Dimension of a certain subspace of $M_n(F).$
$begingroup$
Here is an interesting linear algebra problem:
Let $F$ be any field and $S$ be a symmetric, invertible matrix with entries from $F.$ Then what is the dimension of the vector space:
$$V = {Ain M_n(F)|, A^t = SAS^{-1}}quad$$
?
I believe have solved this problem with the answer being $dim V = dfrac{n(n+1)}{2},$ but my solution is a bit cumbersome to write: I considered a $n^2times n^2$ matrix whose kernel is isomorphic to $V$ and then computed the kernel manually.
I was wondering if there is some slick solution to this (I am almost positive that there is one.) What's troubling is that $F$ is an arbitrary field, considering Jordan normal form let alone diagonalization becomes a bit problematic.
linear-algebra linear-transformations
asked Jan 15 at 23:15
dezdichadodezdichado
6,5551929
$endgroup$
add a comment |
$begingroup$
Here is an interesting linear algebra problem:
Let $F$ be any field and $S$ be a symmetric, invertible matrix with entries from $F.$ Then what is the dimension of the vector space:
$$V = {Ain M_n(F)|, A^t = SAS^{-1}}quad$$
?
I believe have solved this problem with the answer being $dim V = dfrac{n(n+1)}{2},$ but my solution is a bit cumbersome to write: I considered a $n^2times n^2$ matrix whose kernel is isomorphic to $V$ and then computed the kernel manually.
I was wondering if there is some slick solution to this (I am almost positive that there is one.) What's troubling is that $F$ is an arbitrary field, considering Jordan normal form let alone diagonalization becomes a bit problematic.
linear-algebra linear-transformations
asked Jan 15 at 23:15
dezdichadodezdichado
6,5551929
$endgroup$
add a comment |
$begingroup$
Here is an interesting linear algebra problem:
Let $F$ be any field and $S$ be a symmetric, invertible matrix with entries from $F.$ Then what is the dimension of the vector space:
$$V = {Ain M_n(F)|, A^t = SAS^{-1}}quad$$
?
I believe have solved this problem with the answer being $dim V = dfrac{n(n+1)}{2},$ but my solution is a bit cumbersome to write: I considered a $n^2times n^2$ matrix whose kernel is isomorphic to $V$ and then computed the kernel manually.
I was wondering if there is some slick solution to this (I am almost positive that there is one.) What's troubling is that $F$ is an arbitrary field, considering Jordan normal form let alone diagonalization becomes a bit problematic.
linear-algebra linear-transformations
asked Jan 15 at 23:15
dezdichadodezdichado
6,5551929
$endgroup$
Here is an interesting linear algebra problem:
Let $F$ be any field and $S$ be a symmetric, invertible matrix with entries from $F.$ Then what is the dimension of the vector space:
$$V = {Ain M_n(F)|, A^t = SAS^{-1}}quad$$
?
I believe have solved this problem with the answer being $dim V = dfrac{n(n+1)}{2},$ but my solution is a bit cumbersome to write: I considered a $n^2times n^2$ matrix whose kernel is isomorphic to $V$ and then computed the kernel manually.
I was wondering if there is some slick solution to this (I am almost positive that there is one.) What's troubling is that $F$ is an arbitrary field, considering Jordan normal form let alone diagonalization becomes a bit problematic.
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Jan 15 at 23:15
dezdichadodezdichado
6,5551929
asked Jan 15 at 23:15
dezdichadodezdichado
6,5551929
asked Jan 15 at 23:15
dezdichadodezdichado
6,5551929
asked Jan 15 at 23:15
dezdichadodezdichado
6,5551929
asked Jan 15 at 23:15
dezdichadodezdichado
6,5551929
6,5551929
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Consider a linear transform $T: A mapsto SA-A^t S$.
Then the rank of $T$ is at most $(n^2-n)/2$ since $SA-A^t S$ is always skew-symmetric. Here, we need to define 'skew-symmetric' as being an image of $T_0: Amapsto A-A^t$.
