Maximum dimension of space of matrices with a real eigenvalueMaximum dimension of a space of $ntimes n$ real matrices with at least $k$ nonzero eigenvaluesProblems concerning subspaces of $M_n(mathbbC)$Spaces of matrices with same eigenvalue/Great circles in O(n)-orbitsDestroying the structure of a linear system while preserving its maximum eigenvalueExistence of a real eigenvalueMinimum negative eigenvalue of zero-one matricesProving that a certain non-symmetric matrix has an eigenvalue with positive real partBound the eigenvalue of product of matrices?Maximum spectral norm of matrices with given anti-Hermitian part and Hermitian part's spectrumPolynomial Eigenvalue Problem with few non-zero coefficientsMaximum dimension of a space of $ntimes n$ real matrices with at least $k$ nonzero eigenvalues

Maximum dimension of space of matrices with a real eigenvalue


Maximum dimension of a space of $ntimes n$ real matrices with at least $k$ nonzero eigenvaluesProblems concerning subspaces of $M_n(mathbbC)$Spaces of matrices with same eigenvalue/Great circles in O(n)-orbitsDestroying the structure of a linear system while preserving its maximum eigenvalueExistence of a real eigenvalueMinimum negative eigenvalue of zero-one matricesProving that a certain non-symmetric matrix has an eigenvalue with positive real partBound the eigenvalue of product of matrices?Maximum spectral norm of matrices with given anti-Hermitian part and Hermitian part's spectrumPolynomial Eigenvalue Problem with few non-zero coefficientsMaximum dimension of a space of $ntimes n$ real matrices with at least $k$ nonzero eigenvalues













12












$begingroup$


Let $M_n(mathbbR)$ denote the space of all $ntimes n$ real
matrices. What is the maximum dimension $f(n)$ of a subspace $V$ of
$M_n(mathbbR)$ such that every matrix in $V$ has at least one real
eigenvalue? It is easy to see that if $n$ is odd then $f(n)=n^2$,
while if $n$ is even then $n^2-n+1leq f(n)leq n^2-1$.



There is also the ``complementary'' problem: what is the maximum
dimension $g(n)$ of a subspace $W$ of $M_n(mathbbR)$ such that
every nonzero matrix in $W$ has no real eigenvalues? Clearly if $n$ is
odd then $g(n)=0$. When $n$ is even, can one have $g(n)>1$?










share|cite|improve this question











$endgroup$
















    12












    $begingroup$


    Let $M_n(mathbbR)$ denote the space of all $ntimes n$ real
    matrices. What is the maximum dimension $f(n)$ of a subspace $V$ of
    $M_n(mathbbR)$ such that every matrix in $V$ has at least one real
    eigenvalue? It is easy to see that if $n$ is odd then $f(n)=n^2$,
    while if $n$ is even then $n^2-n+1leq f(n)leq n^2-1$.



    There is also the ``complementary'' problem: what is the maximum
    dimension $g(n)$ of a subspace $W$ of $M_n(mathbbR)$ such that
    every nonzero matrix in $W$ has no real eigenvalues? Clearly if $n$ is
    odd then $g(n)=0$. When $n$ is even, can one have $g(n)>1$?










    share|cite|improve this question











    $endgroup$














      12












      12








      12


      4



      $begingroup$


      Let $M_n(mathbbR)$ denote the space of all $ntimes n$ real
      matrices. What is the maximum dimension $f(n)$ of a subspace $V$ of
      $M_n(mathbbR)$ such that every matrix in $V$ has at least one real
      eigenvalue? It is easy to see that if $n$ is odd then $f(n)=n^2$,
      while if $n$ is even then $n^2-n+1leq f(n)leq n^2-1$.



