \lambda_x \ne \lambda_y, this means \langle x, y \rangle = 0, i.e. x and y are orthogonal.

The spectral theorem

According to the spectral theorem, any symmetric matrix is diagonal in its eigenvector basis.

Suppose we have a matrix A with size d x d, A can be written as:

\begin{aligned}  \begin{bmatrix}    & & & & \\    | & | & | & & | \\    v_1 & v_2 & v_3 & ... & v_d \\    | & | & | & & | \\    & & & &   \end{bmatrix}  \begin{bmatrix}    \lambda_1 & & & & \\    & \lambda_2 & & & \\    & & \lambda_3& & \\    & & & ... & \\    & & & & \lambda_d \\  \end{bmatrix}  \begin{bmatrix}    & - & v_1 & - & \\    & - & v_2 & - & \\    & - & v_3 & - & \\    & - & ... & - & \\    & - & v_d & - &  \end{bmatrix}\end{aligned}

With v_1v_d are the orthonormal eigenvectors of A.

If you are looking for proof of this theorem, we suggest Amoeba’s answer on StackExchange, proof by induction on Brilliant, University of Michigan’s proof.

PCA, step-by-step

(Optional) Standardize data
Compute the covariance matrix
Calculate the eigenvalues and eigenvectors
Transform the data

1. (Optional) Standardize data

Suppose we have a dataset (with all numeric variables), the first step is to standardize each feature. Standardization means making each feature have 0-mean and unit variance, i.e.

\begin{aligned}x  = \frac{x - \overline{x}}{\sigma}\end{aligned}

with \overline{x} being the mean and \sigma being the standard deviation. (The terminology for this is: Z-score transformation.) Amongst these 2 transformations, de-mean is highly-recommended for all cases, while dividing the values for the standard deviation is beneficial in most cases.

a. Why 0-centering?

Basically, in the next step, we will compute the covariance matrix of the data, and as you can see from the formula, 0-centering does NOT have any impact on the values of the covariances. Hence, actually, at least in theory, we do not have to center the data. The reasons we should do it in practice are:

  • First, when the means of all variables are 0, the computation of the covariance is simpler, which is:
    \begin{aligned} \text{cov(x, y)} = \frac{\sum_{1 \leq i \leq n} x_i y_i }{n-1} \end{aligned}
    and the covariance matrix:
    \begin{aligned} \text{cov-matrix} = \frac{X^TX}{n-1} \end{aligned}
  • Secondly, in practice, we often use a library to do the PCA for us instead of coding it ourselves, and some libraries actually assume that the input data is 0-centered and use the above, simplified formula to compute the covariance matrix. Thus, if we forget to center the data before calling a PCA package, the result we get might not be what we want to have.

b. Why scaling?

The scaling (i.e. dividing by the standard deviation to make it unit variance) should be used when the variables are different on the unit and/or distribution. For example, you have a dataset of companies, 2 of the features are the total asset (assumed to be in the range 0 – 1 billion $) and the number of employees (1 – 10000 people). The former feature has a much wider range of value and its variance is probably much higher than of the latter. This fact makes PCA focuses mainly on the total asset and overlooks the number of employees when it tries to maximize the retained variance of information.

This situation is like when we want to gauss how rich a person is by the number of telephones and the number of houses he has. Person A has 100 phones and 1 house, while Person B has 1 phone and 10 houses. We end up blindly say that Person A is richer than B because 100 + 1 > 1 + 10, which is obviously wrong as the numbers are in different units.

On the other hand, if the features are measured in the same unit and/or of the same distribution, it is advisable to not scale them, as this act may potentially wideout some useful information.

Also, it is good to know that when the variables all have unit variance, the covariance and correlation matrices are equivalent.

2. Compute the covariance matrix

The covariance matrix of the training data is computed using the previously mentioned formula, i.e.:

\begin{aligned} \text{cov(x, y)} = \frac{\sum_{1 \leq i \leq n} (x_i - \overline{x})(y_i - \overline{y})}{n-1} \end{aligned}

We re-notice it that some libraries use the simplified formula that assumes data is centered around 0:

\begin{aligned} \text{cov-matrix} = \frac{X^TX}{n-1} \end{aligned}

If the number of features in the dataset is d, then the size of the covariance matrix is d x d.

Why do we compute the covariance matrix?

Here we reveal a truth: PCA does not find the optimal transformation that maximizes the variance, it is kind of a greedy algorithm, trying to find the best next-axis that maximizes the variance, locally.