To see if the rank of $T$ is exactly $(n^2-n)/2$, let $B$ be any skew-symmetric matrix with $B=A-A^t$ for some $A$. Since $S$ is invertible, we can find a matrix $C$ such that $A=SC$. Then we have
$$
B=SC-C^tS.
$$
Then we have $T(C) = B$. This proves the claim that the rank of $T$ is $(n^2-n)/2$.
Now, by rank-nullity theorem, we must have
$$
mathrm{dim} mathrm{ker}(T) + (n^2-n)/2 = n^2.
$$
Hence, we have
$$
mathrm{dim} mathrm{ker}(T)= (n^2+n)/2.
$$
Note: After realizing that previous argument fails when the characteristic of the field is $2$, the modification of the argument in this edit works generally for any characteristic.
Alternative simpler method
It is possible to simplify my argument. Let $V_0$ be the space of symmetric matrices
$$
V_0={Bin M_n(F) | B^t= B}.
$$
Then it is easy to see that the dimension of $V_0$ is $(n^2+n)/2$.
We establish an isomorphism (due to $S$ being invertible symmetric) between $V_0$ and $V$ by
$$
f: V_0 rightarrow V
$$
$$ B mapsto S^{-1}B.$$
Theorefore, $V$ has dimension $(n^2+n)/2$.
answered Jan 16 at 0:11
i707107i707107
12.7k21748
$endgroup$
add a comment |
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$begingroup$
Consider a linear transform $T: A mapsto SA-A^t S$.
Then the rank of $T$ is at most $(n^2-n)/2$ since $SA-A^t S$ is always skew-symmetric. Here, we need to define 'skew-symmetric' as being an image of $T_0: Amapsto A-A^t$.
To see if the rank of $T$ is exactly $(n^2-n)/2$, let $B$ be any skew-symmetric matrix with $B=A-A^t$ for some $A$. Since $S$ is invertible, we can find a matrix $C$ such that $A=SC$. Then we have
$$
B=SC-C^tS.
$$
Then we have $T(C) = B$. This proves the claim that the rank of $T$ is $(n^2-n)/2$.
Now, by rank-nullity theorem, we must have
$$
mathrm{dim} mathrm{ker}(T) + (n^2-n)/2 = n^2.
$$
Hence, we have
$$
mathrm{dim} mathrm{ker}(T)= (n^2+n)/2.
$$
Note: After realizing that previous argument fails when the characteristic of the field is $2$, the modification of the argument in this edit works generally for any characteristic.
Alternative simpler method
It is possible to simplify my argument. Let $V_0$ be the space of symmetric matrices
$$
V_0={Bin M_n(F) | B^t= B}.
$$
Then it is easy to see that the dimension of $V_0$ is $(n^2+n)/2$.
We establish an isomorphism (due to $S$ being invertible symmetric) between $V_0$ and $V$ by
$$
f: V_0 rightarrow V
$$
$$ B mapsto S^{-1}B.$$
Theorefore, $V$ has dimension $(n^2+n)/2$.
answered Jan 16 at 0:11
i707107i707107
12.7k21748
$endgroup$
add a comment |
$begingroup$
Consider a linear transform $T: A mapsto SA-A^t S$.
Then the rank of $T$ is at most $(n^2-n)/2$ since $SA-A^t S$ is always skew-symmetric. Here, we need to define 'skew-symmetric' as being an image of $T_0: Amapsto A-A^t$.
To see if the rank of $T$ is exactly $(n^2-n)/2$, let $B$ be any skew-symmetric matrix with $B=A-A^t$ for some $A$. Since $S$ is invertible, we can find a matrix $C$ such that $A=SC$. Then we have
$$
B=SC-C^tS.
$$
Then we have $T(C) = B$. This proves the claim that the rank of $T$ is $(n^2-n)/2$.
Now, by rank-nullity theorem, we must have
$$
mathrm{dim} mathrm{ker}(T) + (n^2-n)/2 = n^2.
$$
Hence, we have
$$
mathrm{dim} mathrm{ker}(T)= (n^2+n)/2.
$$
Note: After realizing that previous argument fails when the characteristic of the field is $2$, the modification of the argument in this edit works generally for any characteristic.
