Is $ f(A) = A + 2A^{T} $ an isomorphism of $ mathbb R^{5,5} $ onto itself?
I have problem with prove or disprove this hypothesis:
Is the linear transformation $ f in L(mathbb R^{5,5},mathbb R^{5,5}) $
$$ f(A) = A + 2A^{T} $$
an isomorphism of the space $ mathbb R^{5,5} $ onto itself?
In order to be an isomorphism, $f$ must be injective and surjective.
Checking if it is surjective:
Assume that as a result we want to get:
$$mathbf{Q} =
begin{bmatrix}
q_{11} & q_{12} & cdots & q_{1n} \
q_{21} & q_{22} & cdots & q_{2n} \
vdots & vdots & ddots & vdots \
q_{m1} & q_{m2} & cdots & q_{mn}
end{bmatrix} $$
and we put as argument $$ mathbf{A} =
begin{bmatrix}
a_{11} & a_{12} & cdots & a_{1n} \
a_{21} & a_{22} & cdots & a_{2n} \
vdots & vdots & ddots & vdots \
a_{m1} & a_{m2} & cdots & a_{mn}
end{bmatrix} $$
then choose $i,j$. Factors $a_{ij}$ $a_{ji}$ $q_{ij}$ $q_{ji}$ are only in this linear system:
$$q_{ij} = a_{ij} + 2a_{ji} wedge q_{ji} = a_{ji} + 2a_{ij}$$
so $$ a_{ij} = frac{2q_{ij}-q_{ji}}{3} $$ and $$ a_{ji} = frac{2q_{ji}-q_{ij}}{3} $$
so the factors $ a_{ij}$ $ a_{ji}$ are determined unambiguously. So it is surjective.
But how to deal with checking injectivity?
Suppose that $$ A+2A^T=B+2B^T $$
$$ A-B = 2(A^T-B^T) $$
and I don't know how to finish that.
linear-algebra matrices vector-space-isomorphism
add a comment |
I have problem with prove or disprove this hypothesis:
Is the linear transformation $ f in L(mathbb R^{5,5},mathbb R^{5,5}) $
$$ f(A) = A + 2A^{T} $$
an isomorphism of the space $ mathbb R^{5,5} $ onto itself?
In order to be an isomorphism, $f$ must be injective and surjective.
Checking if it is surjective:
Assume that as a result we want to get:
$$mathbf{Q} =
begin{bmatrix}
q_{11} & q_{12} & cdots & q_{1n} \
q_{21} & q_{22} & cdots & q_{2n} \
vdots & vdots & ddots & vdots \
q_{m1} & q_{m2} & cdots & q_{mn}
end{bmatrix} $$
and we put as argument $$ mathbf{A} =
begin{bmatrix}
a_{11} & a_{12} & cdots & a_{1n} \
a_{21} & a_{22} & cdots & a_{2n} \
vdots & vdots & ddots & vdots \
a_{m1} & a_{m2} & cdots & a_{mn}
end{bmatrix} $$
then choose $i,j$. Factors $a_{ij}$ $a_{ji}$ $q_{ij}$ $q_{ji}$ are only in this linear system:
$$q_{ij} = a_{ij} + 2a_{ji} wedge q_{ji} = a_{ji} + 2a_{ij}$$
so $$ a_{ij} = frac{2q_{ij}-q_{ji}}{3} $$ and $$ a_{ji} = frac{2q_{ji}-q_{ij}}{3} $$
so the factors $ a_{ij}$ $ a_{ji}$ are determined unambiguously. So it is surjective.
But how to deal with checking injectivity?
Suppose that $$ A+2A^T=B+2B^T $$
$$ A-B = 2(A^T-B^T) $$
and I don't know how to finish that.
linear-algebra matrices vector-space-isomorphism
add a comment |
I have problem with prove or disprove this hypothesis:
Is the linear transformation $ f in L(mathbb R^{5,5},mathbb R^{5,5}) $
$$ f(A) = A + 2A^{T} $$
an isomorphism of the space $ mathbb R^{5,5} $ onto itself?
In order to be an isomorphism, $f$ must be injective and surjective.
