Maximum dimension of space of matrices with a real eigenvalue The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)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
Maximum dimension of space of matrices with a real eigenvalue
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)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
$begingroup$
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 $g(n)>1$?
linear-algebra eigenvalues
$endgroup$
add a comment |
$begingroup$
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 $g(n)>1$?
linear-algebra eigenvalues
$endgroup$
add a comment |
$begingroup$
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 $g(n)>1$?
linear-algebra eigenvalues
$endgroup$
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
linear-algebra eigenvalues
edited Mar 30 at 23:39
Richard Stanley
asked Mar 30 at 18:37
Richard StanleyRichard Stanley
29.3k9117191
29.3k9117191
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
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
Mar 31 at 13:46
add a comment |
$begingroup$
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 $(x^2+a^2+b^2+c^2+d^2+e^2+f^2+g^2)^4$.
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
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.
$endgroup$
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
Mar 31 at 13:46
add a comment |
$begingroup$
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.
$endgroup$
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
Mar 31 at 13:46
add a comment |
$begingroup$
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.
$endgroup$
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.
edited Mar 31 at 9:52
answered Mar 31 at 8:11
S. carmeliS. carmeli
2,495519
2,495519
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
Mar 31 at 13:46
add a comment |
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
Mar 31 at 13:46
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
Mar 31 at 13:46
$begingroup$
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
Mar 31 at 13:46
add a comment |
$begingroup$
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 $(x^2+a^2+b^2+c^2+d^2+e^2+f^2+g^2)^4$.
$endgroup$
add a comment |
$begingroup$
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 $(x^2+a^2+b^2+c^2+d^2+e^2+f^2+g^2)^4$.
$endgroup$
add a comment |
$begingroup$
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 $(x^2+a^2+b^2+c^2+d^2+e^2+f^2+g^2)^4$.
$endgroup$
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 $(x^2+a^2+b^2+c^2+d^2+e^2+f^2+g^2)^4$.
edited Mar 31 at 23:20
answered Mar 31 at 0:58
Noam D. ElkiesNoam D. Elkies
56.7k11200283
56.7k11200283
add a comment |
add a comment |
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