Clarification about this step in matrix proof?
$begingroup$
Suppose $A$ is a matrix of rank $r$ of order $n times m$ and suppose it can be partitioned into pieces $A_{11},A_{12},A_{21},A_{22}$, where the leading minor $A_{11}$ is of size $r times r$ and is non-singular.
In a proof, my textbook uses the following property:
If $A_{11}^-$ is the inverse of $A_{11}$, then by the construction there is a matrix $B$ such that $[A_{21}, A_{22}] = B [ A_{11}, A_{12}]$
It doesn't elaborate/explain, and I'm unsure how exactly this is justified.
Intuitively, I suppose it is because $A$ is of rank $r$, and the sub-matrix $A_{11}$ is also of rank $r$. Could someone explain this step to me?
linear-algebra
asked Jan 14 at 9:20
XiaomiXiaomi
1,071115
$endgroup$
add a comment |
$begingroup$
Suppose $A$ is a matrix of rank $r$ of order $n times m$ and suppose it can be partitioned into pieces $A_{11},A_{12},A_{21},A_{22}$, where the leading minor $A_{11}$ is of size $r times r$ and is non-singular.
In a proof, my textbook uses the following property:
If $A_{11}^-$ is the inverse of $A_{11}$, then by the construction there is a matrix $B$ such that $[A_{21}, A_{22}] = B [ A_{11}, A_{12}]$
It doesn't elaborate/explain, and I'm unsure how exactly this is justified.
Intuitively, I suppose it is because $A$ is of rank $r$, and the sub-matrix $A_{11}$ is also of rank $r$. Could someone explain this step to me?
linear-algebra
asked Jan 14 at 9:20
XiaomiXiaomi
1,071115
$endgroup$
$begingroup$
If it was not the case, either $A_{11}$ would not be full rank or the rank of $A$ would be strictly more than $r$.
$endgroup$
– P. Quinton
Jan 14 at 9:43
add a comment |
$begingroup$
Suppose $A$ is a matrix of rank $r$ of order $n times m$ and suppose it can be partitioned into pieces $A_{11},A_{12},A_{21},A_{22}$, where the leading minor $A_{11}$ is of size $r times r$ and is non-singular.
In a proof, my textbook uses the following property:
If $A_{11}^-$ is the inverse of $A_{11}$, then by the construction there is a matrix $B$ such that $[A_{21}, A_{22}] = B [ A_{11}, A_{12}]$
It doesn't elaborate/explain, and I'm unsure how exactly this is justified.
Intuitively, I suppose it is because $A$ is of rank $r$, and the sub-matrix $A_{11}$ is also of rank $r$. Could someone explain this step to me?
linear-algebra
asked Jan 14 at 9:20
XiaomiXiaomi
1,071115
$endgroup$
Suppose $A$ is a matrix of rank $r$ of order $n times m$ and suppose it can be partitioned into pieces $A_{11},A_{12},A_{21},A_{22}$, where the leading minor $A_{11}$ is of size $r times r$ and is non-singular.
In a proof, my textbook uses the following property:
If $A_{11}^-$ is the inverse of $A_{11}$, then by the construction there is a matrix $B$ such that $[A_{21}, A_{22}] = B [ A_{11}, A_{12}]$
It doesn't elaborate/explain, and I'm unsure how exactly this is justified.
Intuitively, I suppose it is because $A$ is of rank $r$, and the sub-matrix $A_{11}$ is also of rank $r$. Could someone explain this step to me?
linear-algebra
linear-algebra
asked Jan 14 at 9:20
XiaomiXiaomi
1,071115
asked Jan 14 at 9:20
XiaomiXiaomi
1,071115
asked Jan 14 at 9:20
XiaomiXiaomi
1,071115
asked Jan 14 at 9:20
XiaomiXiaomi
1,071115
asked Jan 14 at 9:20
XiaomiXiaomi
1,071115
1,071115
$begingroup$
If it was not the case, either $A_{11}$ would not be full rank or the rank of $A$ would be strictly more than $r$.
$endgroup$
– P. Quinton
Jan 14 at 9:43
add a comment |
$begingroup$
If it was not the case, either $A_{11}$ would not be full rank or the rank of $A$ would be strictly more than $r$.
$endgroup$
– P. Quinton
Jan 14 at 9:43
$begingroup$
If it was not the case, either $A_{11}$ would not be full rank or the rank of $A$ would be strictly more than $r$.
$endgroup$
– P. Quinton
Jan 14 at 9:43
$begingroup$
If it was not the case, either $A_{11}$ would not be full rank or the rank of $A$ would be strictly more than $r$.