Checking if it is surjective:
Assume that as a result we want to get:
$$mathbf{Q} =
begin{bmatrix}
q_{11} & q_{12} & cdots & q_{1n} \
q_{21} & q_{22} & cdots & q_{2n} \
vdots & vdots & ddots & vdots \
q_{m1} & q_{m2} & cdots & q_{mn}
end{bmatrix} $$
and we put as argument $$ mathbf{A} =
begin{bmatrix}
a_{11} & a_{12} & cdots & a_{1n} \
a_{21} & a_{22} & cdots & a_{2n} \
vdots & vdots & ddots & vdots \
a_{m1} & a_{m2} & cdots & a_{mn}
end{bmatrix} $$
then choose $i,j$. Factors $a_{ij}$ $a_{ji}$ $q_{ij}$ $q_{ji}$ are only in this linear system:
$$q_{ij} = a_{ij} + 2a_{ji} wedge q_{ji} = a_{ji} + 2a_{ij}$$
so $$ a_{ij} = frac{2q_{ij}-q_{ji}}{3} $$ and $$ a_{ji} = frac{2q_{ji}-q_{ij}}{3} $$
so the factors $ a_{ij}$ $ a_{ji}$ are determined unambiguously. So it is surjective.
But how to deal with checking injectivity?
Suppose that $$ A+2A^T=B+2B^T $$
$$ A-B = 2(A^T-B^T) $$
and I don't know how to finish that.
linear-algebra matrices vector-space-isomorphism
I have problem with prove or disprove this hypothesis:
Is the linear transformation $ f in L(mathbb R^{5,5},mathbb R^{5,5}) $
$$ f(A) = A + 2A^{T} $$
an isomorphism of the space $ mathbb R^{5,5} $ onto itself?
In order to be an isomorphism, $f$ must be injective and surjective.
Checking if it is surjective:
Assume that as a result we want to get:
$$mathbf{Q} =
begin{bmatrix}
q_{11} & q_{12} & cdots & q_{1n} \
q_{21} & q_{22} & cdots & q_{2n} \
vdots & vdots & ddots & vdots \
q_{m1} & q_{m2} & cdots & q_{mn}
end{bmatrix} $$
and we put as argument $$ mathbf{A} =
begin{bmatrix}
a_{11} & a_{12} & cdots & a_{1n} \
a_{21} & a_{22} & cdots & a_{2n} \
vdots & vdots & ddots & vdots \
a_{m1} & a_{m2} & cdots & a_{mn}
end{bmatrix} $$
then choose $i,j$. Factors $a_{ij}$ $a_{ji}$ $q_{ij}$ $q_{ji}$ are only in this linear system:
$$q_{ij} = a_{ij} + 2a_{ji} wedge q_{ji} = a_{ji} + 2a_{ij}$$
so $$ a_{ij} = frac{2q_{ij}-q_{ji}}{3} $$ and $$ a_{ji} = frac{2q_{ji}-q_{ij}}{3} $$
so the factors $ a_{ij}$ $ a_{ji}$ are determined unambiguously. So it is surjective.
But how to deal with checking injectivity?
Suppose that $$ A+2A^T=B+2B^T $$
$$ A-B = 2(A^T-B^T) $$
and I don't know how to finish that.
linear-algebra matrices vector-space-isomorphism
linear-algebra matrices vector-space-isomorphism
edited Jan 4 at 21:23
greedoid
38.6k114797
38.6k114797
asked Jan 4 at 21:01
VirtualUserVirtualUser
57712
57712
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
Hint:
Check if the kernel of this transformation is trivial. This is enough to establish if it is isomorphism:
$$begin{eqnarray}
f(A) =0
&implies &A=-2A^T \
&implies &A^T=(-2A^T)^T \
&implies &A^T = -2A \
&implies &A =4A \
&implies &A=0
end{eqnarray}
$$
By trivial you mean $ vec{0} $?
– VirtualUser
Jan 4 at 21:06
Yes, that is correct.
– greedoid
Jan 4 at 21:06
Chm.. can you explain this step? $ A^T = -2A rightarrow A =4A $
– VirtualUser
Jan 4 at 21:13
Just plug second line in first
– greedoid
Jan 4 at 21:14
Ahh, that's great, but unfortunately I can't use that because I haven't got this theorem on my lecture... But it is really smart
– VirtualUser
Jan 4 at 21:15
|
show 1 more comment
You have $A-B = 2(A-B)^top$.
