Prove that if $AB = 0$, then rank(A) + rank(B) ≤ p

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Let $A$ be an $m times$ n matrix and $B$ be an $n times p$ matrix.



I understand that since $AB=0$, the column space of $B$ is contained within the nullspace of $A$. Does this mean that $operatornamerank(B) leq operatornamenullity(A)$?



How do I proceed to show that $operatornamerank(A) + operatornamerank(B) leq p$ ?




linear-algebra matrix-rank




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    Hint: Yes. Then use a theorem about rank + nullity. Do you know it?
    – Ethan Bolker
    Oct 26 '15 at 17:06















up vote
2
down vote

favorite
1












Let $A$ be an $m times$ n matrix and $B$ be an $n times p$ matrix.



I understand that since $AB=0$, the column space of $B$ is contained within the nullspace of $A$. Does this mean that $operatornamerank(B) leq operatornamenullity(A)$?



How do I proceed to show that $operatornamerank(A) + operatornamerank(B) leq p$ ?




linear-algebra matrix-rank




share|cite|improve this question


















  • 2




    Hint: Yes. Then use a theorem about rank + nullity. Do you know it?
    – Ethan Bolker
    Oct 26 '15 at 17:06













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





Let $A$ be an $m times$ n matrix and $B$ be an $n times p$ matrix.



I understand that since $AB=0$, the column space of $B$ is contained within the nullspace of $A$. Does this mean that $operatornamerank(B) leq operatornamenullity(A)$?



How do I proceed to show that $operatornamerank(A) + operatornamerank(B) leq p$ ?




linear-algebra matrix-rank




share|cite|improve this question














Let $A$ be an $m times$ n matrix and $B$ be an $n times p$ matrix.



I understand that since $AB=0$, the column space of $B$ is contained within the nullspace of $A$. Does this mean that $operatornamerank(B) leq operatornamenullity(A)$?



How do I proceed to show that $operatornamerank(A) + operatornamerank(B) leq p$ ?





linear-algebra matrix-rank





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 '17 at 22:31









Widawensen

4,23821344




4,23821344










asked Oct 26 '15 at 17:03









Keith Lee

234




234







  • 2




    Hint: Yes. Then use a theorem about rank + nullity. Do you know it?
    – Ethan Bolker
    Oct 26 '15 at 17:06













  • 2




    Hint: Yes. Then use a theorem about rank + nullity. Do you know it?
    – Ethan Bolker
    Oct 26 '15 at 17:06








2




2




Hint: Yes. Then use a theorem about rank + nullity. Do you know it?
– Ethan Bolker
Oct 26 '15 at 17:06





Hint: Yes. Then use a theorem about rank + nullity. Do you know it?
– Ethan Bolker
Oct 26 '15 at 17:06











1 Answer
1






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oldest

votes

















up vote
0
down vote













Yes, you may indeed deduce that the rank of $B$ is less than or equal to the nullity of $A$.



From there, simply apply the rank-nullity theorem (AKA dimension theorem).



Counterexample to question as stated:
$$
A = pmatrix0&1&0\0&0&1\0&0&0
,quad
B = pmatrix1\0\0
$$
$B$ is $3 times 1$ and $AB = 0$, but $operatornamerank(A) + operatornamerank(B) = 3 > 1$.






share|cite|improve this answer






















  • I do know the rank-nullity theorem. However, from rank(B)≤nullity(A), adding rank(A) to both sides, applying the rank-nullity theorem, I arrive at rank(A) + rank(B) ≤ n. I can't seem to arrive at ≤ p.
    – Keith Lee
    Oct 26 '15 at 17:22











  • Hmmm that was an oversight on my part... there could be a typo in the question.
    – Omnomnomnom
    Oct 26 '15 at 17:27










  • In fact, the statement to be proved is incorrect as stated. I'll edit in a counterexample.
    – Omnomnomnom
    Oct 26 '15 at 17:33










  • See my latest edit.
    – Omnomnomnom
    Oct 26 '15 at 17:36






  • 1




    I just clarified and yes it is a typo. So much for spending hours and hours racking my brain over this. Thank you for helping.
    – Keith Lee
    Oct 26 '15 at 22:06











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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
0
down vote













Yes, you may indeed deduce that the rank of $B$ is less than or equal to the nullity of $A$.



From there, simply apply the rank-nullity theorem (AKA dimension theorem).



Counterexample to question as stated:
$$
A = pmatrix0&1&0\0&0&1\0&0&0
,quad
B = pmatrix1\0\0
$$
$B$ is $3 times 1$ and $AB = 0$, but $operatornamerank(A) + operatornamerank(B) = 3 > 1$.






share|cite|improve this answer






















  • I do know the rank-nullity theorem. However, from rank(B)≤nullity(A), adding rank(A) to both sides, applying the rank-nullity theorem, I arrive at rank(A) + rank(B) ≤ n. I can't seem to arrive at ≤ p.
    – Keith Lee
    Oct 26 '15 at 17:22











  • Hmmm that was an oversight on my part... there could be a typo in the question.
    – Omnomnomnom
    Oct 26 '15 at 17:27










  • In fact, the statement to be proved is incorrect as stated. I'll edit in a counterexample.
    – Omnomnomnom
    Oct 26 '15 at 17:33










  • See my latest edit.
    – Omnomnomnom
    Oct 26 '15 at 17:36






  • 1




    I just clarified and yes it is a typo. So much for spending hours and hours racking my brain over this. Thank you for helping.
    – Keith Lee
    Oct 26 '15 at 22:06















up vote
0
down vote













Yes, you may indeed deduce that the rank of $B$ is less than or equal to the nullity of $A$.



From there, simply apply the rank-nullity theorem (AKA dimension theorem).



