Rank of block triangular matrixRank of a lower triangular block matrixThe rank of a block matrix as a function of the rank of its submatrices.Nonsingularity of a block matrixInverse of a certain block matrixWhat is the rank of this matrixRank of the product of two full rank matricesNull space and rank of block matricesAn inequality on the rank of a block matrixRank of a block lower triangular matrixProving a matrix rank equalityFind rank for this special block matrix
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Rank of block triangular matrix
Rank of a lower triangular block matrixThe rank of a block matrix as a function of the rank of its submatrices.Nonsingularity of a block matrixInverse of a certain block matrixWhat is the rank of this matrixRank of the product of two full rank matricesNull space and rank of block matricesAn inequality on the rank of a block matrixRank of a block lower triangular matrixProving a matrix rank equalityFind rank for this special block matrix
$begingroup$
suppose that $C$ is full column rank matrix,
can we say that the following equality is true:
$$operatornamerankbiggl(beginbmatrix
A & 0\B & C
endbmatrixbiggr)= operatornamerankbiggl(beginbmatrix
A \B
endbmatrixbiggr) +operatornamerank(C)$$
I have an idea about it:
$$beginbmatrix
A & 0\B & C
endbmatrix=beginbmatrix
A & 0\B & 0
endbmatrix+beginbmatrix
0 & 0\0 & C
endbmatrix$$.
Now using the inequality, $textrank(A+B)leq textrank(A)+textrank(B)$ results in
$$operatornamerankbiggl(beginbmatrix
A & 0\B & C
endbmatrixbiggr)leq operatornamerankbiggl(beginbmatrix
A \B
endbmatrixbiggr) +operatornamerank(C).$$
The equality holds whenever $Rleft(beginbmatrix
A \B
endbmatrixright) cap R(C)=0$ and $Cleft(beginbmatrix
A \B
endbmatrixright) cap C(C)=0$.
But how to conclude it in this case?!
linear-algebra matrices matrix-rank block-matrices
edited 14 hours ago
Rodrigo de Azevedo
13.1k41960
asked Sep 17 '18 at 12:43
im hsim hs
345
$endgroup$
add a comment |
$begingroup$
suppose that $C$ is full column rank matrix,
can we say that the following equality is true:
$$operatornamerankbiggl(beginbmatrix
A & 0\B & C
endbmatrixbiggr)= operatornamerankbiggl(beginbmatrix
A \B
endbmatrixbiggr) +operatornamerank(C)$$
I have an idea about it:
$$beginbmatrix
A & 0\B & C
endbmatrix=beginbmatrix
A & 0\B & 0
endbmatrix+beginbmatrix
0 & 0\0 & C
endbmatrix$$.
Now using the inequality, $textrank(A+B)leq textrank(A)+textrank(B)$ results in
$$operatornamerankbiggl(beginbmatrix
A & 0\B & C
endbmatrixbiggr)leq operatornamerankbiggl(beginbmatrix
A \B
endbmatrixbiggr) +operatornamerank(C).$$
The equality holds whenever $Rleft(beginbmatrix
A \B
endbmatrixright) cap R(C)=0$ and $Cleft(beginbmatrix
A \B
endbmatrixright) cap C(C)=0$.
But how to conclude it in this case?!
linear-algebra matrices matrix-rank block-matrices
edited 14 hours ago
Rodrigo de Azevedo
13.1k41960
asked Sep 17 '18 at 12:43
im hsim hs
345
$endgroup$
$begingroup$
See this
$endgroup$
– user376343
Sep 17 '18 at 14:55
add a comment |
$begingroup$
suppose that $C$ is full column rank matrix,
can we say that the following equality is true:
$$operatornamerankbiggl(beginbmatrix
A & 0\B & C
endbmatrixbiggr)= operatornamerankbiggl(beginbmatrix
A \B
endbmatrixbiggr) +operatornamerank(C)$$
I have an idea about it:
$$beginbmatrix
A & 0\B & C
endbmatrix=beginbmatrix
A & 0\B & 0
endbmatrix+beginbmatrix
0 & 0\0 & C
endbmatrix$$.
Now using the inequality, $textrank(A+B)leq textrank(A)+textrank(B)$ results in
$$operatornamerankbiggl(beginbmatrix
A & 0\B & C
endbmatrixbiggr)leq operatornamerankbiggl(beginbmatrix
A \B
endbmatrixbiggr) +operatornamerank(C).$$
The equality holds whenever $Rleft(beginbmatrix
A \B
endbmatrixright) cap R(C)=0$ and $Cleft(beginbmatrix
A \B
endbmatrixright) cap C(C)=0$.
But how to conclude it in this case?!
linear-algebra matrices matrix-rank block-matrices
edited 14 hours ago
Rodrigo de Azevedo
13.1k41960
asked Sep 17 '18 at 12:43
im hsim hs
345
$endgroup$
suppose that $C$ is full column rank matrix,
can we say that the following equality is true:
$$operatornamerankbiggl(beginbmatrix
A & 0\B & C
endbmatrixbiggr)= operatornamerankbiggl(beginbmatrix
A \B
endbmatrixbiggr) +operatornamerank(C)$$
I have an idea about it:
$$beginbmatrix
A & 0\B & C
endbmatrix=beginbmatrix
A & 0\B & 0
endbmatrix+beginbmatrix
0 & 0\0 & C
endbmatrix$$.
Now using the inequality, $textrank(A+B)leq textrank(A)+textrank(B)$ results in
$$operatornamerankbiggl(beginbmatrix
A & 0\B & C
endbmatrixbiggr)leq operatornamerankbiggl(beginbmatrix
A \B
endbmatrixbiggr) +operatornamerank(C).$$
The equality holds whenever $Rleft(beginbmatrix
A \B
endbmatrixright) cap R(C)=0$ and $Cleft(beginbmatrix
A \B
endbmatrixright) cap C(C)=0$.
But how to conclude it in this case?!
linear-algebra matrices matrix-rank block-matrices
linear-algebra matrices matrix-rank block-matrices
edited 14 hours ago
Rodrigo de Azevedo
13.1k41960
asked Sep 17 '18 at 12:43
im hsim hs
345
edited 14 hours ago
Rodrigo de Azevedo
13.1k41960
asked Sep 17 '18 at 12:43
im hsim hs
345
edited 14 hours ago
Rodrigo de Azevedo
13.1k41960
edited 14 hours ago
Rodrigo de Azevedo
13.1k41960
edited 14 hours ago
Rodrigo de Azevedo
13.1k41960
13.1k41960
asked Sep 17 '18 at 12:43
im hsim hs
345
asked Sep 17 '18 at 12:43
im hsim hs
345
asked Sep 17 '18 at 12:43
im hsim hs
345
345
$begingroup$
See this
$endgroup$
– user376343
Sep 17 '18 at 14:55
add a comment |
$begingroup$
See this
$endgroup$
– user376343
Sep 17 '18 at 14:55
$begingroup$
See this
$endgroup$
– user376343
Sep 17 '18 at 14:55
$begingroup$
See this
$endgroup$
– user376343
Sep 17 '18 at 14:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You cannot say the equality is true. For example,
$$mathrmrankleft(beginbmatrix0&0\1&1endbmatrixright)neqmathrmrankleft(beginbmatrix0\1endbmatrixright)+mathrmrankleft(beginbmatrix0\1endbmatrixright)$$
answered Sep 17 '18 at 12:45
5xum5xum
91.3k394161
$endgroup$
$begingroup$
Yes, You are right. But under what conditions we can say that?!
$endgroup$
– im hs
Sep 17 '18 at 12:54
$begingroup$
It should be true since I see this in many papers and books! but without any proof and refer! it seems that it should be trivial!!
$endgroup$
– im hs
Sep 17 '18 at 12:54
$begingroup$
and how about if $A$ is invertible?
$endgroup$
– im hs
Sep 17 '18 at 12:56
$begingroup$
@imhs please refer us to an example of a paper or book saying such a thing. "should be true" says nothing if you are given a counter example.
$endgroup$
– GuySa
Sep 17 '18 at 13:30
$begingroup$
@GuySa please look at this for example: robust model-based fault diagnosis for dynamic systems by Chen and Ron Patton on page 75.
$endgroup$
– im hs
Sep 17 '18 at 16:57
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You cannot say the equality is true. For example,
$$mathrmrankleft(beginbmatrix0&0\1&1endbmatrixright)neqmathrmrankleft(beginbmatrix0\1endbmatrixright)+mathrmrankleft(beginbmatrix0\1endbmatrixright)$$
answered Sep 17 '18 at 12:45
5xum5xum
91.3k394161
$endgroup$
$begingroup$
Yes, You are right. But under what conditions we can say that?!
$endgroup$
– im hs
Sep 17 '18 at 12:54
$begingroup$
It should be true since I see this in many papers and books! but without any proof and refer! it seems that it should be trivial!!
$endgroup$
– im hs
Sep 17 '18 at 12:54
$begingroup$
and how about if $A$ is invertible?
$endgroup$
– im hs
Sep 17 '18 at 12:56
$begingroup$
@imhs please refer us to an example of a paper or book saying such a thing. "should be true" says nothing if you are given a counter example.
$endgroup$
– GuySa
Sep 17 '18 at 13:30
$begingroup$
@GuySa please look at this for example: robust model-based fault diagnosis for dynamic systems by Chen and Ron Patton on page 75.
$endgroup$
– im hs
Sep 17 '18 at 16:57
add a comment |
$begingroup$
You cannot say the equality is true. For example,
$$mathrmrankleft(beginbmatrix0&0\1&1endbmatrixright)neqmathrmrankleft(beginbmatrix0\1endbmatrixright)+mathrmrankleft(beginbmatrix0\1endbmatrixright)$$
answered Sep 17 '18 at 12:45
5xum5xum
91.3k394161
$endgroup$
$begingroup$
Yes, You are right. But under what conditions we can say that?!
$endgroup$
– im hs
Sep 17 '18 at 12:54
$begingroup$
It should be true since I see this in many papers and books! but without any proof and refer! it seems that it should be trivial!!
