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If communality = 1, then we have a Heywood case, and if a communality > 1, it is known as an ultra-Heywood case. I read in a SAS manual that an ultra-Heywood case renders a factor solution invalid, and that factor analysts disagree about whether or not a factor solution with a Heywood case can be considered legitimate. I'm particularly interested in the case of a Heywood case within a larger EFA/CFA. Does having a Heywood case (not an ultra-heywood case) renders the whole solution invalid or not? Why? Could someone point me to some literature on this specific topic?

Charlie Glez
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  • I'm remembering to have commented on a same question several days ago, but unable to locate it anymore; was it you question then deleted? – ttnphns Jun 03 '16 at 09:08
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    "Communality=1" case means there is zero unique variance in the variable. It is possible theoretically but unlikely practically. Some programs may set communality to 1 on iterations when they see communality becomes >1, so often "ultra-case" is hidden here. My sheer personal advice would be not to tolerate neither "ultra" nor "non-ultra" case. To struggle Heywood case, try to lessen the number of factors, try other initial communalities (in PAF method), try to drop variables with low KMO, check multicollinearity (see pt 5-6 [here](http://stats.stackexchange.com/a/198684/3277)). – ttnphns Jun 03 '16 at 09:19
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    My opinion is that Heywood case (even "non-ultra") is "degenerate" and "unnatural" and should be treated as invalid. Very high communality, approaching the variable's variance (or 1, in case of analyzing correlations) is dubious. In reality - in psychometrics in particular, - all observed variables should normally have _some_ unique variance – ttnphns Jun 03 '16 at 09:30

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