Is $C(A)+C(B)=C(A+B)$, if $C(A)cap C(B)= {0},$?
$begingroup$
Is $C(A)+C(B)=C(A+B)$, if $C(A)cap C(B)= {0}?$ where $C(A)$ denotes the column space of $A$.
It's true that, $DeclareMathOperator{rank}{rank}rank(A+B)=rank(A)+rank(B),$ if $C(A)cap C(B)= {0}$ so, I guess that may be the above statement is true.So, I tried to prove it, but can't make a reasonable progress.
So, any hints or counterexample?
Edited: As pointed out the above statement is not true.Then, under what conditions the above statement is true?
linear-algebra
asked Jan 16 at 16:14
Tom.Tom.
16619
$endgroup$
add a comment |
$begingroup$
Is $C(A)+C(B)=C(A+B)$, if $C(A)cap C(B)= {0}?$ where $C(A)$ denotes the column space of $A$.
It's true that, $DeclareMathOperator{rank}{rank}rank(A+B)=rank(A)+rank(B),$ if $C(A)cap C(B)= {0}$ so, I guess that may be the above statement is true.So, I tried to prove it, but can't make a reasonable progress.
So, any hints or counterexample?
Edited: As pointed out the above statement is not true.Then, under what conditions the above statement is true?
linear-algebra
asked Jan 16 at 16:14
Tom.Tom.
16619
$endgroup$
$begingroup$
What is meant by $theta$ here? Did you mean $emptyset$? Also, your assertion that rank(A+B)=rank(A)+rank(B) is not true. Consider the fact that $0=A+(-A)$ for any matrix $A$ for an easy counterexample
$endgroup$
– pwerth
Jan 16 at 16:17
$begingroup$
@pwerth I think he means that $operatorname{rank}(A + B) = operatorname{rank}(A) + operatorname{rank}(B)$ if $C(A) cap C(B) = {0}$
$endgroup$
– Omnomnomnom
Jan 16 at 16:18
$begingroup$
Yeah!I edited that.
$endgroup$
– Tom.
Jan 16 at 16:20
1
$begingroup$
Even so, actually, the statement doesn't hold. For instance, with $$ A = pmatrix{1\0}, B = pmatrix{0\1} $$ the column spaces are complimentary but $A + B$ has rank $1$.
$endgroup$
– Omnomnomnom
Jan 16 at 16:21
add a comment |
$begingroup$
Is $C(A)+C(B)=C(A+B)$, if $C(A)cap C(B)= {0}?$ where $C(A)$ denotes the column space of $A$.
It's true that, $DeclareMathOperator{rank}{rank}rank(A+B)=rank(A)+rank(B),$ if $C(A)cap C(B)= {0}$ so, I guess that may be the above statement is true.So, I tried to prove it, but can't make a reasonable progress.
So, any hints or counterexample?
Edited: As pointed out the above statement is not true.Then, under what conditions the above statement is true?
linear-algebra
asked Jan 16 at 16:14
Tom.Tom.
16619
$endgroup$
Is $C(A)+C(B)=C(A+B)$, if $C(A)cap C(B)= {0}?$ where $C(A)$ denotes the column space of $A$.
It's true that, $DeclareMathOperator{rank}{rank}rank(A+B)=rank(A)+rank(B),$ if $C(A)cap C(B)= {0}$ so, I guess that may be the above statement is true.So, I tried to prove it, but can't make a reasonable progress.
So, any hints or counterexample?
Edited: As pointed out the above statement is not true.Then, under what conditions the above statement is true?
linear-algebra
linear-algebra
asked Jan 16 at 16:14
Tom.Tom.
16619
asked Jan 16 at 16:14
Tom.Tom.
16619
edited Jan 16 at 16:26
Tom.
asked Jan 16 at 16:14
Tom.Tom.
16619
asked Jan 16 at 16:14
Tom.Tom.
16619
asked Jan 16 at 16:14
Tom.Tom.
16619
16619
$begingroup$
What is meant by $theta$ here? Did you mean $emptyset$? Also, your assertion that rank(A+B)=rank(A)+rank(B) is not true. Consider the fact that $0=A+(-A)$ for any matrix $A$ for an easy counterexample
$endgroup$
– pwerth
Jan 16 at 16:17
$begingroup$
@pwerth I think he means that $operatorname{rank}(A + B) = operatorname{rank}(A) + operatorname{rank}(B)$ if $C(A) cap C(B) = {0}$
$endgroup$
– Omnomnomnom
Jan 16 at 16:18
$begingroup$
Yeah!I edited that.
$endgroup$
– Tom.
Jan 16 at 16:20
1
$begingroup$
Even so, actually, the statement doesn't hold. For instance, with $$ A = pmatrix{1\0}, B = pmatrix{0\1} $$ the column spaces are complimentary but $A + B$ has rank $1$.
