Kernel Equality












0












$begingroup$


So for my university studies I was given this problem:



Let B ∈ R^k×m and A ∈ Rm×n
. Further assume that Ker(B) ∩ Ran(A) = {o}. Show
that this implies Ker(A) = Ker(BA).^



To this point I have come so far with that problem:



For showing set equality you have to show that



$Ker(A) subset Ker(BA) ,,, land ,,, Ker(BA) subset Ker(A)$



I have managed to show the easier inclusion myself like this:



Show that:



$Ker(A) subset Ker(BA)$



Let $ x in R^n $ be an arb. vector such that $ Ax = 0$



Now look at $BAx$



$BAx = B(Ax) = B (0) = Bcdot 0 = 0$



$ Rightarrow$ $Ker(A)$ is indeed a subset of $Ker(BA)$ Since $x$ was choosen arb.



But now im having a hard time to find a rigorous proof for the other inclusion, I started with this:



Show that:



$Ker(BA) subset Ker(A)$



Now let $ x in R^n$ be an arb. vector such that $BAx = 0$ Now show that also $Ax = 0$.



$BAx = 0$ $y:= Ax, ,,, y in R^m$



$By = 0$



But this is it i haven't come further and I have no clue how to proceed especially how to use that one condition stated in the problem ($ Ker(B) cap Ran(A) =$ {0}).



I would be very glad if someone could help me with this.






linear-algebra proof-verification elementary-set-theory proof-writing







share|cite|improve this question









$endgroup$












  • $begingroup$
    Using ker will produce $ker$ which also has the correct spacing. It is also more appealing to the readers if you put the effort to make the entire question using $rmLaTeX$ and don't use "arb." as a short for "arbitrary".
    $endgroup$
    – Asaf Karagila♦
    Dec 2 '18 at 13:31
















0












$begingroup$


So for my university studies I was given this problem:



Let B ∈ R^k×m and A ∈ Rm×n
. Further assume that Ker(B) ∩ Ran(A) = {o}. Show
that this implies Ker(A) = Ker(BA).^



To this point I have come so far with that problem:



For showing set equality you have to show that



$Ker(A) subset Ker(BA) ,,, land ,,, Ker(BA) subset Ker(A)$



I have managed to show the easier inclusion myself like this:



Show that:



$Ker(A) subset Ker(BA)$



Let $ x in R^n $ be an arb. vector such that $ Ax = 0$



Now look at $BAx$



$BAx = B(Ax) = B (0) = Bcdot 0 = 0$



$ Rightarrow$ $Ker(A)$ is indeed a subset of $Ker(BA)$ Since $x$ was choosen arb.



But now im having a hard time to find a rigorous proof for the other inclusion, I started with this:



Show that:



$Ker(BA) subset Ker(A)$



Now let $ x in R^n$ be an arb. vector such that $BAx = 0$ Now show that also $Ax = 0$.



$BAx = 0$ $y:= Ax, ,,, y in R^m$



$By = 0$



But this is it i haven't come further and I have no clue how to proceed especially how to use that one condition stated in the problem ($ Ker(B) cap Ran(A) =$ {0}).



I would be very glad if someone could help me with this.






linear-algebra proof-verification elementary-set-theory proof-writing







share|cite|improve this question









$endgroup$












  • $begingroup$
    Using ker will produce $ker$ which also has the correct spacing. It is also more appealing to the readers if you put the effort to make the entire question using $rmLaTeX$ and don't use "arb." as a short for "arbitrary".
    $endgroup$
    – Asaf Karagila♦
    Dec 2 '18 at 13:31














0












0








0





$begingroup$


So for my university studies I was given this problem:



Let B ∈ R^k×m and A ∈ Rm×n
. Further assume that Ker(B) ∩ Ran(A) = {o}. Show
that this implies Ker(A) = Ker(BA).^



To this point I have come so far with that problem:



For showing set equality you have to show that



$Ker(A) subset Ker(BA) ,,, land ,,, Ker(BA) subset Ker(A)$



I have managed to show the easier inclusion myself like this:



Show that:



$Ker(A) subset Ker(BA)$



Let $ x in R^n $ be an arb. vector such that $ Ax = 0$



Now look at $BAx$



$BAx = B(Ax) = B (0) = Bcdot 0 = 0$



$ Rightarrow$ $Ker(A)$ is indeed a subset of $Ker(BA)$ Since $x$ was choosen arb.