      There is also the ``complementary'' problem: what is the maximum
      dimension $g(n)$ of a subspace $W$ of $M_n(mathbbR)$ such that
      every nonzero matrix in $W$ has no real eigenvalues? Clearly if $n$ is
      odd then $g(n)=0$. When $n$ is even, can one have $g(n)>1$?










      share|cite|improve this question











      $endgroup$




      Let $M_n(mathbbR)$ denote the space of all $ntimes n$ real
      matrices. What is the maximum dimension $f(n)$ of a subspace $V$ of
      $M_n(mathbbR)$ such that every matrix in $V$ has at least one real
      eigenvalue? It is easy to see that if $n$ is odd then $f(n)=n^2$,
      while if $n$ is even then $n^2-n+1leq f(n)leq n^2-1$.



      There is also the ``complementary'' problem: what is the maximum
      dimension $g(n)$ of a subspace $W$ of $M_n(mathbbR)$ such that
      every nonzero matrix in $W$ has no real eigenvalues? Clearly if $n$ is
      odd then $g(n)=0$. When $n$ is even, can one have $g(n)>1$?







      linear-algebra eigenvalues






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 30 at 23:39







      Richard Stanley

















      asked Mar 30 at 18:37









      Richard StanleyRichard Stanley

      29.2k9116191




      29.2k9116191




















          2 Answers
          2






          active

          oldest

          votes


















          7












          $begingroup$

          For the "complementary" problem, $g(n)<n$ for all positive $n$,
          and the upper bound $g(n) leq n-1$ is sharp at least for $n=1,2,4,8$.



          If $dim W geq n$ then any nonzero vector $v in bf R^n$ is
          an eigenvector of some nonzero matrix in $W$. Indeed if $A_1,ldots,A_n in W$
          are linearly independent then a linear dependence in the $n+1$ vectors
          $v$ and $A_i v$, say $sum_i=1^n c_i A_i v = lambda v$, yields
          nonzero $A = sum_i=1^n c_i A_i in W$ with $Av = lambda v$.



          The equality in $g(n) leq n-1$ is trivial for $n=1$ and easy for $n=2$;
          for $n=4$ and $n=8$ we get examples from the traceless quaternions and
          octonions respectively:
          $$
          left(beginarraycccc
          0 & a & b & c cr
          -a & 0 & -c & b cr
          -b & c & 0 & -a cr
          -c & -b & a & 0
          endarrayright)
          $$

          has characteristic polynomial $(x^2+a^2+b^2+c^2)^2$, and
          $$
          left(beginarraycccccccc
          0 & a & b & c & d & e & f & g cr
          -a & 0 & c &-b & e &-d &-g & f cr
          -b &-c & 0 & a & f & g &-d &-e cr
          -c & b &-a & 0 & g &-f & e &-d cr
          -d &-e &-f &-g & 0 & a & b & c cr
          -e & d &-g & f &-a & 0 &-c & b cr
          -f & g & d &-e &-b & c & 0 &-a cr
          -g &-f & e & d &-c &-b & a & 0
          endarrayright)
          $$

          has characteristic polynomial $(x^2+a^2+b^2+c^2+d^2+e^2+f^2+g^2)^4$.






          share|cite|improve this answer











          $endgroup$




















            6












            $begingroup$

            The "complementary" problem is strongly related to the problem of independent vector fields on $S^n-1$. Indeed, if $W$ is a space of matrices none of them having a real eigen-value and $(A_1,...,A_k)$ is its basis, it means that for every $v$, the vectors $(v,A_1(v),...,A_k(v))$ are linearly independent. Hence, the projections of $(A_1v,...,A_kv)$ to the plane orthogonal to $v$ give rise to $k$ independent vector fields on the sphere $S^n-1$. This problem was solved by Adams using non-trivial algebraic topology, and the exact possible number is $rho(n)-1$ where $rho((2k+1)2^4a+b)=2^b+8a$. Note that this is consistent with the previous answer by Noam, since $S^1,S^3,S^7$ are really the only parallelizable spheres.



