A Bound on the Dimensions of Certain Types of Subspaces
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Let $V$ be a $4$-dimensional vector space over the complex numbers, and let $S$ be a subspace of the endomorphisms of $V$ such that the elements of $S$ commute.
If there exists an element in $S$ that has at least two distinct eigenvalues, is the dimension of $S$ at most $4$? If so (or if not), why?
An example of such a subspace of dimension $5$ that does not have an element with at least two distinct eigenvalues is the set of matrices of the form
$$left(begin{matrix}
a & 0 & c & d \
0 & a & e & f \
0 & 0 & a & 0 \
0 & 0 & 0 & a
end{matrix}right) .$$
linear-algebra abstract-algebra
asked Dec 9 '18 at 19:22
LinearGuyLinearGuy
1511
$endgroup$
add a comment |
$begingroup$
Let $V$ be a $4$-dimensional vector space over the complex numbers, and let $S$ be a subspace of the endomorphisms of $V$ such that the elements of $S$ commute.
If there exists an element in $S$ that has at least two distinct eigenvalues, is the dimension of $S$ at most $4$? If so (or if not), why?
An example of such a subspace of dimension $5$ that does not have an element with at least two distinct eigenvalues is the set of matrices of the form
$$left(begin{matrix}
a & 0 & c & d \
0 & a & e & f \
0 & 0 & a & 0 \
0 & 0 & 0 & a
end{matrix}right) .$$
linear-algebra abstract-algebra
asked Dec 9 '18 at 19:22
LinearGuyLinearGuy
1511
$endgroup$
2
$begingroup$
I see where this is coming from, but I would guess not. Can't find a counterexample at the moment though.
$endgroup$
– Matt Samuel
Dec 10 '18 at 1:55
add a comment |
$begingroup$
Let $V$ be a $4$-dimensional vector space over the complex numbers, and let $S$ be a subspace of the endomorphisms of $V$ such that the elements of $S$ commute.
If there exists an element in $S$ that has at least two distinct eigenvalues, is the dimension of $S$ at most $4$? If so (or if not), why?
An example of such a subspace of dimension $5$ that does not have an element with at least two distinct eigenvalues is the set of matrices of the form
$$left(begin{matrix}
a & 0 & c & d \
0 & a & e & f \
0 & 0 & a & 0 \
0 & 0 & 0 & a
end{matrix}right) .$$
linear-algebra abstract-algebra
asked Dec 9 '18 at 19:22
LinearGuyLinearGuy
1511
$endgroup$
Let $V$ be a $4$-dimensional vector space over the complex numbers, and let $S$ be a subspace of the endomorphisms of $V$ such that the elements of $S$ commute.
If there exists an element in $S$ that has at least two distinct eigenvalues, is the dimension of $S$ at most $4$? If so (or if not), why?
An example of such a subspace of dimension $5$ that does not have an element with at least two distinct eigenvalues is the set of matrices of the form
$$left(begin{matrix}
a & 0 & c & d \
0 & a & e & f \
0 & 0 & a & 0 \
0 & 0 & 0 & a
end{matrix}right) .$$
linear-algebra abstract-algebra
linear-algebra abstract-algebra
asked Dec 9 '18 at 19:22
LinearGuyLinearGuy
1511
asked Dec 9 '18 at 19:22
LinearGuyLinearGuy
1511
asked Dec 9 '18 at 19:22
LinearGuyLinearGuy
1511
asked Dec 9 '18 at 19:22
LinearGuyLinearGuy
1511
asked Dec 9 '18 at 19:22
LinearGuyLinearGuy
1511
1511
2
$begingroup$
I see where this is coming from, but I would guess not. Can't find a counterexample at the moment though.
$endgroup$
– Matt Samuel
Dec 10 '18 at 1:55
add a comment |
2
$begingroup$
I see where this is coming from, but I would guess not. Can't find a counterexample at the moment though.
$endgroup$
– Matt Samuel
Dec 10 '18 at 1:55
2
2
$begingroup$
I see where this is coming from, but I would guess not. Can't find a counterexample at the moment though.
$endgroup$
– Matt Samuel
Dec 10 '18 at 1:55
$begingroup$
I see where this is coming from, but I would guess not. Can't find a counterexample at the moment though.
