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

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite
1












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























    up vote
    2
    down vote

    favorite
    1












    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





















      up vote
      2
      down vote

      favorite
      1









      up vote
      2
      down vote

      favorite
      1






      1





      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





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 23 at 23:31









      vukov

      759414




      759414




















          1 Answer
          1






          active

          oldest

          votes

















          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





















            Your Answer




            StackExchange.ifUsing("editor", function ()
            return StackExchange.using("mathjaxEditing", function ()
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            );
            );
            , "mathjax-editing");

            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "69"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: false,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );








             

            draft saved


            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2860868%2fbasis-for-a-subspace-of-mathbbcn-such-that-the-real-part-is-also-a-basis%23new-answer', 'question_page');

            );

            Post as a guest

















            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            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










              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













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 23 at 23:55









                Theo Bendit

                12k1843




                12k1843






















                     

                    draft saved


                    draft discarded


























                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2860868%2fbasis-for-a-subspace-of-mathbbcn-such-that-the-real-part-is-also-a-basis%23new-answer', 'question_page');

                    );

                    Post as a guest
































































                    Comments

                    Post a Comment

                    Popular posts from this blog

                    What is the equation of a 3D cone with generalised tilt?

                    Image
                    Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an
                    Read more

                    Color the edges and diagonals of a regular polygon

                    Image
                    Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne
                    Read more

                    Relationship between determinant of matrix and determinant of adjoint?

                    Image
                    Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/… – Hector Blandin Nov 17 &
                    Read more