$AA^T = BB^T rightarrow A =BU $
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Is it true that $AA^T = BB^T$ always implies $A =BU $ for some unitary matrix $U$. Or there may exist other scenarios also? All matrices are real
and square.
matrices matrix-equations matrix-decomposition transpose
asked Sep 6 at 10:08
Shanks
17510
add a comment |Â
up vote
0
down vote
favorite
Is it true that $AA^T = BB^T$ always implies $A =BU $ for some unitary matrix $U$. Or there may exist other scenarios also? All matrices are real
and square.
matrices matrix-equations matrix-decomposition transpose
asked Sep 6 at 10:08
Shanks
17510
1
If $A$ is invertible, then $A^T$ is invertible as well and thus $A=BB^T(A^T)^-1=BB^T(A^-1)^T$.
– zzuussee
Sep 6 at 10:12
Can we make a statement on $A$ which is always true?
– Shanks
Sep 6 at 10:16
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is it true that $AA^T = BB^T$ always implies $A =BU $ for some unitary matrix $U$. Or there may exist other scenarios also? All matrices are real
and square.
matrices matrix-equations matrix-decomposition transpose
asked Sep 6 at 10:08
Shanks
17510
Is it true that $AA^T = BB^T$ always implies $A =BU $ for some unitary matrix $U$. Or there may exist other scenarios also? All matrices are real
and square.
matrices matrix-equations matrix-decomposition transpose
matrices matrix-equations matrix-decomposition transpose
asked Sep 6 at 10:08
Shanks
17510
asked Sep 6 at 10:08
Shanks
17510
asked Sep 6 at 10:08
Shanks
17510
asked Sep 6 at 10:08
Shanks
17510
asked Sep 6 at 10:08
Shanks
17510
17510
1
If $A$ is invertible, then $A^T$ is invertible as well and thus $A=BB^T(A^T)^-1=BB^T(A^-1)^T$.
– zzuussee
Sep 6 at 10:12
Can we make a statement on $A$ which is always true?
– Shanks
Sep 6 at 10:16
add a comment |Â
1
If $A$ is invertible, then $A^T$ is invertible as well and thus $A=BB^T(A^T)^-1=BB^T(A^-1)^T$.
– zzuussee
Sep 6 at 10:12
Can we make a statement on $A$ which is always true?
– Shanks
Sep 6 at 10:16
1
1
If $A$ is invertible, then $A^T$ is invertible as well and thus $A=BB^T(A^T)^-1=BB^T(A^-1)^T$.
– zzuussee
Sep 6 at 10:12
If $A$ is invertible, then $A^T$ is invertible as well and thus $A=BB^T(A^T)^-1=BB^T(A^-1)^T$.
– zzuussee
Sep 6 at 10:12
Can we make a statement on $A$ which is always true?
– Shanks
Sep 6 at 10:16
Can we make a statement on $A$ which is always true?
– Shanks
Sep 6 at 10:16
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
This should work in general, although I think it is overkill if $A$ and $B$ are invertible:
The entry at position $(i,j)$ of $AA^T$ is the scalar product of the $i$th row of $A$ with the $j$th row. So the equality of these matrices shows that these scalar products are always the same, both for the rows of $A$ and the rows of $B$. As the scalar product is bilinear, we can extend this result to sums and coefficients.
This gives us a linear map $$varphi : rs(A) to rs(B),$$
where $rs$ denotes the row space such that $varphi$ is compatible with the scalar product. Using that the zero vector is the only vector of length $0$, you can show that $varphi$ is bijective, i.e. an isometry.
Now using a real version of Witt's theorem, you can extend $varphi$ to an isometry $Phi$ on the whole space. But every isometry $Phi$ on the whole space is given by an orthogonal matrix (or a unitary matrix with real entries if you want).
answered Sep 6 at 10:29
Dirk Liebhold
4,151117
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
This should work in general, although I think it is overkill if $A$ and $B$ are invertible:
The entry at position $(i,j)$ of $AA^T$ is the scalar product of the $i$th row of $A$ with the $j$th row. So the equality of these matrices shows that these scalar products are always the same, both for the rows of $A$ and the rows of $B$. As the scalar product is bilinear, we can extend this result to sums and coefficients.