That is,

  1. at first, PCA finds the best vector such that when we project the data points on this vector, the variance of data is the largest, this is the first axis, the first principal component.
  2. Then, amongst all the vectors that are orthogonal to the first axis, PCA finds the one that maximizes the variance of data when being projected on that vector, this is the second axis, the second principal component.
  3. Next, PCA selects from all the vectors that are orthogonal to the 2 selected axes, get the one that maximizes the variance, mark it as the third axis, the third principal component, and so on until there are all d principal components (d is the number of data features).

The point here is that PCA is a greedy method, which tries to maximize the local reward at each point of time.

Let’s call X the 0-centered data matrix (size n x d). Now we attempt to find the first principal component, the vector w that maximizes the variance of data when we project on it.

Call X’ the projected data, we have:

X' = (w^TX^T)^T

This is true because when we want to transform data (X) using a transformation matrix (w), we multiply the transformation matrix on the left of the data (described here). We use the transpose operation to make the operation right. Intuitively, you can observe that when you project data on a 1-dimensional coordinate (w), the resulting point will have only 1 dimension. Here w^T (size 1 x d) multiplies X^T (size d x n) gives a 1 x n matrix, which is the values of n data points after being projected. This 1 x n matrix is then transposed too to make the dimension right (we suppose each row is a data point and each column represents a feature).

Using the transpose property, we simplify the equation:

X' = Xw

X’ (size n x 1) has mean 0 because \sum_{1 \leq i \leq d} (w_i \sum_{1 \leq j \leq n} X_{j,i}) = 0, which, in turn, because X is centered.

Thus, the variance of X’ is:

var(X') = \frac{1}{n-1} \sum_{1 \leq i \leq n} (X'_i)^2

Or, we can say:

\begin{aligned}var(X') &= \frac{1}{n-1} (Xw)^T . Xw \\             &= w^T. (\frac{1}{n-1}X^TX).w \\             &= w^TCw\end{aligned}

with C being the covariance matrix of the original data matrix X.

Hence, we want to find a vector w, whose length is 1 (why unit length?), such that w^TCw is maximal.

That is why we have to find the covariance matrix of the data.

3. Calculate the eigenvalues and eigenvectors of the covariance matrix

Finding the eigenvalues and eigenvectors of a matrix is not a simple task when the size of the matrix is larger than 3 x 3 (you may have known this from your linear algebra course). We are not trying to elaborate on how we do this step ourselves (we can just make use of a library), instead, we attempt to answer the following question as a fundamental part of PCA:

Why do we need the eigenvalues and eigenvectors of the covariance matrix?

As C is symmetric, the spectral theorem tells us that it can be written as Q \lambda Q^T where Q is formed by orthogonal eigenvectors and \lambda is the diagonal matrix of eigenvalues. We also sort the eigenvalues (and the corresponding eigenvectors) so that \lambda_1 \geq \lambda_2 \geq ... \geq \lambda_d.

We have:

\begin{aligned}w^TCw &= w^T Q \lambda Q^T w \\              &= v^T\lambda v \\              &= \sum \lambda_i v_i^2\end{aligned}

Because \| w \|_2 = 1, it is true that \| v \|_2 = 1 since v is nothing more than a transformation of w into a different (but with the same number of dimensions) space using the orthogonal matrix Q^T.

This makes it obvious that the maximum value of \sum \lambda_i v_i^2 is \lambda_1 when v = \begin{bmatrix}1 \\ 0 \\ 0 \\ ... \\ 0 \end{bmatrix}.

As v = Q^T w and Q is orthogonal, v = \begin{bmatrix}1 \\ 0 \\ 0 \\ ... \\ 0 \end{bmatrix} \iff w = Q^T_1, i.e. w = the vector corresponding to \lambda_1, the largest eigenvalue.

That is the reason why we need to compute the eigenvalues and eigenvectors. Above, we explain the process of finding the first principal component, which turns out to be the eigenvector of the most significant eigenvalue. Looking for the subsequent principal components is almost identical since the fact that we are working on orthogonal space makes finding the next component nearly independent of the previous ones.

To conclude this step, we claim that the variance (as an estimation of information) retained by the principal component i-th is equal to the i-th largest eigenvalue (\lambda_i).

The below figure, which we had as an example for the overview section, indicates that the first principal component retains 45% of the total variance, the second retains 28%, etc.