This implies $a_{ij} - b_{ij} = 2 (a_{ji} - b_{ji}) = 4(a_{ij} - b_{ij})$, so...
add a comment |
You're already done. Since the map is from $mathbb R^{5times 5}$ to $mathbb R^{5times 5}$, checking that it's surjective (that it has full rank) also tells you that it's injective (that it has nullity $0$) by the rank-nullity theorem.
add a comment |
Hint:
It suffices to check that $f$ is surjective. For any matrix $B$ we have
$$fleft(-frac13B + frac23B^Tright) = B$$
It you are wondering how we got this, assume $A+2A^T = B$. Then $A^T + 2A = B^T$ so $$B + B^T = 3(A+A^T)$$
Now it follows $$A = B^T - (A+A^T) = B^T - frac13(B + B^T) = -frac13B + frac23B^T$$
This seems to already be contained in the question post.
– Misha Lavrov
Jan 4 at 21:30
@MishaLavrov Oops, I missed it. Still, here it is done slightly cleaner, without coordinates.
– mechanodroid
Jan 4 at 21:31
add a comment |
When you say
$$
a_{ij}=frac{2q_{ji}-q_{ij}}{3}
$$
(I fixed the indices) is “determined unambiguously”, you have proved both surjectivity and injectivity. There is at most one matrix $mathbf{A}$ such that $f(mathbf{A})=mathbf{Q}$. And there is one, determined in the way you did (after fixing the small mistake).
On the other hand, a linear map $Lcolon Vto V$, where $V$ is a finite dimensional vector space, is injective if and only if it is surjective, because of the rank-nullity theorem. Therefore just one of the properties needs to be checked for.
You don't need to go to the level of matrix entries in order to prove surjectivity. Given $Q$ you want to find $A$ such that $f(A)=Q$, that is, $A+2A^T=Q$. If such matrix exists, then also
$$
A^T+2A=Q^T
$$
and so $2A^T+4A=2Q^T$; thus
$$
Q-2Q^T=A+2A^T-2A^T-4A=-3A
$$
so
$$
A=frac{1}{3}(2Q^{T}-Q)
$$
is the only possible one. Since
$$
fleft(frac{1}{3}(2Q^{T}-Q)right)=Q
$$
as it can be readily checked, you have surjectivity and injectivity.
Checking directly for injectivity is easier, though.
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint:
Check if the kernel of this transformation is trivial. This is enough to establish if it is isomorphism:
$$begin{eqnarray}
f(A) =0
&implies &A=-2A^T \
&implies &A^T=(-2A^T)^T \
&implies &A^T = -2A \
&implies &A =4A \
&implies &A=0
end{eqnarray}
$$
By trivial you mean $ vec{0} $?
– VirtualUser
Jan 4 at 21:06
Yes, that is correct.
– greedoid
Jan 4 at 21:06
Chm.. can you explain this step? $ A^T = -2A rightarrow A =4A $
– VirtualUser
Jan 4 at 21:13
Just plug second line in first
– greedoid
Jan 4 at 21:14
Ahh, that's great, but unfortunately I can't use that because I haven't got this theorem on my lecture... But it is really smart
– VirtualUser
Jan 4 at 21:15
|
show 1 more comment
Hint:
Check if the kernel of this transformation is trivial. This is enough to establish if it is isomorphism:
$$begin{eqnarray}
f(A) =0
&implies &A=-2A^T \
&implies &A^T=(-2A^T)^T \
&implies &A^T = -2A \
&implies &A =4A \
&implies &A=0
end{eqnarray}
$$
By trivial you mean $ vec{0} $?
– VirtualUser
Jan 4 at 21:06
Yes, that is correct.
– greedoid
Jan 4 at 21:06
Chm.. can you explain this step? $ A^T = -2A rightarrow A =4A $
– VirtualUser
Jan 4 at 21:13
Just plug second line in first
– greedoid
Jan 4 at 21:14
Ahh, that's great, but unfortunately I can't use that because I haven't got this theorem on my lecture... But it is really smart
– VirtualUser
Jan 4 at 21:15
|
show 1 more comment
Hint:
Check if the kernel of this transformation is trivial. This is enough to establish if it is isomorphism:
$$begin{eqnarray}
f(A) =0
&implies &A=-2A^T \
&implies &A^T=(-2A^T)^T \
&implies &A^T = -2A \
&implies &A =4A \
&implies &A=0
end{eqnarray}
$$
Hint:
Check if the kernel of this transformation is trivial. This is enough to establish if it is isomorphism:
$$begin{eqnarray}
f(A) =0
&implies &A=-2A^T \
&implies &A^T=(-2A^T)^T \
&implies &A^T = -2A \
&implies &A =4A \
&implies &A=0
end{eqnarray}
$$
edited Jan 4 at 21:17
answered Jan 4 at 21:04
greedoidgreedoid
38.6k114797
38.6k114797
By trivial you mean $ vec{0} $?