$endgroup$
– im hs
Sep 17 '18 at 12:54
$begingroup$
and how about if $A$ is invertible?
$endgroup$
– im hs
Sep 17 '18 at 12:56
$begingroup$
@imhs please refer us to an example of a paper or book saying such a thing. "should be true" says nothing if you are given a counter example.
$endgroup$
– GuySa
Sep 17 '18 at 13:30
$begingroup$
@GuySa please look at this for example: robust model-based fault diagnosis for dynamic systems by Chen and Ron Patton on page 75.
$endgroup$
– im hs
Sep 17 '18 at 16:57
add a comment |
$begingroup$
You cannot say the equality is true. For example,
$$mathrmrankleft(beginbmatrix0&0\1&1endbmatrixright)neqmathrmrankleft(beginbmatrix0\1endbmatrixright)+mathrmrankleft(beginbmatrix0\1endbmatrixright)$$
answered Sep 17 '18 at 12:45
5xum5xum
91.3k394161
$endgroup$
You cannot say the equality is true. For example,
$$mathrmrankleft(beginbmatrix0&0\1&1endbmatrixright)neqmathrmrankleft(beginbmatrix0\1endbmatrixright)+mathrmrankleft(beginbmatrix0\1endbmatrixright)$$
answered Sep 17 '18 at 12:45
5xum5xum
91.3k394161
answered Sep 17 '18 at 12:45
5xum5xum
91.3k394161
answered Sep 17 '18 at 12:45
5xum5xum
91.3k394161
answered Sep 17 '18 at 12:45
5xum5xum
91.3k394161
91.3k394161
$begingroup$
Yes, You are right. But under what conditions we can say that?!
$endgroup$
– im hs
Sep 17 '18 at 12:54
$begingroup$
It should be true since I see this in many papers and books! but without any proof and refer! it seems that it should be trivial!!
$endgroup$
– im hs
Sep 17 '18 at 12:54
$begingroup$
and how about if $A$ is invertible?
$endgroup$
– im hs
Sep 17 '18 at 12:56
$begingroup$
@imhs please refer us to an example of a paper or book saying such a thing. "should be true" says nothing if you are given a counter example.
$endgroup$
– GuySa
Sep 17 '18 at 13:30
$begingroup$
@GuySa please look at this for example: robust model-based fault diagnosis for dynamic systems by Chen and Ron Patton on page 75.
$endgroup$
– im hs
Sep 17 '18 at 16:57
add a comment |
$begingroup$
Yes, You are right. But under what conditions we can say that?!
$endgroup$
– im hs
Sep 17 '18 at 12:54
$begingroup$
It should be true since I see this in many papers and books! but without any proof and refer! it seems that it should be trivial!!
$endgroup$
– im hs
Sep 17 '18 at 12:54
$begingroup$
and how about if $A$ is invertible?
$endgroup$
– im hs
Sep 17 '18 at 12:56
$begingroup$
@imhs please refer us to an example of a paper or book saying such a thing. "should be true" says nothing if you are given a counter example.
$endgroup$
– GuySa
Sep 17 '18 at 13:30
$begingroup$
@GuySa please look at this for example: robust model-based fault diagnosis for dynamic systems by Chen and Ron Patton on page 75.
$endgroup$
– im hs
Sep 17 '18 at 16:57
$begingroup$
Yes, You are right. But under what conditions we can say that?!
$endgroup$
– im hs
Sep 17 '18 at 12:54
$begingroup$
Yes, You are right. But under what conditions we can say that?!
$endgroup$
– im hs
Sep 17 '18 at 12:54
$begingroup$
It should be true since I see this in many papers and books! but without any proof and refer! it seems that it should be trivial!!
$endgroup$
– im hs
Sep 17 '18 at 12:54
$begingroup$
It should be true since I see this in many papers and books! but without any proof and refer! it seems that it should be trivial!!
$endgroup$
– im hs
Sep 17 '18 at 12:54
$begingroup$
and how about if $A$ is invertible?
$endgroup$
– im hs
Sep 17 '18 at 12:56
$begingroup$
and how about if $A$ is invertible?
$endgroup$
– im hs
Sep 17 '18 at 12:56
$begingroup$
@imhs please refer us to an example of a paper or book saying such a thing. "should be true" says nothing if you are given a counter example.
$endgroup$
– GuySa
Sep 17 '18 at 13:30
$begingroup$
@imhs please refer us to an example of a paper or book saying such a thing. "should be true" says nothing if you are given a counter example.
$endgroup$
– GuySa
Sep 17 '18 at 13:30
$begingroup$
@GuySa please look at this for example: robust model-based fault diagnosis for dynamic systems by Chen and Ron Patton on page 75.
$endgroup$
– im hs
Sep 17 '18 at 16:57
$begingroup$
@GuySa please look at this for example: robust model-based fault diagnosis for dynamic systems by Chen and Ron Patton on page 75.
$endgroup$
– im hs
Sep 17 '18 at 16:57
add a comment |
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$begingroup$
See this
$endgroup$
– user376343
Sep 17 '18 at 14:55