A Bound on the Dimensions of Certain Types of Subspaces
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 at 19:22
LinearGuy
1437
add a comment |
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 at 19:22
LinearGuy
1437
2
I see where this is coming from, but I would guess not. Can't find a counterexample at the moment though.
– Matt Samuel
Dec 10 at 1:55
add a comment |
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 at 19:22
LinearGuy
1437
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 at 19:22
LinearGuy
1437
asked Dec 9 at 19:22
LinearGuy
1437
asked Dec 9 at 19:22
LinearGuy
1437
asked Dec 9 at 19:22
LinearGuy
1437
asked Dec 9 at 19:22
LinearGuy
1437
1437
2
I see where this is coming from, but I would guess not. Can't find a counterexample at the moment though.
– Matt Samuel
Dec 10 at 1:55
add a comment |
2
I see where this is coming from, but I would guess not. Can't find a counterexample at the moment though.
– Matt Samuel
Dec 10 at 1:55
2
2
I see where this is coming from, but I would guess not. Can't find a counterexample at the moment though.
– Matt Samuel
Dec 10 at 1:55
I see where this is coming from, but I would guess not. Can't find a counterexample at the moment though.
– Matt Samuel
Dec 10 at 1:55
add a comment |
1 Answer
1
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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 at 20:53
Lukas Geyer
13.3k1454
+1 Thanks for the theorem and for this nice application.
– Tengu
Dec 12 at 12:30
add a comment |
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1 Answer
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1 Answer
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active
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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 at 20:53
Lukas Geyer
13.3k1454
+1 Thanks for the theorem and for this nice application.
– Tengu
Dec 12 at 12:30
add a comment |
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 at 20:53
Lukas Geyer
13.3k1454
+1 Thanks for the theorem and for this nice application.
– Tengu
Dec 12 at 12:30
add a comment |
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 at 20:53
Lukas Geyer
13.3k1454
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 at 20:53
Lukas Geyer
13.3k1454
answered Dec 10 at 20:53
Lukas Geyer
13.3k1454
answered Dec 10 at 20:53
Lukas Geyer
13.3k1454
answered Dec 10 at 20:53
Lukas Geyer
13.3k1454
13.3k1454
+1 Thanks for the theorem and for this nice application.
– Tengu
Dec 12 at 12:30
add a comment |
+1 Thanks for the theorem and for this nice application.
– Tengu
Dec 12 at 12:30
+1 Thanks for the theorem and for this nice application.
– Tengu
Dec 12 at 12:30
+1 Thanks for the theorem and for this nice application.
– Tengu
Dec 12 at 12:30
add a comment |
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2
I see where this is coming from, but I would guess not. Can't find a counterexample at the moment though.
– Matt Samuel
Dec 10 at 1:55