Counterexample to question as stated:
$$
A = pmatrix0&1&0\0&0&1\0&0&0
,quad
B = pmatrix1\0\0
$$
$B$ is $3 times 1$ and $AB = 0$, but $operatornamerank(A) + operatornamerank(B) = 3 > 1$.






share|cite|improve this answer






















  • I do know the rank-nullity theorem. However, from rank(B)≤nullity(A), adding rank(A) to both sides, applying the rank-nullity theorem, I arrive at rank(A) + rank(B) ≤ n. I can't seem to arrive at ≤ p.
    – Keith Lee
    Oct 26 '15 at 17:22











  • Hmmm that was an oversight on my part... there could be a typo in the question.
    – Omnomnomnom
    Oct 26 '15 at 17:27










  • In fact, the statement to be proved is incorrect as stated. I'll edit in a counterexample.
    – Omnomnomnom
    Oct 26 '15 at 17:33










  • See my latest edit.
    – Omnomnomnom
    Oct 26 '15 at 17:36






  • 1




    I just clarified and yes it is a typo. So much for spending hours and hours racking my brain over this. Thank you for helping.
    – Keith Lee
    Oct 26 '15 at 22:06













up vote
0
down vote










up vote
0
down vote









Yes, you may indeed deduce that the rank of $B$ is less than or equal to the nullity of $A$.



From there, simply apply the rank-nullity theorem (AKA dimension theorem).



Counterexample to question as stated:
$$
A = pmatrix0&1&0\0&0&1\0&0&0
,quad
B = pmatrix1\0\0
$$
$B$ is $3 times 1$ and $AB = 0$, but $operatornamerank(A) + operatornamerank(B) = 3 > 1$.






share|cite|improve this answer














Yes, you may indeed deduce that the rank of $B$ is less than or equal to the nullity of $A$.



From there, simply apply the rank-nullity theorem (AKA dimension theorem).



Counterexample to question as stated:
$$
A = pmatrix0&1&0\0&0&1\0&0&0
,quad
B = pmatrix1\0\0
$$
$B$ is $3 times 1$ and $AB = 0$, but $operatornamerank(A) + operatornamerank(B) = 3 > 1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Oct 26 '15 at 17:36

























answered Oct 26 '15 at 17:06









Omnomnomnom

122k784170




122k784170











  • I do know the rank-nullity theorem. However, from rank(B)≤nullity(A), adding rank(A) to both sides, applying the rank-nullity theorem, I arrive at rank(A) + rank(B) ≤ n. I can't seem to arrive at ≤ p.
    – Keith Lee
    Oct 26 '15 at 17:22











  • Hmmm that was an oversight on my part... there could be a typo in the question.
    – Omnomnomnom
    Oct 26 '15 at 17:27










  • In fact, the statement to be proved is incorrect as stated. I'll edit in a counterexample.
    – Omnomnomnom
    Oct 26 '15 at 17:33










  • See my latest edit.
    – Omnomnomnom
    Oct 26 '15 at 17:36






  • 1




    I just clarified and yes it is a typo. So much for spending hours and hours racking my brain over this. Thank you for helping.
    – Keith Lee
    Oct 26 '15 at 22:06

















  • I do know the rank-nullity theorem. However, from rank(B)≤nullity(A), adding rank(A) to both sides, applying the rank-nullity theorem, I arrive at rank(A) + rank(B) ≤ n. I can't seem to arrive at ≤ p.
    – Keith Lee
    Oct 26 '15 at 17:22











  • Hmmm that was an oversight on my part... there could be a typo in the question.
    – Omnomnomnom
    Oct 26 '15 at 17:27










  • In fact, the statement to be proved is incorrect as stated. I'll edit in a counterexample.
    – Omnomnomnom
    Oct 26 '15 at 17:33










  • See my latest edit.
    – Omnomnomnom
    Oct 26 '15 at 17:36






  • 1




    I just clarified and yes it is a typo. So much for spending hours and hours racking my brain over this. Thank you for helping.
    – Keith Lee
    Oct 26 '15 at 22:06
















I do know the rank-nullity theorem. However, from rank(B)≤nullity(A), adding rank(A) to both sides, applying the rank-nullity theorem, I arrive at rank(A) + rank(B) ≤ n. I can't seem to arrive at ≤ p.
– Keith Lee
Oct 26 '15 at 17:22





I do know the rank-nullity theorem. However, from rank(B)≤nullity(A), adding rank(A) to both sides, applying the rank-nullity theorem, I arrive at rank(A) + rank(B) ≤ n. I can't seem to arrive at ≤ p.
– Keith Lee
Oct 26 '15 at 17:22













Hmmm that was an oversight on my part... there could be a typo in the question.
– Omnomnomnom
Oct 26 '15 at 17:27




Hmmm that was an oversight on my part... there could be a typo in the question.
– Omnomnomnom
Oct 26 '15 at 17:27












In fact, the statement to be proved is incorrect as stated. I'll edit in a counterexample.
– Omnomnomnom
Oct 26 '15 at 17:33




In fact, the statement to be proved is incorrect as stated. I'll edit in a counterexample.
– Omnomnomnom
Oct 26 '15 at 17:33












See my latest edit.
– Omnomnomnom
Oct 26 '15 at 17:36




See my latest edit.
– Omnomnomnom
Oct 26 '15 at 17:36




1




1




I just clarified and yes it is a typo. So much for spending hours and hours racking my brain over this. Thank you for helping.
– Keith Lee
Oct 26 '15 at 22:06





I just clarified and yes it is a typo. So much for spending hours and hours racking my brain over this. Thank you for helping.
– Keith Lee
Oct 26 '15 at 22:06


















 

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