Mixing
The congruence of two partitioned matrices
$begingroup$
Two $(m+n)times (m+n)$ real symmetric matrices $X$ and $Y$ can be partitioned as follows:
$X=left( begin{matrix} A_{ntimes n} & O_{ntimes m} \ O_{mtimes n} & C_{mtimes m} end{matrix} right)$
and
$Y=left( begin{matrix} B_{ntimes n} & O_{ntimes m} \ O_{mtimes n} & D_{mtimes m} end{matrix} right)$.
I am wondering whether $C$ is congruent to $D$ if we know $X$ is congruent to $Y$, and $A$ is congruent to $B$. Many thanks.
linear-algebra matrices
asked Jan 9 at 13:47
Kangping YanKangping Yan
1767
$endgroup$
add a comment |
$begingroup$
Two $(m+n)times (m+n)$ real symmetric matrices $X$ and $Y$ can be partitioned as follows:
$X=left( begin{matrix} A_{ntimes n} & O_{ntimes m} \ O_{mtimes n} & C_{mtimes m} end{matrix} right)$
and
$Y=left( begin{matrix} B_{ntimes n} & O_{ntimes m} \ O_{mtimes n} & D_{mtimes m} end{matrix} right)$.
I am wondering whether $C$ is congruent to $D$ if we know $X$ is congruent to $Y$, and $A$ is congruent to $B$. Many thanks.
linear-algebra matrices
asked Jan 9 at 13:47
Kangping YanKangping Yan
1767
$endgroup$
$begingroup$
Do you mean congruent in the sense defined here?
$endgroup$
– Omnomnomnom
Jan 9 at 15:43
add a comment |
$begingroup$
Two $(m+n)times (m+n)$ real symmetric matrices $X$ and $Y$ can be partitioned as follows:
$X=left( begin{matrix} A_{ntimes n} & O_{ntimes m} \ O_{mtimes n} & C_{mtimes m} end{matrix} right)$
and
$Y=left( begin{matrix} B_{ntimes n} & O_{ntimes m} \ O_{mtimes n} & D_{mtimes m} end{matrix} right)$.
I am wondering whether $C$ is congruent to $D$ if we know $X$ is congruent to $Y$, and $A$ is congruent to $B$. Many thanks.
linear-algebra matrices
asked Jan 9 at 13:47
Kangping YanKangping Yan
1767
$endgroup$
Two $(m+n)times (m+n)$ real symmetric matrices $X$ and $Y$ can be partitioned as follows:
$X=left( begin{matrix} A_{ntimes n} & O_{ntimes m} \ O_{mtimes n} & C_{mtimes m} end{matrix} right)$
and
$Y=left( begin{matrix} B_{ntimes n} & O_{ntimes m} \ O_{mtimes n} & D_{mtimes m} end{matrix} right)$.
I am wondering whether $C$ is congruent to $D$ if we know $X$ is congruent to $Y$, and $A$ is congruent to $B$. Many thanks.
linear-algebra matrices
linear-algebra matrices
asked Jan 9 at 13:47
Kangping YanKangping Yan
1767
asked Jan 9 at 13:47
Kangping YanKangping Yan
1767
edited Jan 10 at 1:19
Kangping Yan
asked Jan 9 at 13:47
Kangping YanKangping Yan
1767
asked Jan 9 at 13:47
Kangping YanKangping Yan
1767
asked Jan 9 at 13:47
Kangping YanKangping Yan
1767
1767
$begingroup$
Do you mean congruent in the sense defined here?
$endgroup$
– Omnomnomnom
Jan 9 at 15:43
add a comment |
$begingroup$
Do you mean congruent in the sense defined here?
$endgroup$
– Omnomnomnom
Jan 9 at 15:43
$begingroup$
Do you mean congruent in the sense defined here?
$endgroup$
– Omnomnomnom
Jan 9 at 15:43
$begingroup$
Do you mean congruent in the sense defined here?
$endgroup$
– Omnomnomnom
Jan 9 at 15:43
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The answer to your question is yes, and it can be seen as a straight-forward consequence of Sylvester's law of inertia.