$endgroup$
– Omnomnomnom
Jan 16 at 16:21
add a comment |
$begingroup$
What is meant by $theta$ here? Did you mean $emptyset$? Also, your assertion that rank(A+B)=rank(A)+rank(B) is not true. Consider the fact that $0=A+(-A)$ for any matrix $A$ for an easy counterexample
$endgroup$
– pwerth
Jan 16 at 16:17
$begingroup$
@pwerth I think he means that $operatorname{rank}(A + B) = operatorname{rank}(A) + operatorname{rank}(B)$ if $C(A) cap C(B) = {0}$
$endgroup$
– Omnomnomnom
Jan 16 at 16:18
$begingroup$
Yeah!I edited that.
$endgroup$
– Tom.
Jan 16 at 16:20
1
$begingroup$
Even so, actually, the statement doesn't hold. For instance, with $$ A = pmatrix{1\0}, B = pmatrix{0\1} $$ the column spaces are complimentary but $A + B$ has rank $1$.
$endgroup$
– Omnomnomnom
Jan 16 at 16:21
$begingroup$
What is meant by $theta$ here? Did you mean $emptyset$? Also, your assertion that rank(A+B)=rank(A)+rank(B) is not true. Consider the fact that $0=A+(-A)$ for any matrix $A$ for an easy counterexample
$endgroup$
– pwerth
Jan 16 at 16:17
$begingroup$
What is meant by $theta$ here? Did you mean $emptyset$? Also, your assertion that rank(A+B)=rank(A)+rank(B) is not true. Consider the fact that $0=A+(-A)$ for any matrix $A$ for an easy counterexample
$endgroup$
– pwerth
Jan 16 at 16:17
$begingroup$
@pwerth I think he means that $operatorname{rank}(A + B) = operatorname{rank}(A) + operatorname{rank}(B)$ if $C(A) cap C(B) = {0}$
$endgroup$
– Omnomnomnom
Jan 16 at 16:18
$begingroup$
@pwerth I think he means that $operatorname{rank}(A + B) = operatorname{rank}(A) + operatorname{rank}(B)$ if $C(A) cap C(B) = {0}$
$endgroup$
– Omnomnomnom
Jan 16 at 16:18
$begingroup$
Yeah!I edited that.
$endgroup$
– Tom.
Jan 16 at 16:20
$begingroup$
Yeah!I edited that.
$endgroup$
– Tom.
Jan 16 at 16:20
1
1
$begingroup$
Even so, actually, the statement doesn't hold. For instance, with $$ A = pmatrix{1\0}, B = pmatrix{0\1} $$ the column spaces are complimentary but $A + B$ has rank $1$.
$endgroup$
– Omnomnomnom
Jan 16 at 16:21
$begingroup$
Even so, actually, the statement doesn't hold. For instance, with $$ A = pmatrix{1\0}, B = pmatrix{0\1} $$ the column spaces are complimentary but $A + B$ has rank $1$.
$endgroup$
– Omnomnomnom
Jan 16 at 16:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assuming that $C(A) cap C(B) = {0}$, it is not necessarily true that $operatorname{rank}(A+B) = operatorname{rank}(A) + operatorname{rank}(B)$, or that $C(A+B) = C(A) + C(B)$.
As a counterexample for both of these, consider the matrices
$$
A = pmatrix{1\0}, quad B = pmatrix{0\1}
$$
It is true, however, that $C(A+B) subset C(A) + C(B)$, and $operatorname{rank}(A+B) leq operatorname{rank}(A) + operatorname{rank}(B)$.
answered Jan 16 at 16:23
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
$begingroup$
Then, under what conditions the given statement is true? i.e. $C(A)+C(B)=C(A+B)$
$endgroup$
– Tom.
Jan 16 at 16:27
$begingroup$
Can you give a bit more context? Is there a specific situation where you're expecting this situation to hold? Why are you looking for matrices satisfying this condition in the first place?
$endgroup$
– Omnomnomnom
Jan 16 at 16:29
add a comment |
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$begingroup$
Assuming that $C(A) cap C(B) = {0}$, it is not necessarily true that $operatorname{rank}(A+B) = operatorname{rank}(A) + operatorname{rank}(B)$, or that $C(A+B) = C(A) + C(B)$.
As a counterexample for both of these, consider the matrices
$$
A = pmatrix{1\0}, quad B = pmatrix{0\1}
$$
It is true, however, that $C(A+B) subset C(A) + C(B)$, and $operatorname{rank}(A+B) leq operatorname{rank}(A) + operatorname{rank}(B)$.
answered Jan 16 at 16:23
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
$begingroup$
Then, under what conditions the given statement is true? i.e. $C(A)+C(B)=C(A+B)$
$endgroup$
– Tom.
Jan 16 at 16:27
$begingroup$
Can you give a bit more context? Is there a specific situation where you're expecting this situation to hold? Why are you looking for matrices satisfying this condition in the first place?
$endgroup$
– Omnomnomnom
Jan 16 at 16:29
add a comment |
$begingroup$
Assuming that $C(A) cap C(B) = {0}$, it is not necessarily true that $operatorname{rank}(A+B) = operatorname{rank}(A) + operatorname{rank}(B)$, or that $C(A+B) = C(A) + C(B)$.
