Span of vectors with more entries in each vector than the amount of vectors

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I can easily visualize why $beginbmatrix
1\ 0
endbmatrix$ and $beginbmatrix
0\ 1
endbmatrix$ spans $R^2$. I have learned that we need at least two linearly independent vectors to span $R^2$, $3$ to span $R^3$ and so on... This is easy to visualize when the entries in the vectors are the same as the amount of total vectors.



But what happens if we for instance have the two vectors $beginbmatrix
1\ 0 \ 1 \ 0
endbmatrix$ and $beginbmatrix
0\1\0\1
endbmatrix$.



Clearly, the two vectors are linearly independent, but do they still only span $R^2$? If we add the two vectors we produce $beginbmatrix
1\ 1\1\1
endbmatrix$ which is a vector in $R^4$, right?



So my question boils down to what is the span of $beginbmatrix
1\ 0 \ 1 \ 0
endbmatrix$ and $beginbmatrix
0\1\0\1
endbmatrix$, and in general what is the span of a set of vectors with more entries in each vector than vectors in total?






linear-algebra vector-spaces







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    up vote
    2
    down vote

    favorite












    I can easily visualize why $beginbmatrix
    1\ 0
    endbmatrix$     and $beginbmatrix
    0\ 1
    endbmatrix$     spans $R^2$. I have learned that we need at least two linearly independent vectors to span $R^2$, $3$ to span $R^3$ and so on... This is easy to visualize when the entries in the vectors are the same as the amount of total vectors.



    But what happens if we for instance have the two vectors $beginbmatrix
    1\ 0 \ 1 \ 0
    endbmatrix$     and $beginbmatrix
    0\1\0\1
    endbmatrix$    .



    Clearly, the two vectors are linearly independent, but do they still only span $R^2$? If we add the two vectors we produce $beginbmatrix
    1\ 1\1\1
    endbmatrix$     which is a vector in $R^4$, right?



    So my question boils down to what is the span of $beginbmatrix
    1\ 0 \ 1 \ 0
    endbmatrix$     and $beginbmatrix
    0\1\0\1
    endbmatrix$    , and in general what is the span of a set of vectors with more entries in each vector than vectors in total?






    linear-algebra vector-spaces







    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I can easily visualize why $beginbmatrix
      1\ 0
      endbmatrix$       and $beginbmatrix
      0\ 1
      endbmatrix$       spans $R^2$. I have learned that we need at least two linearly independent vectors to span $R^2$, $3$ to span $R^3$ and so on... This is easy to visualize when the entries in the vectors are the same as the amount of total vectors.



      But what happens if we for instance have the two vectors $beginbmatrix
      1\ 0 \ 1 \ 0
      endbmatrix$       and $beginbmatrix
      0\1\0\1
      endbmatrix$      .



      Clearly, the two vectors are linearly independent, but do they still only span $R^2$? If we add the two vectors we produce $beginbmatrix
      1\ 1\1\1
      endbmatrix$       which is a vector in $R^4$, right?



      So my question boils down to what is the span of $beginbmatrix
      1\ 0 \ 1 \ 0
      endbmatrix$       and $beginbmatrix
      0\1\0\1
      endbmatrix$      , and in general what is the span of a set of vectors with more entries in each vector than vectors in total?






      linear-algebra vector-spaces







      share|cite|improve this question















      I can easily visualize why $beginbmatrix
      1\ 0
      endbmatrix$       and $beginbmatrix
      0\ 1
      endbmatrix$       spans $R^2$. I have learned that we need at least two linearly independent vectors to span $R^2$, $3$ to span $R^3$ and so on... This is easy to visualize when the entries in the vectors are the same as the amount of total vectors.



      But what happens if we for instance have the two vectors $beginbmatrix
      1\ 0 \ 1 \ 0
      endbmatrix$       and $beginbmatrix
      0\1\0\1
      endbmatrix$      .



      Clearly, the two vectors are linearly independent, but do they still only span $R^2$? If we add the two vectors we produce $beginbmatrix
      1\ 1\1\1
      endbmatrix$       which is a vector in $R^4$, right?



      So my question boils down to what is the span of $beginbmatrix
      1\ 0 \ 1 \ 0
      endbmatrix$       and $beginbmatrix
      0\1\0\1
      endbmatrix$      , and in general what is the span of a set of vectors with more entries in each vector than vectors in total?






      linear-algebra vector-spaces




      linear-algebra vector-spaces






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      edited 21 mins ago









      gimusi

      76k73889




      76k73889










      asked 32 mins ago









      novo

      40818




      40818




















          3 Answers
          3






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          up vote
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          down vote



          accepted










          The span of the vectors you mention is a subspace of the vector space to which they belong. So $(1,0,1,0)$ and $(0,1,0,1)$ span a two dimensional subspace (since as you note the vectors are linearly independent) of $mathbbR^4$. It is not $mathbbR^2$ exactly as this subspace contains $4$-tuples, but it is `equivalen' (in some sense) to $mathbbR^2$ if this is of interest to you.