            Now, to see that this is the answer in our case, it suffices to note that the construction of the examples in Adams work actually arrises from a solution to the question you ask: they all come from construction of linear examples. Namely, consider a Clifford algebra $Cl_k$ having a representation of dimension $n$. Then the imaginary clifford elements $(e_1,...,e_k)$ give example to a space of matrices as you wish, and it is known that this is a maximal example for independent vector fields on on the $n-1$-sphere as well.






            share|cite|improve this answer











            $endgroup$








            • 1




              $begingroup$
              Thanks! In fact, Adams' result was the motivation for my question, as well as mathoverflow.net/questions/309135. Note that $dim V+dim Wleq n^2$, since otherwise $Vcap Wneq 0$. Thus Noam's answer solves my first question for $n=1,2,4,8$, and S. carmeli's answer improves the upper bound in general.
              $endgroup$
              – Richard Stanley
              2 days ago











            Your Answer





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            2 Answers
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            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            7












            $begingroup$

            For the "complementary" problem, $g(n)<n$ for all positive $n$,
            and the upper bound $g(n) leq n-1$ is sharp at least for $n=1,2,4,8$.



            If $dim W geq n$ then any nonzero vector $v in bf R^n$ is
            an eigenvector of some nonzero matrix in $W$. Indeed if $A_1,ldots,A_n in W$
            are linearly independent then a linear dependence in the $n+1$ vectors
            $v$ and $A_i v$, say $sum_i=1^n c_i A_i v = lambda v$, yields
            nonzero $A = sum_i=1^n c_i A_i in W$ with $Av = lambda v$.



            The equality in $g(n) leq n-1$ is trivial for $n=1$ and easy for $n=2$;
            for $n=4$ and $n=8$ we get examples from the traceless quaternions and
            octonions respectively:
            $$
            left(beginarraycccc
            0 & a & b & c cr
            -a & 0 & -c & b cr
            -b & c & 0 & -a cr
            -c & -b & a & 0
            endarrayright)
            $$

            has characteristic polynomial $(x^2+a^2+b^2+c^2)^2$, and
            $$
            left(beginarraycccccccc
            0 & a & b & c & d & e & f & g cr
            -a & 0 & c &-b & e &-d &-g & f cr
            -b &-c & 0 & a & f & g &-d &-e cr
            -c & b &-a & 0 & g &-f & e &-d cr
            -d &-e &-f &-g & 0 & a & b & c cr
            -e & d &-g & f &-a & 0 &-c & b cr
            -f & g & d &-e &-b & c & 0 &-a cr
            -g &-f & e & d &-c &-b & a & 0
            endarrayright)
            $$

            has characteristic polynomial $(x^2+a^2+b^2+c^2+d^2+e^2+f^2+g^2)^4$.






            share|cite|improve this answer











            $endgroup$

















              7












              $begingroup$

              For the "complementary" problem, $g(n)<n$ for all positive $n$,
              and the upper bound $g(n) leq n-1$ is sharp at least for $n=1,2,4,8$.



              If $dim W geq n$ then any nonzero vector $v in bf R^n$ is
              an eigenvector of some nonzero matrix in $W$. Indeed if $A_1,ldots,A_n in W$
              are linearly independent then a linear dependence in the $n+1$ vectors
              $v$ and $A_i v$, say $sum_i=1^n c_i A_i v = lambda v$, yields
              nonzero $A = sum_i=1^n c_i A_i in W$ with $Av = lambda v$.



              The equality in $g(n) leq n-1$ is trivial for $n=1$ and easy for $n=2$;
              for $n=4$ and $n=8$ we get examples from the traceless quaternions and
              octonions respectively:
              $$
              left(beginarraycccc
              0 & a & b & c cr
              -a & 0 & -c & b cr
              -b & c & 0 & -a cr
              -c & -b & a & 0
              endarrayright)
              $$

              has characteristic polynomial $(x^2+a^2+b^2+c^2)^2$, and
              $$
              left(beginarraycccccccc
              0 & a & b & c & d & e & f & g cr
              -a & 0 & c &-b & e &-d &-g & f cr
              -b &-c & 0 & a & f & g &-d &-e cr
              -c & b &-a & 0 & g &-f & e &-d cr
              -d &-e &-f &-g & 0 & a & b & c cr
              -e & d &-g & f &-a & 0 &-c & b cr
              -f & g & d &-e &-b & c & 0 &-a cr
              -g &-f & e & d &-c &-b & a & 0
              endarrayright)
              $$

              has characteristic polynomial $(x^2+a^2+b^2+c^2+d^2+e^2+f^2+g^2)^4$.






