How many stereoisomers are possible for hepta-1,6-diene? I am asked to identify the number of stereoisomers that are possible for hepta-1,6-diene $(\ce{H2C=CHCH2CH2CH2CH=CH2})$.

I drew the bond-line structure as follows, and I think that there can be cis and trans isomers, but my answer is wrong. The correct answer states that since neither of the double bonds exhibit stereoisomerism, this compound does not have any stereoisomers. 

Originally, I thought that single bonds can be considered too. But I think I my error is that by looking at the single bonds, we are considering the constitutional isomers of this compound, rather than the stereoisomers (which are determined by the 'fixed' double bonds). Since double bonds lack the free rotation that single bonds have, for any compound that may exhibit stereoisomerism, we can only consider the double bonds. Also, are we considering only one double bond at a time? 

Is this reasoning correct? Can someone clarify and expound more on this please?
In the case of your original drawing with the terminal double bonds, there are no E or Z (trans or cis) isomers!



E and Z isomers are only possible if both ends of the $\ce{C=C}$ double bond each carry a substituent other than a hydrogen atom.


As far as the other structure is concerned, E and Z isomers look as follows: