# How can eigenfaces (PCA eigenvectors on face image data) be displayed as images?

I am trying to clarify some concepts for face recognition. According to my understanding, given a training set of images with each image measuring 225 x 255 pixels, we will have a matrix of training images, n x (255 x 255).

Using PCA, we would be reducing the high dimensions of 255 x 255 to something smaller say 200.

However, i have seen cases when blogs display the eigenfaces. I would assume that the eigenfaces would have a dimension of 200. How would it be possible that the resulting eigenfaces image have the same dimensions as the original image? Although it seems that the eignfaces are much blurred.

• – amoeba May 8 '17 at 11:18

PCA does dimensional reduction by expressing $D$ dimensional vectors on an $M$ dimensional subspace, with $M<D.$ The vector itself can be written as a linear combination of $M$ eigenvectors, where the eigenvector is itself a unit vector that lives in the $D$ dimensional space.
Consider, for example, a two dimensional space which we reduce to one dimension using PCA. We find that the principal eigenvector is the unit vector that points equally in the positive $\hat{x}$ and $\hat{y}$ direction, i.e. $$\hat{v} = \frac{1}{\sqrt{2}} (\hat{x} + \hat{y}).$$ In this case I'm using the hat ($\hat{x}$) symbol to indicate that it's a unit vector. You can think of this as a one-dimensional line going through a two-dimensional plane. In our reduced space, we can express any point $w$ in the two dimensional space as a one-dimensional (or scalar) value by projecting it onto the eigenvector, i.e. by calculating $w \cdot \hat{v}.$ So the point $(3,2)$ becomes $5/\sqrt{2},$ etc. But the eigenvector $\hat{v}$ is still expressed in the original two dimensions.
In general, we express a $D$ dimensional vector, $x,$ as a reduced $M$ dimensional vector $a$, where each component $a_i$ of $a$ is given by, $$a_i = \sum_j x_j V_{i j}$$ where $V_{i j}$ is the $j$th component of the $i$'th eigenvector, and $i = 1, \dots, M$ and $j = 1, \dots, D.$ For that to work, the $i$th eigenvector must have $D$ components to take an inner product with $x$.
In your case, you can express a "reduced" vector of 200 components by taking the original image, a vector of 65025 components, and taking its inner product with each of the 200 images, each of which has 65025 components. Each inner product result is a component of your 200-dimensional vector. We expect each eigenvector to have the same number of dimensions as the original space. That is, we expect $M$ eigenvectors, each of which are $D$-dimensional.
• Because an eigenvector is a vector that lives in the larger space. Just like in my $D=2$ example, I wrote an eigenvector having two components. Similarly, in your $D=65025$ example, an eigenvector has $65025$ components. The reduced space is simply expressed by the number of eigenvectors you take. If you reduce it to $200$ dimensions, then you get $200$ eigenfaces, each of which have $65025$ components. Then you produce a $200$ dimensional vector by taking $200$ inner products of a test image, one with each eigenface. – Bridgeburners May 7 '17 at 7:31