              share|cite|improve this answer











              $endgroup$















                7












                7








                7





                $begingroup$

                For the "complementary" problem, $g(n)<n$ for all positive $n$,
                and the upper bound $g(n) leq n-1$ is sharp at least for $n=1,2,4,8$.



                If $dim W geq n$ then any nonzero vector $v in bf R^n$ is
                an eigenvector of some nonzero matrix in $W$. Indeed if $A_1,ldots,A_n in W$
                are linearly independent then a linear dependence in the $n+1$ vectors
                $v$ and $A_i v$, say $sum_i=1^n c_i A_i v = lambda v$, yields
                nonzero $A = sum_i=1^n c_i A_i in W$ with $Av = lambda v$.



                The equality in $g(n) leq n-1$ is trivial for $n=1$ and easy for $n=2$;
                for $n=4$ and $n=8$ we get examples from the traceless quaternions and
                octonions respectively:
                $$
                left(beginarraycccc
                0 & a & b & c cr
                -a & 0 & -c & b cr
                -b & c & 0 & -a cr
                -c & -b & a & 0
                endarrayright)
                $$

                has characteristic polynomial $(x^2+a^2+b^2+c^2)^2$, and
                $$
                left(beginarraycccccccc
                0 & a & b & c & d & e & f & g cr
                -a & 0 & c &-b & e &-d &-g & f cr
                -b &-c & 0 & a & f & g &-d &-e cr
                -c & b &-a & 0 & g &-f & e &-d cr
                -d &-e &-f &-g & 0 & a & b & c cr
                -e & d &-g & f &-a & 0 &-c & b cr
                -f & g & d &-e &-b & c & 0 &-a cr
                -g &-f & e & d &-c &-b & a & 0
                endarrayright)
                $$

                has characteristic polynomial $(x^2+a^2+b^2+c^2+d^2+e^2+f^2+g^2)^4$.






                share|cite|improve this answer











                $endgroup$



                For the "complementary" problem, $g(n)<n$ for all positive $n$,
                and the upper bound $g(n) leq n-1$ is sharp at least for $n=1,2,4,8$.



                If $dim W geq n$ then any nonzero vector $v in bf R^n$ is
                an eigenvector of some nonzero matrix in $W$. Indeed if $A_1,ldots,A_n in W$
                are linearly independent then a linear dependence in the $n+1$ vectors
                $v$ and $A_i v$, say $sum_i=1^n c_i A_i v = lambda v$, yields
                nonzero $A = sum_i=1^n c_i A_i in W$ with $Av = lambda v$.



                The equality in $g(n) leq n-1$ is trivial for $n=1$ and easy for $n=2$;
                for $n=4$ and $n=8$ we get examples from the traceless quaternions and
                octonions respectively:
                $$
                left(beginarraycccc
                0 & a & b & c cr
                -a & 0 & -c & b cr
                -b & c & 0 & -a cr
                -c & -b & a & 0
                endarrayright)
                $$

                has characteristic polynomial $(x^2+a^2+b^2+c^2)^2$, and
                $$
                left(beginarraycccccccc
                0 & a & b & c & d & e & f & g cr
                -a & 0 & c &-b & e &-d &-g & f cr
                -b &-c & 0 & a & f & g &-d &-e cr
                -c & b &-a & 0 & g &-f & e &-d cr
                -d &-e &-f &-g & 0 & a & b & c cr
                -e & d &-g & f &-a & 0 &-c & b cr
                -f & g & d &-e &-b & c & 0 &-a cr
                -g &-f & e & d &-c &-b & a & 0
                endarrayright)
                $$