Alternative simpler method
It is possible to simplify my argument. Let $V_0$ be the space of symmetric matrices
$$
V_0={Bin M_n(F) | B^t= B}.
$$
Then it is easy to see that the dimension of $V_0$ is $(n^2+n)/2$.
We establish an isomorphism (due to $S$ being invertible symmetric) between $V_0$ and $V$ by
$$
f: V_0 rightarrow V
$$
$$ B mapsto S^{-1}B.$$
Theorefore, $V$ has dimension $(n^2+n)/2$.
answered Jan 16 at 0:11
i707107i707107
12.7k21748
$endgroup$
add a comment |
$begingroup$
Consider a linear transform $T: A mapsto SA-A^t S$.
Then the rank of $T$ is at most $(n^2-n)/2$ since $SA-A^t S$ is always skew-symmetric. Here, we need to define 'skew-symmetric' as being an image of $T_0: Amapsto A-A^t$.
To see if the rank of $T$ is exactly $(n^2-n)/2$, let $B$ be any skew-symmetric matrix with $B=A-A^t$ for some $A$. Since $S$ is invertible, we can find a matrix $C$ such that $A=SC$. Then we have
$$
B=SC-C^tS.
$$
Then we have $T(C) = B$. This proves the claim that the rank of $T$ is $(n^2-n)/2$.
Now, by rank-nullity theorem, we must have
$$
mathrm{dim} mathrm{ker}(T) + (n^2-n)/2 = n^2.
$$
Hence, we have
$$
mathrm{dim} mathrm{ker}(T)= (n^2+n)/2.
$$
Note: After realizing that previous argument fails when the characteristic of the field is $2$, the modification of the argument in this edit works generally for any characteristic.
Alternative simpler method
It is possible to simplify my argument. Let $V_0$ be the space of symmetric matrices
$$
V_0={Bin M_n(F) | B^t= B}.
$$
Then it is easy to see that the dimension of $V_0$ is $(n^2+n)/2$.
We establish an isomorphism (due to $S$ being invertible symmetric) between $V_0$ and $V$ by
$$
f: V_0 rightarrow V
$$
$$ B mapsto S^{-1}B.$$
Theorefore, $V$ has dimension $(n^2+n)/2$.
answered Jan 16 at 0:11
i707107i707107
12.7k21748
$endgroup$
Consider a linear transform $T: A mapsto SA-A^t S$.
Then the rank of $T$ is at most $(n^2-n)/2$ since $SA-A^t S$ is always skew-symmetric. Here, we need to define 'skew-symmetric' as being an image of $T_0: Amapsto A-A^t$.
To see if the rank of $T$ is exactly $(n^2-n)/2$, let $B$ be any skew-symmetric matrix with $B=A-A^t$ for some $A$. Since $S$ is invertible, we can find a matrix $C$ such that $A=SC$. Then we have
$$
B=SC-C^tS.
$$
Then we have $T(C) = B$. This proves the claim that the rank of $T$ is $(n^2-n)/2$.
Now, by rank-nullity theorem, we must have
$$
mathrm{dim} mathrm{ker}(T) + (n^2-n)/2 = n^2.
$$
Hence, we have
$$
mathrm{dim} mathrm{ker}(T)= (n^2+n)/2.
$$
Note: After realizing that previous argument fails when the characteristic of the field is $2$, the modification of the argument in this edit works generally for any characteristic.
Alternative simpler method
It is possible to simplify my argument. Let $V_0$ be the space of symmetric matrices
$$
V_0={Bin M_n(F) | B^t= B}.
$$
Then it is easy to see that the dimension of $V_0$ is $(n^2+n)/2$.
We establish an isomorphism (due to $S$ being invertible symmetric) between $V_0$ and $V$ by
$$
f: V_0 rightarrow V
$$
$$ B mapsto S^{-1}B.$$
Theorefore, $V$ has dimension $(n^2+n)/2$.
answered Jan 16 at 0:11
i707107i707107
12.7k21748
edited Jan 16 at 21:54
answered Jan 16 at 0:11
i707107i707107
12.7k21748
answered Jan 16 at 0:11
i707107i707107
12.7k21748
answered Jan 16 at 0:11
i707107i707107
12.7k21748
12.7k21748
add a comment |
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