Timeline for How to interpret PCA loadings?
Current License: CC BY-SA 3.0
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| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
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| Jan 21, 2016 at 11:20 | history | edited | ttnphns | CC BY-SA 3.0 |
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| Dec 6, 2014 at 20:59 | comment | added | amoeba | Good point, @Nick, this is indeed not possible, as the total variance of a $4\times4$ correlation matrix must be $4$, so two PCs both with eigenvalues $1$ must account for $50\%$ of the variability. I am not explaining this to you, of course, but for other possible readers of this thread. The ttnphns's answer remains correct though (+1), we just have no other choice as to ignore the number $90\%$ reported by the OP. | |
| Dec 6, 2014 at 19:46 | history | edited | ttnphns | CC BY-SA 3.0 |
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| Dec 6, 2014 at 13:30 | history | edited | ttnphns | CC BY-SA 3.0 |
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| Dec 6, 2014 at 12:23 | comment | added | ttnphns | Nick, I believe this a question to the OP. He didn't give the data or covariance/correlation matrix. All we had from him is a (rather unrealistic) loading matrix of 2 first PCs. | |
| Dec 6, 2014 at 12:20 | history | edited | ttnphns | CC BY-SA 3.0 |
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| Dec 6, 2014 at 12:14 | comment | added | Nick Cox | If 2 components out of 4 account for 90% of variability how come their eigenvalues sum to 2? | |
| Dec 6, 2014 at 12:06 | history | edited | ttnphns | CC BY-SA 3.0 |
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| Apr 4, 2014 at 8:47 | vote | accept | priyanka | ||
| Apr 4, 2014 at 8:43 | history | edited | ttnphns | CC BY-SA 3.0 |
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| Apr 4, 2014 at 8:36 | history | edited | ttnphns | CC BY-SA 3.0 |
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| Apr 4, 2014 at 8:28 | history | answered | ttnphns | CC BY-SA 3.0 |