                has characteristic polynomial $(x^2+a^2+b^2+c^2+d^2+e^2+f^2+g^2)^4$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 2 days ago

























                answered Mar 31 at 0:58









                Noam D. ElkiesNoam D. Elkies

                56.6k11199282




                56.6k11199282





















                    6












                    $begingroup$

                    The "complementary" problem is strongly related to the problem of independent vector fields on $S^n-1$. Indeed, if $W$ is a space of matrices none of them having a real eigen-value and $(A_1,...,A_k)$ is its basis, it means that for every $v$, the vectors $(v,A_1(v),...,A_k(v))$ are linearly independent. Hence, the projections of $(A_1v,...,A_kv)$ to the plane orthogonal to $v$ give rise to $k$ independent vector fields on the sphere $S^n-1$. This problem was solved by Adams using non-trivial algebraic topology, and the exact possible number is $rho(n)-1$ where $rho((2k+1)2^4a+b)=2^b+8a$. Note that this is consistent with the previous answer by Noam, since $S^1,S^3,S^7$ are really the only parallelizable spheres.



                    Now, to see that this is the answer in our case, it suffices to note that the construction of the examples in Adams work actually arrises from a solution to the question you ask: they all come from construction of linear examples. Namely, consider a Clifford algebra $Cl_k$ having a representation of dimension $n$. Then the imaginary clifford elements $(e_1,...,e_k)$ give example to a space of matrices as you wish, and it is known that this is a maximal example for independent vector fields on on the $n-1$-sphere as well.






                    share|cite|improve this answer











                    $endgroup$








                    • 1




                      $begingroup$
                      Thanks! In fact, Adams' result was the motivation for my question, as well as mathoverflow.net/questions/309135. Note that $dim V+dim Wleq n^2$, since otherwise $Vcap Wneq 0$. Thus Noam's answer solves my first question for $n=1,2,4,8$, and S. carmeli's answer improves the upper bound in general.
                      $endgroup$
                      – Richard Stanley
                      2 days ago















                    6












                    $begingroup$

                    The "complementary" problem is strongly related to the problem of independent vector fields on $S^n-1$. Indeed, if $W$ is a space of matrices none of them having a real eigen-value and $(A_1,...,A_k)$ is its basis, it means that for every $v$, the vectors $(v,A_1(v),...,A_k(v))$ are linearly independent. Hence, the projections of $(A_1v,...,A_kv)$ to the plane orthogonal to $v$ give rise to $k$ independent vector fields on the sphere $S^n-1$. This problem was solved by Adams using non-trivial algebraic topology, and the exact possible number is $rho(n)-1$ where $rho((2k+1)2^4a+b)=2^b+8a$. Note that this is consistent with the previous answer by Noam, since $S^1,S^3,S^7$ are really the only parallelizable spheres.



                    Now, to see that this is the answer in our case, it suffices to note that the construction of the examples in Adams work actually arrises from a solution to the question you ask: they all come from construction of linear examples. Namely, consider a Clifford algebra $Cl_k$ having a representation of dimension $n$. Then the imaginary clifford elements $(e_1,...,e_k)$ give example to a space of matrices as you wish, and it is known that this is a maximal example for independent vector fields on on the $n-1$-sphere as well.






                    share|cite|improve this answer











                    $endgroup$








                    • 1




                      $begingroup$
                      Thanks! In fact, Adams' result was the motivation for my question, as well as mathoverflow.net/questions/309135. Note that $dim V+dim Wleq n^2$, since otherwise $Vcap Wneq 0$. Thus Noam's answer solves my first question for $n=1,2,4,8$, and S. carmeli's answer improves the upper bound in general.
                      $endgroup$
                      – Richard Stanley
                      2 days ago













                    6












                    6








                    6





                    $begingroup$

                    The "complementary" problem is strongly related to the problem of independent vector fields on $S^n-1$. Indeed, if $W$ is a space of matrices none of them having a real eigen-value and $(A_1,...,A_k)$ is its basis, it means that for every $v$, the vectors $(v,A_1(v),...,A_k(v))$ are linearly independent. Hence, the projections of $(A_1v,...,A_kv)$ to the plane orthogonal to $v$ give rise to $k$ independent vector fields on the sphere $S^n-1$. This problem was solved by Adams using non-trivial algebraic topology, and the exact possible number is $rho(n)-1$ where $rho((2k+1)2^4a+b)=2^b+8a$. Note that this is consistent with the previous answer by Noam, since $S^1,S^3,S^7$ are really the only parallelizable spheres.