It is also equivalent to:

\begin{aligned} \frac{\lambda_1}{\sum \lambda_i} = 0.45  \end{aligned}
\begin{aligned} \frac{\lambda_2}{\sum \lambda_i} = 0.28  \end{aligned}

Information retained by each new feature (principal component)
Information retained by each new feature (principal component)

4. Transform the data

We have gone through the difficult part, the last step is the easy one.

We have an original dataset X (n x d), an orthonormal basis of eigenvectors Q (d x d) with the corresponding diagonal matrix of eigenvalues \lambda (d x d), in which the eigenvalues are sorted in decreasing order.

\begin{aligned}X = \begin{bmatrix}  - & \text{data point 1} & - \\  - & \text{data point 2} & - \\    & ... & \\  - & \text{data point n} & - \\\end{bmatrix}, Q =  \begin{bmatrix}    & & & & \\    | & | & | & & | \\    v_1 & v_2 & v_3 & ... & v_d \\    | & | & | & & | \\    & & & &  \end{bmatrix}, \lambda =  \begin{bmatrix}    \lambda_1 & & & & \\    & \lambda_2 & & & \\    & & \lambda_3& & \\    & & & ... & \\    & & & & \lambda_d \\  \end{bmatrix}\end{aligned}

There are 2 general options for transformation:

  • We want to reduce the d dimensional data down to a specific k (k < d).
  • We want to reduce the dimensionality as much as possible but at least p % of the variance must be kept. For this, get the smallest number k so that \begin{aligned}\frac{\lambda_1 + \lambda_2 + ... + \lambda_k}{\lambda_1 + \lambda_2 + ... + \lambda_d} >= p \end{aligned}

For either case, we have a specific value of k. Q is the transformation matrix that is supposed to transform our data points (the matrix X) to the new basis (or coordinate) in a lossless way (all variance is kept, but the number of dimensions also remains d). However, as we want the new basis to have only k most-meaningful dimensions, we trim the matrix Q by taking the first k columns of it, call this matrix T.

\begin{aligned}T =  \begin{bmatrix}    & & & \\    | & | & | & | \\    v_1 & v_2 & ... & v_k \\    | & | & |  & | \\    & & &  \end{bmatrix}\end{aligned}

We transform our data points (matrix X) using this transformation matrix T.

\begin{aligned}X_\text{PCA} &= (T^T X^T)^T                &= XT\end{aligned}

The result we get is a matrix X_\text{PCA} of size n x k, with each data point occupies a row of k features.

Test your understanding

Principal Component Analysis fully explained - Quiz 3

1 / 10

Can non-square matrices have eigenvalues?

2 / 10

The PCA works best when ...

3 / 10

Suppose we have variables A and B. A has unit a while B has unit b. What is the unit of their correlation?

4 / 10

Look at

Screenshot From 2020 03 20 09 59 25

Which is the transpose of the eigenvector?

5 / 10

Suppose we have variables A and B. A has unit a while B has unit b. What is the unit of their covariance?

6 / 10

Look at

Screenshot From 2020 03 20 09 59 25

Which is the eigenvalue?

7 / 10

The transformation of vector v using a transformation matrix A is ...

8 / 10

In which cases should we scale the data (make it unit variance) when doing PCA?

9 / 10

What is the relationship between covariance and correlation?

10 / 10

Given a dataset with d dimensions and a number k (k < d). Can PCA guarantee to find the best set of k coordinate axes that maximizes the variance of data?

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So we are done with the intuition and maths behind the so-called Principal Component Analysis (PCA). Actually, the maths part is more complex than it seems, and we are sure it is normal to have to re-read the article several times in order to fully grasp its content. In case you are still in doubt, you might want to take a look at the reference list or drop me a comment (or message).

As a final note, we want to emphasize that PCA does linear transformations, which means it works best when some of our features are linearly correlated to each other. In case the correlations are not linear, we may try pre-transforming to make them linear before applying PCA. Another option is to use a nonlinear dimensionality reduction method (e.g. Isomap).

Thanks for reading!


  • CS229’s lecture note 10 from Andrew NG: link
  • Q&A about applying PCA on nonlinear data: link
  • Lindsay I Smith’s tutorial on PCA: link
  • The nice answers on StackExchange: link, link, and link
  • Spectral Theorem explained by University of Michigan: link
  • Arturo explains why eigenvectors of a symmetric matrix are orthogonal: link
  • Wiki’s page about Transformation matrix: link
  • Q&A about centering in PCA: link and link

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