$endgroup$
– P. Quinton
Jan 14 at 9:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The matrix $A$ is split into two rows $[A_{11}, A_{12}]$ (top) and $[A_{21}, A_{22}]$ (bottom). Since the total rank is $r$ and the top row already has rank $r$, each of the bottom rows is a linear combination of some of the first $r$ rows. Denoting $A_j$ for the $j$th row of $A$, this means for each $i > r$ there are coefficients $b_{ij}$ such that
$$A_i = sum_{1leq j leq r}b_{ij}A_j$$
These coefficients populate a $(n-r)times r$ matrix $B = (b_{ij})$, with $BA_{11} = A_{21}$ and $BA_{12} = A_{22}$.
answered Jan 14 at 9:51
BenBen
4,283617
$endgroup$
$begingroup$
thanks a lot, that makes a lot of sense
$endgroup$
– Xiaomi
Jan 14 at 10:34
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The matrix $A$ is split into two rows $[A_{11}, A_{12}]$ (top) and $[A_{21}, A_{22}]$ (bottom). Since the total rank is $r$ and the top row already has rank $r$, each of the bottom rows is a linear combination of some of the first $r$ rows. Denoting $A_j$ for the $j$th row of $A$, this means for each $i > r$ there are coefficients $b_{ij}$ such that
$$A_i = sum_{1leq j leq r}b_{ij}A_j$$
These coefficients populate a $(n-r)times r$ matrix $B = (b_{ij})$, with $BA_{11} = A_{21}$ and $BA_{12} = A_{22}$.
answered Jan 14 at 9:51
BenBen
4,283617
$endgroup$
$begingroup$
thanks a lot, that makes a lot of sense
$endgroup$
– Xiaomi
Jan 14 at 10:34
add a comment |
$begingroup$
The matrix $A$ is split into two rows $[A_{11}, A_{12}]$ (top) and $[A_{21}, A_{22}]$ (bottom). Since the total rank is $r$ and the top row already has rank $r$, each of the bottom rows is a linear combination of some of the first $r$ rows. Denoting $A_j$ for the $j$th row of $A$, this means for each $i > r$ there are coefficients $b_{ij}$ such that
$$A_i = sum_{1leq j leq r}b_{ij}A_j$$
These coefficients populate a $(n-r)times r$ matrix $B = (b_{ij})$, with $BA_{11} = A_{21}$ and $BA_{12} = A_{22}$.
answered Jan 14 at 9:51
BenBen
4,283617
$endgroup$
$begingroup$
thanks a lot, that makes a lot of sense
$endgroup$
– Xiaomi
Jan 14 at 10:34
add a comment |
$begingroup$
The matrix $A$ is split into two rows $[A_{11}, A_{12}]$ (top) and $[A_{21}, A_{22}]$ (bottom). Since the total rank is $r$ and the top row already has rank $r$, each of the bottom rows is a linear combination of some of the first $r$ rows. Denoting $A_j$ for the $j$th row of $A$, this means for each $i > r$ there are coefficients $b_{ij}$ such that
$$A_i = sum_{1leq j leq r}b_{ij}A_j$$
These coefficients populate a $(n-r)times r$ matrix $B = (b_{ij})$, with $BA_{11} = A_{21}$ and $BA_{12} = A_{22}$.
answered Jan 14 at 9:51
BenBen
4,283617
$endgroup$
The matrix $A$ is split into two rows $[A_{11}, A_{12}]$ (top) and $[A_{21}, A_{22}]$ (bottom). Since the total rank is $r$ and the top row already has rank $r$, each of the bottom rows is a linear combination of some of the first $r$ rows. Denoting $A_j$ for the $j$th row of $A$, this means for each $i > r$ there are coefficients $b_{ij}$ such that
$$A_i = sum_{1leq j leq r}b_{ij}A_j$$
These coefficients populate a $(n-r)times r$ matrix $B = (b_{ij})$, with $BA_{11} = A_{21}$ and $BA_{12} = A_{22}$.
answered Jan 14 at 9:51
BenBen
4,283617
answered Jan 14 at 9:51
BenBen
4,283617
answered Jan 14 at 9:51
BenBen
4,283617
answered Jan 14 at 9:51
BenBen
4,283617
4,283617
$begingroup$
thanks a lot, that makes a lot of sense
$endgroup$
– Xiaomi
Jan 14 at 10:34
add a comment |
$begingroup$
thanks a lot, that makes a lot of sense
$endgroup$
– Xiaomi
Jan 14 at 10:34
$begingroup$
thanks a lot, that makes a lot of sense
$endgroup$
– Xiaomi
Jan 14 at 10:34
$begingroup$
thanks a lot, that makes a lot of sense
$endgroup$
– Xiaomi
Jan 14 at 10:34
add a comment |
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$begingroup$
If it was not the case, either $A_{11}$ would not be full rank or the rank of $A$ would be strictly more than $r$.
$endgroup$
– P. Quinton
Jan 14 at 9:43