Prove rank(BAC)=rank(BA)=rank(AC)=rank(A)How to prove and interpret $operatornamerank(AB) leq operatornamemin(operatornamerank(A), operatornamerank(B))$?Prove or disprove that Y = AX-CRank of a matrix when adding new columnsProof concerning matrix composition.Rank of $I_m - X_m times m$ given rank of $X$Rank one decomposition or elementary tensor decomposition of matrices over commutative ringsVisualizing transpose of matricesMatrix proof on the zero matrixShowing Orthogonal Projection Matrix Multiplied by Full-Rank Matrices is Positive-DefiniteSize of a Maximal Non-Zero Minor in a Linear Map equals RankRank, nullity and consistency for two matrices
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Prove rank(BAC)=rank(BA)=rank(AC)=rank(A)
How to prove and interpret $operatornamerank(AB) leq operatornamemin(operatornamerank(A), operatornamerank(B))$?Prove or disprove that Y = AX-CRank of a matrix when adding new columnsProof concerning matrix composition.Rank of $I_m - X_m times m$ given rank of $X$Rank one decomposition or elementary tensor decomposition of matrices over commutative ringsVisualizing transpose of matricesMatrix proof on the zero matrixShowing Orthogonal Projection Matrix Multiplied by Full-Rank Matrices is Positive-DefiniteSize of a Maximal Non-Zero Minor in a Linear Map equals RankRank, nullity and consistency for two matrices
$begingroup$
Would anyone know how to prove the following?
It is stated as a theorem in the textbook without further explanations.
Let $A$ be an $m times n$ matrix, $B$ an $mtimes m$ matrix, and $C$ a $ntimes n$ matrix. Then if $B$ and $C$ are nonsingular matrices, it follows that:
$$ rank(BAC)=rank(BA)=rank(AC)=rank(A)$$
I have searched for other similar questions but the proofs seem to all rely on some notion of fields, and dim(), but the textbook has yet to touch on such concepts at this point.
Your help would be greatly appreciated.
linear-algebra matrices matrix-rank
edited Mar 10 at 17:29
Rodrigo de Azevedo
13k41960
asked May 17 '16 at 8:24
KenKen
11
$endgroup$
add a comment |
$begingroup$
Would anyone know how to prove the following?
It is stated as a theorem in the textbook without further explanations.
Let $A$ be an $m times n$ matrix, $B$ an $mtimes m$ matrix, and $C$ a $ntimes n$ matrix. Then if $B$ and $C$ are nonsingular matrices, it follows that:
$$ rank(BAC)=rank(BA)=rank(AC)=rank(A)$$
I have searched for other similar questions but the proofs seem to all rely on some notion of fields, and dim(), but the textbook has yet to touch on such concepts at this point.
Your help would be greatly appreciated.
linear-algebra matrices matrix-rank
edited Mar 10 at 17:29
Rodrigo de Azevedo
13k41960
asked May 17 '16 at 8:24
KenKen
11
$endgroup$
1
$begingroup$
Use $textrank(B^-1BA) leq textrank(BA) leq textrank(A)$
$endgroup$
– Anon
May 17 '16 at 8:28
$begingroup$
How is rank defined if the textbook has not introduced dimension yet?
$endgroup$
– David
May 17 '16 at 8:54
add a comment |
$begingroup$
Would anyone know how to prove the following?
It is stated as a theorem in the textbook without further explanations.
Let $A$ be an $m times n$ matrix, $B$ an $mtimes m$ matrix, and $C$ a $ntimes n$ matrix. Then if $B$ and $C$ are nonsingular matrices, it follows that:
$$ rank(BAC)=rank(BA)=rank(AC)=rank(A)$$
I have searched for other similar questions but the proofs seem to all rely on some notion of fields, and dim(), but the textbook has yet to touch on such concepts at this point.
Your help would be greatly appreciated.
linear-algebra matrices matrix-rank
edited Mar 10 at 17:29
Rodrigo de Azevedo
13k41960
asked May 17 '16 at 8:24
KenKen
11
$endgroup$
Would anyone know how to prove the following?
It is stated as a theorem in the textbook without further explanations.