$endgroup$
– Matt Samuel
Dec 10 '18 at 1:55
add a comment |
1 Answer
1
active
oldest
votes
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By an old theorem of Schur (see this simple proof in an early paper by the late great Maryam Mirzakhani), the maximal number of linearly independent linear endomorphisms of $mathbb{C}^n$ is $N(n) = lfloor n^2/4 rfloor + 1$. Your example gives the maximum dimension $N(4)=5$. However, if some matrix $A in S$ has two different eigenvalues, then $S$ will be reducible since every matrix in $S$ has these different eigenspaces of $A$ as invariant subspaces. For $n<4$, one always has $N(n) = n$, so the maximum dimension in that case is $4$.
answered Dec 10 '18 at 20:53
Lukas GeyerLukas Geyer
13.8k1556
$endgroup$
$begingroup$
+1 Thanks for the theorem and for this nice application.
$endgroup$
– Tengu
Dec 12 '18 at 12:30
add a comment |
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1 Answer
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1 Answer
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$begingroup$
By an old theorem of Schur (see this simple proof in an early paper by the late great Maryam Mirzakhani), the maximal number of linearly independent linear endomorphisms of $mathbb{C}^n$ is $N(n) = lfloor n^2/4 rfloor + 1$. Your example gives the maximum dimension $N(4)=5$. However, if some matrix $A in S$ has two different eigenvalues, then $S$ will be reducible since every matrix in $S$ has these different eigenspaces of $A$ as invariant subspaces. For $n<4$, one always has $N(n) = n$, so the maximum dimension in that case is $4$.
answered Dec 10 '18 at 20:53
Lukas GeyerLukas Geyer
13.8k1556
$endgroup$
$begingroup$
+1 Thanks for the theorem and for this nice application.
$endgroup$
– Tengu
Dec 12 '18 at 12:30
add a comment |
$begingroup$
By an old theorem of Schur (see this simple proof in an early paper by the late great Maryam Mirzakhani), the maximal number of linearly independent linear endomorphisms of $mathbb{C}^n$ is $N(n) = lfloor n^2/4 rfloor + 1$. Your example gives the maximum dimension $N(4)=5$. However, if some matrix $A in S$ has two different eigenvalues, then $S$ will be reducible since every matrix in $S$ has these different eigenspaces of $A$ as invariant subspaces. For $n<4$, one always has $N(n) = n$, so the maximum dimension in that case is $4$.
answered Dec 10 '18 at 20:53
Lukas GeyerLukas Geyer
13.8k1556
$endgroup$
$begingroup$
+1 Thanks for the theorem and for this nice application.
$endgroup$
– Tengu
Dec 12 '18 at 12:30
add a comment |
$begingroup$
By an old theorem of Schur (see this simple proof in an early paper by the late great Maryam Mirzakhani), the maximal number of linearly independent linear endomorphisms of $mathbb{C}^n$ is $N(n) = lfloor n^2/4 rfloor + 1$. Your example gives the maximum dimension $N(4)=5$. However, if some matrix $A in S$ has two different eigenvalues, then $S$ will be reducible since every matrix in $S$ has these different eigenspaces of $A$ as invariant subspaces. For $n<4$, one always has $N(n) = n$, so the maximum dimension in that case is $4$.
answered Dec 10 '18 at 20:53
Lukas GeyerLukas Geyer
13.8k1556
$endgroup$
By an old theorem of Schur (see this simple proof in an early paper by the late great Maryam Mirzakhani), the maximal number of linearly independent linear endomorphisms of $mathbb{C}^n$ is $N(n) = lfloor n^2/4 rfloor + 1$. Your example gives the maximum dimension $N(4)=5$. However, if some matrix $A in S$ has two different eigenvalues, then $S$ will be reducible since every matrix in $S$ has these different eigenspaces of $A$ as invariant subspaces. For $n<4$, one always has $N(n) = n$, so the maximum dimension in that case is $4$.
answered Dec 10 '18 at 20:53
Lukas GeyerLukas Geyer
13.8k1556
answered Dec 10 '18 at 20:53
Lukas GeyerLukas Geyer
13.8k1556
answered Dec 10 '18 at 20:53
Lukas GeyerLukas Geyer
13.8k1556
answered Dec 10 '18 at 20:53
Lukas GeyerLukas Geyer
13.8k1556
13.8k1556
$begingroup$
+1 Thanks for the theorem and for this nice application.
$endgroup$
– Tengu
Dec 12 '18 at 12:30
add a comment |
$begingroup$
+1 Thanks for the theorem and for this nice application.
$endgroup$
– Tengu
Dec 12 '18 at 12:30
$begingroup$
+1 Thanks for the theorem and for this nice application.
$endgroup$
– Tengu
Dec 12 '18 at 12:30
$begingroup$
+1 Thanks for the theorem and for this nice application.
$endgroup$
– Tengu
Dec 12 '18 at 12:30
add a comment |
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$begingroup$
I see where this is coming from, but I would guess not. Can't find a counterexample at the moment though.
$endgroup$
– Matt Samuel
Dec 10 '18 at 1:55