A direct proof might look like this: let $P,Q$ be such that $P^TXP = Y$ and $Q^TAQ = B$. It follows that
$$
left[P pmatrix{Q^{-1} & 0\0 & I}right]^T pmatrix{A&0\0&C} left[P pmatrix{Q^{-1} & 0\0 & I}right] = pmatrix{A&0\0&D}
$$
That is, $operatorname{diag}(A,C)$ is congruent to $operatorname{diag}(A,D)$. From here, we would need to deduce that $C$ is congruent to $D$.
answered Jan 9 at 15:57
OmnomnomnomOmnomnomnom
127k790180
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer to your question is yes, and it can be seen as a straight-forward consequence of Sylvester's law of inertia.
A direct proof might look like this: let $P,Q$ be such that $P^TXP = Y$ and $Q^TAQ = B$. It follows that
$$
left[P pmatrix{Q^{-1} & 0\0 & I}right]^T pmatrix{A&0\0&C} left[P pmatrix{Q^{-1} & 0\0 & I}right] = pmatrix{A&0\0&D}
$$
That is, $operatorname{diag}(A,C)$ is congruent to $operatorname{diag}(A,D)$. From here, we would need to deduce that $C$ is congruent to $D$.
answered Jan 9 at 15:57
OmnomnomnomOmnomnomnom
127k790180
$endgroup$
add a comment |
$begingroup$
The answer to your question is yes, and it can be seen as a straight-forward consequence of Sylvester's law of inertia.
A direct proof might look like this: let $P,Q$ be such that $P^TXP = Y$ and $Q^TAQ = B$. It follows that
$$
left[P pmatrix{Q^{-1} & 0\0 & I}right]^T pmatrix{A&0\0&C} left[P pmatrix{Q^{-1} & 0\0 & I}right] = pmatrix{A&0\0&D}
$$
That is, $operatorname{diag}(A,C)$ is congruent to $operatorname{diag}(A,D)$. From here, we would need to deduce that $C$ is congruent to $D$.
answered Jan 9 at 15:57
OmnomnomnomOmnomnomnom
127k790180
$endgroup$
add a comment |
$begingroup$
The answer to your question is yes, and it can be seen as a straight-forward consequence of Sylvester's law of inertia.
A direct proof might look like this: let $P,Q$ be such that $P^TXP = Y$ and $Q^TAQ = B$. It follows that
$$
left[P pmatrix{Q^{-1} & 0\0 & I}right]^T pmatrix{A&0\0&C} left[P pmatrix{Q^{-1} & 0\0 & I}right] = pmatrix{A&0\0&D}
$$
That is, $operatorname{diag}(A,C)$ is congruent to $operatorname{diag}(A,D)$. From here, we would need to deduce that $C$ is congruent to $D$.
answered Jan 9 at 15:57
OmnomnomnomOmnomnomnom
127k790180
$endgroup$
The answer to your question is yes, and it can be seen as a straight-forward consequence of Sylvester's law of inertia.
A direct proof might look like this: let $P,Q$ be such that $P^TXP = Y$ and $Q^TAQ = B$. It follows that
$$
left[P pmatrix{Q^{-1} & 0\0 & I}right]^T pmatrix{A&0\0&C} left[P pmatrix{Q^{-1} & 0\0 & I}right] = pmatrix{A&0\0&D}
$$
That is, $operatorname{diag}(A,C)$ is congruent to $operatorname{diag}(A,D)$. From here, we would need to deduce that $C$ is congruent to $D$.
answered Jan 9 at 15:57
OmnomnomnomOmnomnomnom
127k790180
answered Jan 9 at 15:57
OmnomnomnomOmnomnomnom
127k790180
answered Jan 9 at 15:57
OmnomnomnomOmnomnomnom
127k790180
answered Jan 9 at 15:57
OmnomnomnomOmnomnomnom
127k790180
127k790180
add a comment |
add a comment |
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$begingroup$
Do you mean congruent in the sense defined here?
$endgroup$
– Omnomnomnom
Jan 9 at 15:43