# How to interpret eigenvectors in PCA analysis?

I'm trying to apply the output from PCA analysis I've run on some yield curve history and am getting a bit confused. I have followed the steps below,

1. From a history of the yield curve ($$m \times n$$ where $$m$$ are the days and $$n$$ are the tenor points) I derive the daily returns
2. Calculate a covariance matrix from the daily returns
3. Calculate the eigenvectors and eigenvalues that correspond to the covariance matrix. The first two eigenvectors (principal components) are,

Tenor      PC2          PC1
1Y         -0.15028     0.05154
2Y         -0.32138     0.13310
3Y         -0.41856     0.20852
5Y         -0.41465     0.26116
7Y         -0.33315     0.29555
10Y        -0.19980     0.31919
15Y        -0.04874     0.33416
20Y         0.08760     0.34173


PC1 explains 85% of the variance and PC2 explains 11%, so these two components are enough for the purposes of my analysis.

What I would like to do is measure the amount the yield curve moves if I assume the 5Y point moves by exactly one unit with every point moving in the same direction (the move exactly described by PC1).

To do this I calculate a shift to the yield curve using the following expansion, $$\textbf{PC1} = \Bigg[ \textbf{SD} \cdot \bigg[ \textbf{P} \cdot \big[ \textbf{Sc} \cdot \textbf{SD} \cdot \textbf{P} \big] ^ {-1} \bigg] \Bigg] \cdot \big[1\big]$$

where,

• $$\textbf{SD}$$ is the standard deviation matrix of size $$n \times n$$ where $$n$$ is the number of tenors, the diagonal value is the square root of the diagonal of the covariance matrix and zeros everywhere
• $$\textbf{P}$$ is the first Principal Component vector, which is simply the eigenvalues of the covariance matrix of the daily returns of size $$n \times 1$$
• $$\textbf{Sc}$$ is a scalar vector that represents a unit shift at the 5Y point, so is a $$n \times 1$$ vector with 0 at each element apart from 1 at the corresponding 5Y tenor

The output of the above expansion does indeed lead to an $$n \times 1$$ vector that has a shift for every tenor and exactly 1 at the 5Y point, so the result does appear to be correct.

This is the output,

0.071660453
0.336473214
0.707324248
1
1.204548309
1.340536029
1.429582409
1.489268877


However, some other references suggest that I can use the eigenvectors as weights directly. If I do that and scale the results of the PC1 vector so that the 5Y element is exactly 1, I get this result,

0.197343561
0.509633359
0.798451041
1
1.131677872
1.222203892
1.279510769
1.308506297


So my question simply is - which result is correct? Is it the first one or the second, which is simply a linear scaling of the eigenvector?

• Welcome to CV. I've a few questions to clarify what you are doing. what is SD? The standard deviation of a matrix I am familiar with is a scalar (see allexamshelps.com/…) . Do you mean the covariance matrix with element by element square root? Or maybe the eigenvalues? You allude to references, could you provide some? – ReneBt Feb 15 at 14:15
• Hi Rene, I've edited my question about. In specific answer to your questions SD is square root of the diagonal of the covariance matrix. I've also linked to a reference. – insomniac Feb 15 at 15:02

$$Data = Scores * PCs$$