– VirtualUser
Jan 4 at 21:06
Yes, that is correct.
– greedoid
Jan 4 at 21:06
Chm.. can you explain this step? $ A^T = -2A rightarrow A =4A $
– VirtualUser
Jan 4 at 21:13
Just plug second line in first
– greedoid
Jan 4 at 21:14
Ahh, that's great, but unfortunately I can't use that because I haven't got this theorem on my lecture... But it is really smart
– VirtualUser
Jan 4 at 21:15
|
show 1 more comment
By trivial you mean $ vec{0} $?
– VirtualUser
Jan 4 at 21:06
Yes, that is correct.
– greedoid
Jan 4 at 21:06
Chm.. can you explain this step? $ A^T = -2A rightarrow A =4A $
– VirtualUser
Jan 4 at 21:13
Just plug second line in first
– greedoid
Jan 4 at 21:14
Ahh, that's great, but unfortunately I can't use that because I haven't got this theorem on my lecture... But it is really smart
– VirtualUser
Jan 4 at 21:15
By trivial you mean $ vec{0} $?
– VirtualUser
Jan 4 at 21:06
By trivial you mean $ vec{0} $?
– VirtualUser
Jan 4 at 21:06
Yes, that is correct.
– greedoid
Jan 4 at 21:06
Yes, that is correct.
– greedoid
Jan 4 at 21:06
Chm.. can you explain this step? $ A^T = -2A rightarrow A =4A $
– VirtualUser
Jan 4 at 21:13
Chm.. can you explain this step? $ A^T = -2A rightarrow A =4A $
– VirtualUser
Jan 4 at 21:13
Just plug second line in first
– greedoid
Jan 4 at 21:14
Just plug second line in first
– greedoid
Jan 4 at 21:14
Ahh, that's great, but unfortunately I can't use that because I haven't got this theorem on my lecture... But it is really smart
– VirtualUser
Jan 4 at 21:15
Ahh, that's great, but unfortunately I can't use that because I haven't got this theorem on my lecture... But it is really smart
– VirtualUser
Jan 4 at 21:15
|
show 1 more comment
You have $A-B = 2(A-B)^top$.
This implies $a_{ij} - b_{ij} = 2 (a_{ji} - b_{ji}) = 4(a_{ij} - b_{ij})$, so...
add a comment |
You have $A-B = 2(A-B)^top$.
This implies $a_{ij} - b_{ij} = 2 (a_{ji} - b_{ji}) = 4(a_{ij} - b_{ij})$, so...
add a comment |
You have $A-B = 2(A-B)^top$.
This implies $a_{ij} - b_{ij} = 2 (a_{ji} - b_{ji}) = 4(a_{ij} - b_{ij})$, so...
You have $A-B = 2(A-B)^top$.
This implies $a_{ij} - b_{ij} = 2 (a_{ji} - b_{ji}) = 4(a_{ij} - b_{ij})$, so...
answered Jan 4 at 21:05
angryavianangryavian
39.4k23280
39.4k23280
add a comment |
add a comment |
You're already done. Since the map is from $mathbb R^{5times 5}$ to $mathbb R^{5times 5}$, checking that it's surjective (that it has full rank) also tells you that it's injective (that it has nullity $0$) by the rank-nullity theorem.
add a comment |
You're already done. Since the map is from $mathbb R^{5times 5}$ to $mathbb R^{5times 5}$, checking that it's surjective (that it has full rank) also tells you that it's injective (that it has nullity $0$) by the rank-nullity theorem.
add a comment |
You're already done. Since the map is from $mathbb R^{5times 5}$ to $mathbb R^{5times 5}$, checking that it's surjective (that it has full rank) also tells you that it's injective (that it has nullity $0$) by the rank-nullity theorem.