          So in general a span of $m<n$ linearly independent vectors in $mathbbR^n$ is a $m$-dimensional subspace of $mathbbR^n$.






          share|cite|improve this answer



























            up vote
            3
            down vote













            Your vectors live in $mathbbR^4$, not in $mathbbR^2$, and therefore it is out of the question that they span $mathbbR^2$. Since they live in $mathbbR^4$, they span a subspace of $mathbbR^4$, whish happens to be$$leftbeginbmatrixx&y&x&yendbmatrix^T,middle.$$






            share|cite|improve this answer




















            • I really don't understand downvotes for such kind of OP and answers!
              – gimusi
              23 mins ago










            • @gimusi Me neither.
              – José Carlos Santos
              21 mins ago

















            up vote
            2
            down vote













            You are right the two vectors are linearly independent and therefore they span a $2$ dimensional subspace in $mathbbR^4$ that is in parametric form



            $$beginbmatrix
            x\ y \ z \ w
            endbmatrix=scdot beginbmatrix
            1\ 0 \ 1 \ 0
            endbmatrix+tcdot beginbmatrix
            0\ 1 \ 0 \ 1
            endbmatrix=beginbmatrix
            s\ t \ s \ t
            endbmatrix$$



            with $s,t in mathbbR$, or in cartesian form



            • $x=z$

            • $y=w$





            share|cite|improve this answer




















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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              1
              down vote



              accepted










              The span of the vectors you mention is a subspace of the vector space to which they belong. So $(1,0,1,0)$ and $(0,1,0,1)$ span a two dimensional subspace (since as you note the vectors are linearly independent) of $mathbbR^4$. It is not $mathbbR^2$ exactly as this subspace contains $4$-tuples, but it is `equivalen' (in some sense) to $mathbbR^2$ if this is of interest to you.



              So in general a span of $m<n$ linearly independent vectors in $mathbbR^n$ is a $m$-dimensional subspace of $mathbbR^n$.






              share|cite|improve this answer
























                up vote
                1
                down vote



                accepted










                The span of the vectors you mention is a subspace of the vector space to which they belong. So $(1,0,1,0)$ and $(0,1,0,1)$ span a two dimensional subspace (since as you note the vectors are linearly independent) of $mathbbR^4$. It is not $mathbbR^2$ exactly as this subspace contains $4$-tuples, but it is `equivalen' (in some sense) to $mathbbR^2$ if this is of interest to you.



                So in general a span of $m<n$ linearly independent vectors in $mathbbR^n$ is a $m$-dimensional subspace of $mathbbR^n$.






                share|cite|improve this answer






















                  up vote
                  1
                  down vote



                  accepted







                  up vote
                  1
                  down vote



                  accepted






                  The span of the vectors you mention is a subspace of the vector space to which they belong. So $(1,0,1,0)$ and $(0,1,0,1)$ span a two dimensional subspace (since as you note the vectors are linearly independent) of $mathbbR^4$. It is not $mathbbR^2$ exactly as this subspace contains $4$-tuples, but it is `equivalen' (in some sense) to $mathbbR^2$ if this is of interest to you.



                  So in general a span of $m<n$ linearly independent vectors in $mathbbR^n$ is a $m$-dimensional subspace of $mathbbR^n$.






                  share|cite|improve this answer












                  The span of the vectors you mention is a subspace of the vector space to which they belong. So $(1,0,1,0)$ and $(0,1,0,1)$ span a two dimensional subspace (since as you note the vectors are linearly independent) of $mathbbR^4$. It is not $mathbbR^2$ exactly as this subspace contains $4$-tuples, but it is `equivalen' (in some sense) to $mathbbR^2$ if this is of interest to you.



                  So in general a span of $m<n$ linearly independent vectors in $mathbbR^n$ is a $m$-dimensional subspace of $mathbbR^n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 24 mins ago









                  AnyAD

                  1,601711




                  1,601711




















                      up vote
                      3
                      down vote













                      Your vectors live in $mathbbR^4$, not in $mathbbR^2$, and therefore it is out of the question that they span $mathbbR^2$. Since they live in $mathbbR^4$, they span a subspace of $mathbbR^4$, whish happens to be$$leftbeginbmatrixx&y&x&yendbmatrix^T,middle.$$






                      share|cite|improve this answer




















                      • I really don't understand downvotes for such kind of OP and answers!
                        – gimusi
                        23 mins ago










