# Can I use combination of eigenvectors as a single vector to explain most of variance?

I have a problem trying to find a combination (or weighted average) of variables (statistics) that best explains the sample statistics. A – n x p matrix (n: observations p: variables, here are statistics for each observation)

The steps used are:

1. Use correlation matrix C of A to get eigenvectors $V (V1, V2,\cdots,Vp)$ and eigenvalues $D(D1,D2,\cdots,Dp)$.

2. Then pick up top 3 eigenvalues and eigenvectors $(V1, V2, V3)$ and $(D1,D2,D3)$ which explain 85% of the variance for the example. Transform $V1$ to $V1^*$ as: $V1^* = (\frac{v11}{\sum_1^pvp1}, \frac{v21}{\sum_1^pvp1}, \frac{vp1}{\sum_1^pvp1})$

3. Then form a vector using these 3 as: $V*=\frac{D1}{(D1+D2+D3)}\cdot V1^*+\frac{D2}{(D1+D2+D3)}\cdot V2^*+\frac{D3}{(D1+D2+D3)}\cdot V3^*$

So $V^*$ is a $p \times 1$ vector and used as weights for these variables (statistics) assuming this will give the best explanation of the variance.

Please see attached image if the formula is not clear. Thanks

• What's the question? I don't see a question mark anywhere Jun 15, 2018 at 9:51
• At the end of title, maybe? Jun 15, 2018 at 10:56
• Please format the question properly. Use bold or slant style for text to highlight the important point to be visible at one sight. Jun 15, 2018 at 10:58

So, any other linear combination (including $V^*$, you described) has variance lower or equal to that given by eigenvector coresponding to largest eigenvalue.