You're already done. Since the map is from $mathbb R^{5times 5}$ to $mathbb R^{5times 5}$, checking that it's surjective (that it has full rank) also tells you that it's injective (that it has nullity $0$) by the rank-nullity theorem.
answered Jan 4 at 21:28
Misha LavrovMisha Lavrov
44.3k555106
44.3k555106
add a comment |
add a comment |
Hint:
It suffices to check that $f$ is surjective. For any matrix $B$ we have
$$fleft(-frac13B + frac23B^Tright) = B$$
It you are wondering how we got this, assume $A+2A^T = B$. Then $A^T + 2A = B^T$ so $$B + B^T = 3(A+A^T)$$
Now it follows $$A = B^T - (A+A^T) = B^T - frac13(B + B^T) = -frac13B + frac23B^T$$
This seems to already be contained in the question post.
– Misha Lavrov
Jan 4 at 21:30
@MishaLavrov Oops, I missed it. Still, here it is done slightly cleaner, without coordinates.
– mechanodroid
Jan 4 at 21:31
add a comment |
Hint:
It suffices to check that $f$ is surjective. For any matrix $B$ we have
$$fleft(-frac13B + frac23B^Tright) = B$$
It you are wondering how we got this, assume $A+2A^T = B$. Then $A^T + 2A = B^T$ so $$B + B^T = 3(A+A^T)$$
Now it follows $$A = B^T - (A+A^T) = B^T - frac13(B + B^T) = -frac13B + frac23B^T$$
This seems to already be contained in the question post.
– Misha Lavrov
Jan 4 at 21:30
@MishaLavrov Oops, I missed it. Still, here it is done slightly cleaner, without coordinates.
– mechanodroid
Jan 4 at 21:31
add a comment |
Hint:
It suffices to check that $f$ is surjective. For any matrix $B$ we have
$$fleft(-frac13B + frac23B^Tright) = B$$
It you are wondering how we got this, assume $A+2A^T = B$. Then $A^T + 2A = B^T$ so $$B + B^T = 3(A+A^T)$$
Now it follows $$A = B^T - (A+A^T) = B^T - frac13(B + B^T) = -frac13B + frac23B^T$$
Hint:
It suffices to check that $f$ is surjective. For any matrix $B$ we have
$$fleft(-frac13B + frac23B^Tright) = B$$
It you are wondering how we got this, assume $A+2A^T = B$. Then $A^T + 2A = B^T$ so $$B + B^T = 3(A+A^T)$$
Now it follows $$A = B^T - (A+A^T) = B^T - frac13(B + B^T) = -frac13B + frac23B^T$$
answered Jan 4 at 21:28
mechanodroidmechanodroid
27k62446
27k62446
This seems to already be contained in the question post.
– Misha Lavrov
Jan 4 at 21:30
@MishaLavrov Oops, I missed it. Still, here it is done slightly cleaner, without coordinates.
– mechanodroid
Jan 4 at 21:31
add a comment |
This seems to already be contained in the question post.
– Misha Lavrov
Jan 4 at 21:30
@MishaLavrov Oops, I missed it. Still, here it is done slightly cleaner, without coordinates.
– mechanodroid
Jan 4 at 21:31
This seems to already be contained in the question post.
– Misha Lavrov
Jan 4 at 21:30
This seems to already be contained in the question post.
– Misha Lavrov
Jan 4 at 21:30
@MishaLavrov Oops, I missed it. Still, here it is done slightly cleaner, without coordinates.
– mechanodroid
Jan 4 at 21:31
@MishaLavrov Oops, I missed it. Still, here it is done slightly cleaner, without coordinates.
– mechanodroid
Jan 4 at 21:31
add a comment |
When you say
$$
a_{ij}=frac{2q_{ji}-q_{ij}}{3}
$$
(I fixed the indices) is “determined unambiguously”, you have proved both surjectivity and injectivity. There is at most one matrix $mathbf{A}$ such that $f(mathbf{A})=mathbf{Q}$. And there is one, determined in the way you did (after fixing the small mistake).
On the other hand, a linear map $Lcolon Vto V$, where $V$ is a finite dimensional vector space, is injective if and only if it is surjective, because of the rank-nullity theorem. Therefore just one of the properties needs to be checked for.