                      • @gimusi Me neither.
                        – José Carlos Santos
                        21 mins ago














                      up vote
                      3
                      down vote













                      Your vectors live in $mathbbR^4$, not in $mathbbR^2$, and therefore it is out of the question that they span $mathbbR^2$. Since they live in $mathbbR^4$, they span a subspace of $mathbbR^4$, whish happens to be$$leftbeginbmatrixx&y&x&yendbmatrix^T,middle.$$






                      share|cite|improve this answer




















                      • I really don't understand downvotes for such kind of OP and answers!
                        – gimusi
                        23 mins ago










                      • @gimusi Me neither.
                        – José Carlos Santos
                        21 mins ago












                      up vote
                      3
                      down vote










                      up vote
                      3
                      down vote









                      Your vectors live in $mathbbR^4$, not in $mathbbR^2$, and therefore it is out of the question that they span $mathbbR^2$. Since they live in $mathbbR^4$, they span a subspace of $mathbbR^4$, whish happens to be$$leftbeginbmatrixx&y&x&yendbmatrix^T,middle.$$






                      share|cite|improve this answer












                      Your vectors live in $mathbbR^4$, not in $mathbbR^2$, and therefore it is out of the question that they span $mathbbR^2$. Since they live in $mathbbR^4$, they span a subspace of $mathbbR^4$, whish happens to be$$leftbeginbmatrixx&y&x&yendbmatrix^T,middle.$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 28 mins ago









                      José Carlos Santos

                      125k17101188




                      125k17101188











                      • I really don't understand downvotes for such kind of OP and answers!
                        – gimusi
                        23 mins ago










                      • @gimusi Me neither.
                        – José Carlos Santos
                        21 mins ago
















                      • I really don't understand downvotes for such kind of OP and answers!
                        – gimusi
                        23 mins ago










                      • @gimusi Me neither.
                        – José Carlos Santos
                        21 mins ago















                      I really don't understand downvotes for such kind of OP and answers!
                      – gimusi
                      23 mins ago




                      I really don't understand downvotes for such kind of OP and answers!
                      – gimusi
                      23 mins ago












                      @gimusi Me neither.
                      – José Carlos Santos
                      21 mins ago




                      @gimusi Me neither.
                      – José Carlos Santos
                      21 mins ago










                      up vote
                      2
                      down vote













                      You are right the two vectors are linearly independent and therefore they span a $2$ dimensional subspace in $mathbbR^4$ that is in parametric form



                      $$beginbmatrix
                      x\ y \ z \ w
                      endbmatrix=scdot beginbmatrix
                      1\ 0 \ 1 \ 0
                      endbmatrix+tcdot beginbmatrix
                      0\ 1 \ 0 \ 1
                      endbmatrix=beginbmatrix
                      s\ t \ s \ t
                      endbmatrix$$



                      with $s,t in mathbbR$, or in cartesian form



                      • $x=z$

                      • $y=w$





                      share|cite|improve this answer
























                        up vote
                        2
                        down vote













                        You are right the two vectors are linearly independent and therefore they span a $2$ dimensional subspace in $mathbbR^4$ that is in parametric form



                        $$beginbmatrix
                        x\ y \ z \ w
                        endbmatrix=scdot beginbmatrix
                        1\ 0 \ 1 \ 0
                        endbmatrix+tcdot beginbmatrix
                        0\ 1 \ 0 \ 1
                        endbmatrix=beginbmatrix
                        s\ t \ s \ t
                        endbmatrix$$



                        with $s,t in mathbbR$, or in cartesian form



                        • $x=z$

                        • $y=w$





                        share|cite|improve this answer






















                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          You are right the two vectors are linearly independent and therefore they span a $2$ dimensional subspace in $mathbbR^4$ that is in parametric form



                          $$beginbmatrix
                          x\ y \ z \ w
                          endbmatrix=scdot beginbmatrix
                          1\ 0 \ 1 \ 0
                          endbmatrix+tcdot beginbmatrix
                          0\ 1 \ 0 \ 1
                          endbmatrix=beginbmatrix
                          s\ t \ s \ t
                          endbmatrix$$



                          with $s,t in mathbbR$, or in cartesian form



                          • $x=z$

                          • $y=w$





                          share|cite|improve this answer












                          You are right the two vectors are linearly independent and therefore they span a $2$ dimensional subspace in $mathbbR^4$ that is in parametric form



                          $$beginbmatrix
                          x\ y \ z \ w
                          endbmatrix=scdot beginbmatrix
                          1\ 0 \ 1 \ 0
                          endbmatrix+tcdot beginbmatrix
                          0\ 1 \ 0 \ 1
                          endbmatrix=beginbmatrix
                          s\ t \ s \ t
                          endbmatrix$$



                          with $s,t in mathbbR$, or in cartesian form



                          • $x=z$

                          • $y=w$






                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 30 mins ago









                          gimusi

                          76k73889




                          76k73889



























                               

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