As a counterexample for both of these, consider the matrices
$$
A = pmatrix{1\0}, quad B = pmatrix{0\1}
$$
It is true, however, that $C(A+B) subset C(A) + C(B)$, and $operatorname{rank}(A+B) leq operatorname{rank}(A) + operatorname{rank}(B)$.
answered Jan 16 at 16:23
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
$begingroup$
Then, under what conditions the given statement is true? i.e. $C(A)+C(B)=C(A+B)$
$endgroup$
– Tom.
Jan 16 at 16:27
$begingroup$
Can you give a bit more context? Is there a specific situation where you're expecting this situation to hold? Why are you looking for matrices satisfying this condition in the first place?
$endgroup$
– Omnomnomnom
Jan 16 at 16:29
add a comment |
$begingroup$
Assuming that $C(A) cap C(B) = {0}$, it is not necessarily true that $operatorname{rank}(A+B) = operatorname{rank}(A) + operatorname{rank}(B)$, or that $C(A+B) = C(A) + C(B)$.
As a counterexample for both of these, consider the matrices
$$
A = pmatrix{1\0}, quad B = pmatrix{0\1}
$$
It is true, however, that $C(A+B) subset C(A) + C(B)$, and $operatorname{rank}(A+B) leq operatorname{rank}(A) + operatorname{rank}(B)$.
answered Jan 16 at 16:23
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
Assuming that $C(A) cap C(B) = {0}$, it is not necessarily true that $operatorname{rank}(A+B) = operatorname{rank}(A) + operatorname{rank}(B)$, or that $C(A+B) = C(A) + C(B)$.
As a counterexample for both of these, consider the matrices
$$
A = pmatrix{1\0}, quad B = pmatrix{0\1}
$$
It is true, however, that $C(A+B) subset C(A) + C(B)$, and $operatorname{rank}(A+B) leq operatorname{rank}(A) + operatorname{rank}(B)$.
answered Jan 16 at 16:23
OmnomnomnomOmnomnomnom
129k794188
answered Jan 16 at 16:23
OmnomnomnomOmnomnomnom
129k794188
answered Jan 16 at 16:23
OmnomnomnomOmnomnomnom
129k794188
answered Jan 16 at 16:23
OmnomnomnomOmnomnomnom
129k794188
129k794188
$begingroup$
Then, under what conditions the given statement is true? i.e. $C(A)+C(B)=C(A+B)$
$endgroup$
– Tom.
Jan 16 at 16:27
$begingroup$
Can you give a bit more context? Is there a specific situation where you're expecting this situation to hold? Why are you looking for matrices satisfying this condition in the first place?
$endgroup$
– Omnomnomnom
Jan 16 at 16:29
add a comment |
$begingroup$
Then, under what conditions the given statement is true? i.e. $C(A)+C(B)=C(A+B)$
$endgroup$
– Tom.
Jan 16 at 16:27
$begingroup$
Can you give a bit more context? Is there a specific situation where you're expecting this situation to hold? Why are you looking for matrices satisfying this condition in the first place?
$endgroup$
– Omnomnomnom
Jan 16 at 16:29
$begingroup$
Then, under what conditions the given statement is true? i.e. $C(A)+C(B)=C(A+B)$
$endgroup$
– Tom.
Jan 16 at 16:27
$begingroup$
Then, under what conditions the given statement is true? i.e. $C(A)+C(B)=C(A+B)$
$endgroup$
– Tom.
Jan 16 at 16:27
$begingroup$
Can you give a bit more context? Is there a specific situation where you're expecting this situation to hold? Why are you looking for matrices satisfying this condition in the first place?
$endgroup$
– Omnomnomnom
Jan 16 at 16:29
$begingroup$
Can you give a bit more context? Is there a specific situation where you're expecting this situation to hold? Why are you looking for matrices satisfying this condition in the first place?
$endgroup$
– Omnomnomnom
Jan 16 at 16:29
add a comment |
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$begingroup$
What is meant by $theta$ here? Did you mean $emptyset$? Also, your assertion that rank(A+B)=rank(A)+rank(B) is not true. Consider the fact that $0=A+(-A)$ for any matrix $A$ for an easy counterexample
$endgroup$
– pwerth
Jan 16 at 16:17
$begingroup$
@pwerth I think he means that $operatorname{rank}(A + B) = operatorname{rank}(A) + operatorname{rank}(B)$ if $C(A) cap C(B) = {0}$
$endgroup$
– Omnomnomnom
Jan 16 at 16:18
$begingroup$
Yeah!I edited that.
$endgroup$
– Tom.
Jan 16 at 16:20
1
$begingroup$
Even so, actually, the statement doesn't hold. For instance, with $$ A = pmatrix{1\0}, B = pmatrix{0\1} $$ the column spaces are complimentary but $A + B$ has rank $1$.
$endgroup$
– Omnomnomnom
Jan 16 at 16:21