Basis for a subspace of $mathbbC^n$ such that the real part is also a basis

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Let $V subseteq mathbbC^n$ be a subspace of dimension $k$, viewing $mathbbC^n$ as a complex vector space. For $v in V$, we can take the "real part" of $v$ by just taking the real part of each entry. Then $mathrmRe(v) enspace vert enspace v in V$ is a real vector space.



Is the dimension of this vector space necessarily $k$? In other words, is there a basis of $V$ such that the real part of the basis is a basis of this real vector space?



I feel like the answer must be yes, and that this should be a generic property. However, this taking real parts operation doesn't seem to have very nice properties.




linear-algebra




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    Let $V subseteq mathbbC^n$ be a subspace of dimension $k$, viewing $mathbbC^n$ as a complex vector space. For $v in V$, we can take the "real part" of $v$ by just taking the real part of each entry. Then $mathrmRe(v) enspace vert enspace v in V$ is a real vector space.



    Is the dimension of this vector space necessarily $k$? In other words, is there a basis of $V$ such that the real part of the basis is a basis of this real vector space?



    I feel like the answer must be yes, and that this should be a generic property. However, this taking real parts operation doesn't seem to have very nice properties.




    linear-algebra




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      Let $V subseteq mathbbC^n$ be a subspace of dimension $k$, viewing $mathbbC^n$ as a complex vector space. For $v in V$, we can take the "real part" of $v$ by just taking the real part of each entry. Then $mathrmRe(v) enspace vert enspace v in V$ is a real vector space.



      Is the dimension of this vector space necessarily $k$? In other words, is there a basis of $V$ such that the real part of the basis is a basis of this real vector space?



      I feel like the answer must be yes, and that this should be a generic property. However, this taking real parts operation doesn't seem to have very nice properties.




      linear-algebra




      share|cite|improve this question











      Let $V subseteq mathbbC^n$ be a subspace of dimension $k$, viewing $mathbbC^n$ as a complex vector space. For $v in V$, we can take the "real part" of $v$ by just taking the real part of each entry. Then $mathrmRe(v) enspace vert enspace v in V$ is a real vector space.



      Is the dimension of this vector space necessarily $k$? In other words, is there a basis of $V$ such that the real part of the basis is a basis of this real vector space?



      I feel like the answer must be yes, and that this should be a generic property. However, this taking real parts operation doesn't seem to have very nice properties.





      linear-algebra





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      asked Jul 23 at 23:31









      vukov

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          No. For example, consider $V = operatornamespan(1, 1 + i)$. We compute the space directly:
          beginalign*
          operatornameRe((x + iy)(1, 1 + i)) &= (operatornameRe(x + iy),operatornameRe(x - y + (x + y)i)) \
          &= (x, x - y) \
          &= x(1, 1) + y(0, -1),
          endalign*
          which yields all of $mathbbR^2$.






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            up vote
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            accepted










            No. For example, consider $V = operatornamespan(1, 1 + i)$. We compute the space directly:
            beginalign*
            operatornameRe((x + iy)(1, 1 + i)) &= (operatornameRe(x + iy),operatornameRe(x - y + (x + y)i)) \
            &= (x, x - y) \
            &= x(1, 1) + y(0, -1),
            endalign*
            which yields all of $mathbbR^2$.






            share|cite|improve this answer

























              up vote
              3
              down vote



              accepted










              No. For example, consider $V = operatornamespan(1, 1 + i)$. We compute the space directly:
              beginalign*
              operatornameRe((x + iy)(1, 1 + i)) &= (operatornameRe(x + iy),operatornameRe(x - y + (x + y)i)) \
              &= (x, x - y) \
              &= x(1, 1) + y(0, -1),
              endalign*
              which yields all of $mathbbR^2$.






              share|cite|improve this answer























                up vote
                3
                down vote



                accepted







                up vote
                3
                down vote



                accepted






                No. For example, consider $V = operatornamespan(1, 1 + i)$. We compute the space directly:
                beginalign*
                operatornameRe((x + iy)(1, 1 + i)) &= (operatornameRe(x + iy),operatornameRe(x - y + (x + y)i)) \
                &= (x, x - y) \
                &= x(1, 1) + y(0, -1),
                endalign*
                which yields all of $mathbbR^2$.






                share|cite|improve this answer













                No. For example, consider $V = operatornamespan(1, 1 + i)$. We compute the space directly:
                beginalign*
                operatornameRe((x + iy)(1, 1 + i)) &= (operatornameRe(x + iy),operatornameRe(x - y + (x + y)i)) \
                &= (x, x - y) \
                &= x(1, 1) + y(0, -1),
                endalign*
                which yields all of $mathbbR^2$.







                share|cite|improve this answer













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                share|cite|improve this answer











                answered Jul 23 at 23:55









                Theo Bendit

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