But now im having a hard time to find a rigorous proof for the other inclusion, I started with this:



Show that:



$Ker(BA) subset Ker(A)$



Now let $ x in R^n$ be an arb. vector such that $BAx = 0$ Now show that also $Ax = 0$.



$BAx = 0$ $y:= Ax, ,,, y in R^m$



$By = 0$



But this is it i haven't come further and I have no clue how to proceed especially how to use that one condition stated in the problem ($ Ker(B) cap Ran(A) =$ {0}).



I would be very glad if someone could help me with this.






linear-algebra proof-verification elementary-set-theory proof-writing







share|cite|improve this question









$endgroup$




So for my university studies I was given this problem:



Let B ∈ R^k×m and A ∈ Rm×n
. Further assume that Ker(B) ∩ Ran(A) = {o}. Show
that this implies Ker(A) = Ker(BA).^



To this point I have come so far with that problem:



For showing set equality you have to show that



$Ker(A) subset Ker(BA) ,,, land ,,, Ker(BA) subset Ker(A)$



I have managed to show the easier inclusion myself like this:



Show that:



$Ker(A) subset Ker(BA)$



Let $ x in R^n $ be an arb. vector such that $ Ax = 0$



Now look at $BAx$



$BAx = B(Ax) = B (0) = Bcdot 0 = 0$



$ Rightarrow$ $Ker(A)$ is indeed a subset of $Ker(BA)$ Since $x$ was choosen arb.



But now im having a hard time to find a rigorous proof for the other inclusion, I started with this:



Show that:



$Ker(BA) subset Ker(A)$



Now let $ x in R^n$ be an arb. vector such that $BAx = 0$ Now show that also $Ax = 0$.



$BAx = 0$ $y:= Ax, ,,, y in R^m$



$By = 0$



But this is it i haven't come further and I have no clue how to proceed especially how to use that one condition stated in the problem ($ Ker(B) cap Ran(A) =$ {0}).



I would be very glad if someone could help me with this.






linear-algebra proof-verification elementary-set-theory proof-writing




linear-algebra proof-verification elementary-set-theory proof-writing






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 '18 at 12:56









MathmeeeeenMathmeeeeen

193




193












  • $begingroup$
    Using ker will produce $ker$ which also has the correct spacing. It is also more appealing to the readers if you put the effort to make the entire question using $rmLaTeX$ and don't use "arb." as a short for "arbitrary".
    $endgroup$
    – Asaf Karagila♦
    Dec 2 '18 at 13:31


















  • $begingroup$
    Using ker will produce $ker$ which also has the correct spacing. It is also more appealing to the readers if you put the effort to make the entire question using $rmLaTeX$ and don't use "arb." as a short for "arbitrary".
    $endgroup$
    – Asaf Karagila♦
    Dec 2 '18 at 13:31
















$begingroup$
Using ker will produce $ker$ which also has the correct spacing. It is also more appealing to the readers if you put the effort to make the entire question using $rmLaTeX$ and don't use "arb." as a short for "arbitrary".
$endgroup$
– Asaf Karagila♦
Dec 2 '18 at 13:31




$begingroup$
Using ker will produce $ker$ which also has the correct spacing. It is also more appealing to the readers if you put the effort to make the entire question using $rmLaTeX$ and don't use "arb." as a short for "arbitrary".
$endgroup$
– Asaf Karagila♦
Dec 2 '18 at 13:31










1 Answer
1






active

oldest

votes


















1












$begingroup$

$mathbf xinmathsf{Ker}BA$ or equivalently $BAmathbf x=mathbf0$ implies that $Amathbf xinmathsf{Ker}B$.



Also we have $Amathbf xinmathsf{Ran}A$, so $Amathbf xinmathsf{Ran}Acapmathsf{Ker}B={mathbf0} $.