Let $A$ be an $m times n$ matrix, $B$ an $mtimes m$ matrix, and $C$ a $ntimes n$ matrix. Then if $B$ and $C$ are nonsingular matrices, it follows that:
$$ rank(BAC)=rank(BA)=rank(AC)=rank(A)$$
I have searched for other similar questions but the proofs seem to all rely on some notion of fields, and dim(), but the textbook has yet to touch on such concepts at this point.
Your help would be greatly appreciated.
linear-algebra matrices matrix-rank
linear-algebra matrices matrix-rank
edited Mar 10 at 17:29
Rodrigo de Azevedo
13k41960
asked May 17 '16 at 8:24
KenKen
11
edited Mar 10 at 17:29
Rodrigo de Azevedo
13k41960
asked May 17 '16 at 8:24
KenKen
11
edited Mar 10 at 17:29
Rodrigo de Azevedo
13k41960
edited Mar 10 at 17:29
Rodrigo de Azevedo
13k41960
edited Mar 10 at 17:29
Rodrigo de Azevedo
13k41960
13k41960
asked May 17 '16 at 8:24
KenKen
11
asked May 17 '16 at 8:24
KenKen
11
asked May 17 '16 at 8:24
KenKen
11
11
1
$begingroup$
Use $textrank(B^-1BA) leq textrank(BA) leq textrank(A)$
$endgroup$
– Anon
May 17 '16 at 8:28
$begingroup$
How is rank defined if the textbook has not introduced dimension yet?
$endgroup$
– David
May 17 '16 at 8:54
add a comment |
1
$begingroup$
Use $textrank(B^-1BA) leq textrank(BA) leq textrank(A)$
$endgroup$
– Anon
May 17 '16 at 8:28
$begingroup$
How is rank defined if the textbook has not introduced dimension yet?
$endgroup$
– David
May 17 '16 at 8:54
1
1
$begingroup$
Use $textrank(B^-1BA) leq textrank(BA) leq textrank(A)$
$endgroup$
– Anon
May 17 '16 at 8:28
$begingroup$
Use $textrank(B^-1BA) leq textrank(BA) leq textrank(A)$
$endgroup$
– Anon
May 17 '16 at 8:28
$begingroup$
How is rank defined if the textbook has not introduced dimension yet?
$endgroup$
– David
May 17 '16 at 8:54
$begingroup$
How is rank defined if the textbook has not introduced dimension yet?
$endgroup$
– David
May 17 '16 at 8:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
HINT
It is a known fact: $textRank(AB)leq min(textRank(A), textRank(B))tag 1$
Using (1) you can prove $textRank(AB)= textRank(A)$ if $B$ nonsingular.
See this and this for proofs.
edited Apr 13 '17 at 12:20
Community♦
1
$endgroup$
add a comment |
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$begingroup$
HINT
It is a known fact: $textRank(AB)leq min(textRank(A), textRank(B))tag 1$
Using (1) you can prove $textRank(AB)= textRank(A)$ if $B$ nonsingular.
See this and this for proofs.
edited Apr 13 '17 at 12:20
Community♦
1
$endgroup$
add a comment |
$begingroup$
HINT
It is a known fact: $textRank(AB)leq min(textRank(A), textRank(B))tag 1$
Using (1) you can prove $textRank(AB)= textRank(A)$ if $B$ nonsingular.
See this and this for proofs.
edited Apr 13 '17 at 12:20
Community♦
1
$endgroup$
add a comment |
$begingroup$
HINT
It is a known fact: $textRank(AB)leq min(textRank(A), textRank(B))tag 1$
Using (1) you can prove $textRank(AB)= textRank(A)$ if $B$ nonsingular.
See this and this for proofs.
edited Apr 13 '17 at 12:20
Community♦
1
$endgroup$
HINT
It is a known fact: $textRank(AB)leq min(textRank(A), textRank(B))tag 1$
Using (1) you can prove $textRank(AB)= textRank(A)$ if $B$ nonsingular.
See this and this for proofs.
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered May 17 '16 at 8:28
user261263
add a comment |
add a comment |
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1
$begingroup$
Use $textrank(B^-1BA) leq textrank(BA) leq textrank(A)$
$endgroup$
– Anon
May 17 '16 at 8:28
$begingroup$
How is rank defined if the textbook has not introduced dimension yet?
$endgroup$
– David
May 17 '16 at 8:54