Is $C(A)+C(B)=C(A+B)$, if $C(A)cap C(B)= {0},$?












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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 16:26







Tom.

















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


















  • $begingroup$
    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










1 Answer
1






active

oldest

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2












$begingroup$

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 $operatorname{rank}(A+B) leq operatorname{rank}(A) + operatorname{rank}(B)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












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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 $operatorname{rank}(A+B) leq operatorname{rank}(A) + operatorname{rank}(B)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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 $operatorname{rank}(A+B) leq operatorname{rank}(A) + operatorname{rank}(B)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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 $operatorname{rank}(A+B) leq operatorname{rank}(A) + operatorname{rank}(B)$.






share|cite|improve this answer









$endgroup$



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 $operatorname{rank}(A+B) leq operatorname{rank}(A) + operatorname{rank}(B)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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