This gives us a linear map $$varphi : rs(A) to rs(B),$$
where $rs$ denotes the row space such that $varphi$ is compatible with the scalar product. Using that the zero vector is the only vector of length $0$, you can show that $varphi$ is bijective, i.e. an isometry.
Now using a real version of Witt's theorem, you can extend $varphi$ to an isometry $Phi$ on the whole space. But every isometry $Phi$ on the whole space is given by an orthogonal matrix (or a unitary matrix with real entries if you want).
answered Sep 6 at 10:29
Dirk Liebhold
4,151117
add a comment |Â
up vote
1
down vote
This should work in general, although I think it is overkill if $A$ and $B$ are invertible:
The entry at position $(i,j)$ of $AA^T$ is the scalar product of the $i$th row of $A$ with the $j$th row. So the equality of these matrices shows that these scalar products are always the same, both for the rows of $A$ and the rows of $B$. As the scalar product is bilinear, we can extend this result to sums and coefficients.
This gives us a linear map $$varphi : rs(A) to rs(B),$$
where $rs$ denotes the row space such that $varphi$ is compatible with the scalar product. Using that the zero vector is the only vector of length $0$, you can show that $varphi$ is bijective, i.e. an isometry.
Now using a real version of Witt's theorem, you can extend $varphi$ to an isometry $Phi$ on the whole space. But every isometry $Phi$ on the whole space is given by an orthogonal matrix (or a unitary matrix with real entries if you want).
answered Sep 6 at 10:29
Dirk Liebhold
4,151117
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This should work in general, although I think it is overkill if $A$ and $B$ are invertible:
The entry at position $(i,j)$ of $AA^T$ is the scalar product of the $i$th row of $A$ with the $j$th row. So the equality of these matrices shows that these scalar products are always the same, both for the rows of $A$ and the rows of $B$. As the scalar product is bilinear, we can extend this result to sums and coefficients.
This gives us a linear map $$varphi : rs(A) to rs(B),$$
where $rs$ denotes the row space such that $varphi$ is compatible with the scalar product. Using that the zero vector is the only vector of length $0$, you can show that $varphi$ is bijective, i.e. an isometry.
Now using a real version of Witt's theorem, you can extend $varphi$ to an isometry $Phi$ on the whole space. But every isometry $Phi$ on the whole space is given by an orthogonal matrix (or a unitary matrix with real entries if you want).
answered Sep 6 at 10:29
Dirk Liebhold
4,151117
This should work in general, although I think it is overkill if $A$ and $B$ are invertible:
The entry at position $(i,j)$ of $AA^T$ is the scalar product of the $i$th row of $A$ with the $j$th row. So the equality of these matrices shows that these scalar products are always the same, both for the rows of $A$ and the rows of $B$. As the scalar product is bilinear, we can extend this result to sums and coefficients.
This gives us a linear map $$varphi : rs(A) to rs(B),$$
where $rs$ denotes the row space such that $varphi$ is compatible with the scalar product. Using that the zero vector is the only vector of length $0$, you can show that $varphi$ is bijective, i.e. an isometry.
Now using a real version of Witt's theorem, you can extend $varphi$ to an isometry $Phi$ on the whole space. But every isometry $Phi$ on the whole space is given by an orthogonal matrix (or a unitary matrix with real entries if you want).
answered Sep 6 at 10:29
Dirk Liebhold
4,151117
answered Sep 6 at 10:29
Dirk Liebhold
4,151117
answered Sep 6 at 10:29
Dirk Liebhold
4,151117
answered Sep 6 at 10:29
Dirk Liebhold
4,151117
4,151117
add a comment |Â
add a comment |Â
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1
If $A$ is invertible, then $A^T$ is invertible as well and thus $A=BB^T(A^T)^-1=BB^T(A^-1)^T$.
– zzuussee
Sep 6 at 10:12
Can we make a statement on $A$ which is always true?
– Shanks
Sep 6 at 10:16