Standard basis vectors of C^n?

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SUMMARY

The standard basis vectors of Cn are indeed the same as those of Rn, with the primary distinction being that the scalars in Cn are complex numbers. The standard basis vectors for Cn are represented as <1, 0, 0, ..., 0>, <0, 1, 0, ..., 0>, and so forth. When considering Cn as a vector space over the real numbers, the dimension doubles to 2n, and the standard basis includes vectors such as <1, 0, 0, ..., 0> and . This principle applies universally to any vector space Fn, where F denotes any field.

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pivoxa15
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I take it that the standard basis vectors of C^n is the same as the standard basis vectors of R^n?

It would seem so as scalars in C^n are complex numbers.
 
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Yes, the "standard" basis vectors for Cn over the complex numbers, since you say " scalars in Cn are complex numbers", are <1, 0, 0, ..., 0>, <0, 1, 0, ..., 0>, etc. just as for Rn.

Of course, you can also think of Cn as a vector space over the real numbers, in which case the dimension is 2n and the "standard" basis is <1, 0, 0, ..., 0>, <i, 0, 0, ..., 0>, <0, 1, 0, ..., 0>, <0, i, 0,..., 0>, etc.
 
The standard basis for any F^n, where F is any field, is the same, with the scalars being from F
 

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