You don't need to go to the level of matrix entries in order to prove surjectivity. Given $Q$ you want to find $A$ such that $f(A)=Q$, that is, $A+2A^T=Q$. If such matrix exists, then also
$$
A^T+2A=Q^T
$$
and so $2A^T+4A=2Q^T$; thus
$$
Q-2Q^T=A+2A^T-2A^T-4A=-3A
$$
so
$$
A=frac{1}{3}(2Q^{T}-Q)
$$
is the only possible one. Since
$$
fleft(frac{1}{3}(2Q^{T}-Q)right)=Q
$$
as it can be readily checked, you have surjectivity and injectivity.
Checking directly for injectivity is easier, though.
add a comment |
When you say
$$
a_{ij}=frac{2q_{ji}-q_{ij}}{3}
$$
(I fixed the indices) is “determined unambiguously”, you have proved both surjectivity and injectivity. There is at most one matrix $mathbf{A}$ such that $f(mathbf{A})=mathbf{Q}$. And there is one, determined in the way you did (after fixing the small mistake).
On the other hand, a linear map $Lcolon Vto V$, where $V$ is a finite dimensional vector space, is injective if and only if it is surjective, because of the rank-nullity theorem. Therefore just one of the properties needs to be checked for.
You don't need to go to the level of matrix entries in order to prove surjectivity. Given $Q$ you want to find $A$ such that $f(A)=Q$, that is, $A+2A^T=Q$. If such matrix exists, then also
$$
A^T+2A=Q^T
$$
and so $2A^T+4A=2Q^T$; thus
$$
Q-2Q^T=A+2A^T-2A^T-4A=-3A
$$
so
$$
A=frac{1}{3}(2Q^{T}-Q)
$$
is the only possible one. Since
$$
fleft(frac{1}{3}(2Q^{T}-Q)right)=Q
$$
as it can be readily checked, you have surjectivity and injectivity.
Checking directly for injectivity is easier, though.
add a comment |
When you say
$$
a_{ij}=frac{2q_{ji}-q_{ij}}{3}
$$
(I fixed the indices) is “determined unambiguously”, you have proved both surjectivity and injectivity. There is at most one matrix $mathbf{A}$ such that $f(mathbf{A})=mathbf{Q}$. And there is one, determined in the way you did (after fixing the small mistake).
On the other hand, a linear map $Lcolon Vto V$, where $V$ is a finite dimensional vector space, is injective if and only if it is surjective, because of the rank-nullity theorem. Therefore just one of the properties needs to be checked for.
You don't need to go to the level of matrix entries in order to prove surjectivity. Given $Q$ you want to find $A$ such that $f(A)=Q$, that is, $A+2A^T=Q$. If such matrix exists, then also
$$
A^T+2A=Q^T
$$
and so $2A^T+4A=2Q^T$; thus
$$
Q-2Q^T=A+2A^T-2A^T-4A=-3A
$$
so
$$
A=frac{1}{3}(2Q^{T}-Q)
$$
is the only possible one. Since
$$
fleft(frac{1}{3}(2Q^{T}-Q)right)=Q
$$
as it can be readily checked, you have surjectivity and injectivity.
Checking directly for injectivity is easier, though.
When you say
$$
a_{ij}=frac{2q_{ji}-q_{ij}}{3}
$$
(I fixed the indices) is “determined unambiguously”, you have proved both surjectivity and injectivity. There is at most one matrix $mathbf{A}$ such that $f(mathbf{A})=mathbf{Q}$. And there is one, determined in the way you did (after fixing the small mistake).
On the other hand, a linear map $Lcolon Vto V$, where $V$ is a finite dimensional vector space, is injective if and only if it is surjective, because of the rank-nullity theorem. Therefore just one of the properties needs to be checked for.
You don't need to go to the level of matrix entries in order to prove surjectivity. Given $Q$ you want to find $A$ such that $f(A)=Q$, that is, $A+2A^T=Q$. If such matrix exists, then also
$$
A^T+2A=Q^T
$$
and so $2A^T+4A=2Q^T$; thus
$$
Q-2Q^T=A+2A^T-2A^T-4A=-3A
$$
so
$$
A=frac{1}{3}(2Q^{T}-Q)
$$
is the only possible one. Since
$$
fleft(frac{1}{3}(2Q^{T}-Q)right)=Q
$$
as it can be readily checked, you have surjectivity and injectivity.
Checking directly for injectivity is easier, though.
answered Jan 4 at 21:31
egregegreg
179k1485202
179k1485202
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