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The PCA algorithm can be formulated in terms of the correlation matrix (assume the data $X$ has already been normalized and we are only considering projection onto the first PC). The objective function can be written as:

$$ \max_w (Xw)^T(Xw)\; \: \text{s.t.} \: \:w^Tw = 1. $$

This is fine, and we use Lagrangian multipliers to solve it, i.e. rewriting it as:

$$ \max_w [(Xw)^T(Xw) - \lambda w^Tw], $$

which is equivalent to

$$ \max_w \frac{ (Xw)^T(Xw) }{w^Tw},$$

and hence (see here on Mathworld) seems to be equal to $$\max_w \sum_{i=1}^n \text{(distance from point $x_i$ to line $w$)}^2.$$

But this is saying to maximize the distance between point and line, and from what I've read here, this is incorrect -- it should be $\min$, not $\max$. Where is my error?

Or, can someone show me the link between maximizing variance in projected space and minimizing distance between point and line?

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  • $\begingroup$ I think minimum distance is used to meet the criterion of orthogonality for the components. The points are projected into the PCs that are orthogonal to each other but in each successive component the remaining variance is maximized. $\endgroup$
    – Michael R. Chernick
    Commented Jul 12, 2012 at 15:42
  • $\begingroup$ Hint: What happens when you consider the smallest eigenvalue first, rather than the largest one? $\endgroup$
    – whuber ♦
    Commented Jul 12, 2012 at 15:45
  • $\begingroup$ @whuber The smallest eigenvalue probably has the PC that is the solution to the final objective function. But this PC does not maximixe the original objective function. $\endgroup$
    – Cam.Davidson.Pilon
    Commented Jul 12, 2012 at 17:06
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    $\begingroup$ I'm not sure what you mean by "final" and "original" objective function, Cam. PCA is not (conceptually) an optimization program. Its output is a set of principal directions, not just one. It is an (interesting) mathematical theorem that these directions can be found by solving a sequence of constrained quadratic programs, but that's not basic to the concepts or the practice of PCA. I am only suggesting that, by focusing on the smallest eigenvalue rather than on the largest one, you can reconcile the two ideas of (1) minimizing distances and (2) taking an optimization view of PCA. $\endgroup$
    – whuber ♦
    Commented Jul 12, 2012 at 18:32
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    $\begingroup$ That's okay - your answer was the non-mistake version of what I was trying to do. $\endgroup$
    – Cam.Davidson.Pilon
    Commented Feb 3, 2015 at 13:22

1 Answer 1

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Let $\newcommand{\X}{\mathbf X}\X$ be a centered data matrix with $n$ observations in rows. Let $\newcommand{\S}{\boldsymbol \Sigma}\S=\X^\top\X/(n-1)$ be its covariance matrix. Let $\newcommand{\w}{\mathbf w}\w$ be a unit vector specifying an axis in the variable space. We want $\w$ to be the first principal axis.

According to the first approach, first principal axis maximizes the variance of the projection $\X \w$ (variance of the first principal component). This variance is given by the $$\mathrm{Var}(\X\w)=\w^\top\X^\top \X \w/(n-1)=\w^\top\S\w.$$

According to the second approach, first principal axis minimizes the reconstruction error between $\X$ and its reconstruction $\X\w\w^\top$, i.e. the sum of squared distances between the original points and their projections onto $\w$. The square of the reconstruction error is given by \begin{align}\newcommand{\tr}{\mathrm{tr}} \|\X-\X\w\w^\top\|^2 &=\tr\left((\X-\X\w\w^\top)(\X-\X\w\w^\top)^\top\right) \\ &=\tr\left((\X-\X\w\w^\top)(\X^\top-\w\w^\top\X^\top)\right) \\ &=\tr(\X\X^\top)-2\tr(\X\w\w^\top\X^\top)+\tr(\X\w\w^\top\w\w^\top\X^\top) \\ &=\mathrm{const}-\tr(\X\w\w^\top\X^\top) \\ &=\mathrm{const}-\tr(\w^\top\X^\top\X\w) \\ &=\mathrm{const} - \mathrm{const} \cdot \w^\top \S \w. \end{align}

Notice the minus sign before the main term. Because of that, minimizing the reconstruction error amounts to maximizing $\w^\top \S \w$, which is the variance. So minimizing reconstruction error is equivalent to maximizing the variance; both formulations yield the same $\w$.

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    $\begingroup$ @alberto What is inside the trace is a number (1x1 matrix); a trace of a number is this number itself, so the trace can be removed. The constant appears because $\Sigma$ is equal to $X^\top X/n$, so there is this $1/n$ factor. $\endgroup$
    – amoeba
    Commented Jan 6, 2017 at 16:00
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    $\begingroup$ @Leullame The calculation will hold verbatim for $W$ if it is a matrix with orthonormal columns. You need $W^\top W = I$ to go from line #3 to line #4. If matrix $W$ has orthonormal columns, then indeed $xWW^\top$ will be a projection of $x$ onto the subspace spanned by the columns of $W$ (here $x$ is a row vector). $\endgroup$
    – amoeba
    Commented Feb 18, 2017 at 18:51
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    $\begingroup$ Can you describe how this changes if the data is not centered? $\endgroup$
    – elliotp
    Commented Feb 13, 2018 at 21:44
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    $\begingroup$ @elliotp I am not deriving the solution in this answer. I am only proving that maximum variance direction coincides with minimum error direction. No need for Lagrangian here. $\endgroup$
    – amoeba
    Commented Feb 18, 2018 at 16:37
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    $\begingroup$ @DanielLópez Well, we are looking for a 1-dimensional subspace minimizing reconstruction error. A 1-dimensional subspace can be defined by a unit-norm vector pointing into its direction, which is what $w$ is taken to be. It has unit norm by construction. $\endgroup$
    – amoeba
    Commented Feb 22, 2018 at 11:54

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