                    Now, to see that this is the answer in our case, it suffices to note that the construction of the examples in Adams work actually arrises from a solution to the question you ask: they all come from construction of linear examples. Namely, consider a Clifford algebra $Cl_k$ having a representation of dimension $n$. Then the imaginary clifford elements $(e_1,...,e_k)$ give example to a space of matrices as you wish, and it is known that this is a maximal example for independent vector fields on on the $n-1$-sphere as well.






                    share|cite|improve this answer











                    $endgroup$



                    The "complementary" problem is strongly related to the problem of independent vector fields on $S^n-1$. Indeed, if $W$ is a space of matrices none of them having a real eigen-value and $(A_1,...,A_k)$ is its basis, it means that for every $v$, the vectors $(v,A_1(v),...,A_k(v))$ are linearly independent. Hence, the projections of $(A_1v,...,A_kv)$ to the plane orthogonal to $v$ give rise to $k$ independent vector fields on the sphere $S^n-1$. This problem was solved by Adams using non-trivial algebraic topology, and the exact possible number is $rho(n)-1$ where $rho((2k+1)2^4a+b)=2^b+8a$. Note that this is consistent with the previous answer by Noam, since $S^1,S^3,S^7$ are really the only parallelizable spheres.



                    Now, to see that this is the answer in our case, it suffices to note that the construction of the examples in Adams work actually arrises from a solution to the question you ask: they all come from construction of linear examples. Namely, consider a Clifford algebra $Cl_k$ having a representation of dimension $n$. Then the imaginary clifford elements $(e_1,...,e_k)$ give example to a space of matrices as you wish, and it is known that this is a maximal example for independent vector fields on on the $n-1$-sphere as well.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 2 days ago

























                    answered 2 days ago









                    S. carmeliS. carmeli

                    2,400519




                    2,400519







                    • 1




                      $begingroup$
                      Thanks! In fact, Adams' result was the motivation for my question, as well as mathoverflow.net/questions/309135. Note that $dim V+dim Wleq n^2$, since otherwise $Vcap Wneq 0$. Thus Noam's answer solves my first question for $n=1,2,4,8$, and S. carmeli's answer improves the upper bound in general.
                      $endgroup$
                      – Richard Stanley
                      2 days ago












                    • 1




                      $begingroup$
                      Thanks! In fact, Adams' result was the motivation for my question, as well as mathoverflow.net/questions/309135. Note that $dim V+dim Wleq n^2$, since otherwise $Vcap Wneq 0$. Thus Noam's answer solves my first question for $n=1,2,4,8$, and S. carmeli's answer improves the upper bound in general.
                      $endgroup$
                      – Richard Stanley
                      2 days ago







                    1




                    1




                    $begingroup$
                    Thanks! In fact, Adams' result was the motivation for my question, as well as mathoverflow.net/questions/309135. Note that $dim V+dim Wleq n^2$, since otherwise $Vcap Wneq 0$. Thus Noam's answer solves my first question for $n=1,2,4,8$, and S. carmeli's answer improves the upper bound in general.
                    $endgroup$
                    – Richard Stanley
                    2 days ago




                    $begingroup$
                    Thanks! In fact, Adams' result was the motivation for my question, as well as mathoverflow.net/questions/309135. Note that $dim V+dim Wleq n^2$, since otherwise $Vcap Wneq 0$. Thus Noam's answer solves my first question for $n=1,2,4,8$, and S. carmeli's answer improves the upper bound in general.
                    $endgroup$
                    – Richard Stanley
                    2 days ago

















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