So actually we have $Amathbf x=mathbf0$ or equivalently $mathbf xinmathsf{Ker}A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you maybe explain this a bit more in detail?
    $endgroup$
    – Mathmeeeeen
    Dec 2 '18 at 13:09










  • $begingroup$
    Do you understand and agree with the first line?
    $endgroup$
    – drhab
    Dec 2 '18 at 13:10












  • $begingroup$
    Yes I do understand it.
    $endgroup$
    – Mathmeeeeen
    Dec 2 '18 at 13:15










  • $begingroup$
    What I actually prove is that for every vector $x$ in $mathsf{Ker}BA$ the vector $Ax$ is an element of $mathsf{Ker}B$ and an element of $mathsf{Ran}A$. Then we can draw the conclusion that $Ax=0$. This comes to the same as $xinmathsf{Ker}A$. Proved is then that every vector that is an element of $mathsf{Ker}BA$ is also an element of $mathsf{Ker}A$, q.e.d..
    $endgroup$
    – drhab
    Dec 2 '18 at 13:16












  • $begingroup$
    Thank you very much that helped me a lot!
    $endgroup$
    – Mathmeeeeen
    Dec 2 '18 at 13:26











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

$mathbf xinmathsf{Ker}BA$ or equivalently $BAmathbf x=mathbf0$ implies that $Amathbf xinmathsf{Ker}B$.



Also we have $Amathbf xinmathsf{Ran}A$, so $Amathbf xinmathsf{Ran}Acapmathsf{Ker}B={mathbf0} $.



So actually we have $Amathbf x=mathbf0$ or equivalently $mathbf xinmathsf{Ker}A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you maybe explain this a bit more in detail?
    $endgroup$
    – Mathmeeeeen
    Dec 2 '18 at 13:09










  • $begingroup$
    Do you understand and agree with the first line?
    $endgroup$
    – drhab
    Dec 2 '18 at 13:10












  • $begingroup$
    Yes I do understand it.
    $endgroup$
    – Mathmeeeeen
    Dec 2 '18 at 13:15










  • $begingroup$
    What I actually prove is that for every vector $x$ in $mathsf{Ker}BA$ the vector $Ax$ is an element of $mathsf{Ker}B$ and an element of $mathsf{Ran}A$. Then we can draw the conclusion that $Ax=0$. This comes to the same as $xinmathsf{Ker}A$. Proved is then that every vector that is an element of $mathsf{Ker}BA$ is also an element of $mathsf{Ker}A$, q.e.d..
    $endgroup$
    – drhab
    Dec 2 '18 at 13:16












  • $begingroup$
    Thank you very much that helped me a lot!
    $endgroup$
    – Mathmeeeeen
    Dec 2 '18 at 13:26
















1












$begingroup$

$mathbf xinmathsf{Ker}BA$ or equivalently $BAmathbf x=mathbf0$ implies that $Amathbf xinmathsf{Ker}B$.



Also we have $Amathbf xinmathsf{Ran}A$, so $Amathbf xinmathsf{Ran}Acapmathsf{Ker}B={mathbf0} $.



So actually we have $Amathbf x=mathbf0$ or equivalently $mathbf xinmathsf{Ker}A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you maybe explain this a bit more in detail?
    $endgroup$
    – Mathmeeeeen
    Dec 2 '18 at 13:09










  • $begingroup$
    Do you understand and agree with the first line?
    $endgroup$
    – drhab
    Dec 2 '18 at 13:10












  • $begingroup$
    Yes I do understand it.
    $endgroup$
    – Mathmeeeeen
    Dec 2 '18 at 13:15










  • $begingroup$
    What I actually prove is that for every vector $x$ in $mathsf{Ker}BA$ the vector $Ax$ is an element of $mathsf{Ker}B$ and an element of $mathsf{Ran}A$. Then we can draw the conclusion that $Ax=0$. This comes to the same as $xinmathsf{Ker}A$. Proved is then that every vector that is an element of $mathsf{Ker}BA$ is also an element of $mathsf{Ker}A$, q.e.d..
    $endgroup$
    – drhab
    Dec 2 '18 at 13:16












  • $begingroup$
    Thank you very much that helped me a lot!
    $endgroup$
    – Mathmeeeeen
    Dec 2 '18 at 13:26














1












1








1





$begingroup$

$mathbf xinmathsf{Ker}BA$ or equivalently $BAmathbf x=mathbf0$ implies that $Amathbf xinmathsf{Ker}B$.



Also we have $Amathbf xinmathsf{Ran}A$, so $Amathbf xinmathsf{Ran}Acapmathsf{Ker}B={mathbf0} $.



So actually we have $Amathbf x=mathbf0$ or equivalently $mathbf xinmathsf{Ker}A$.






share|cite|improve this answer









$endgroup$



$mathbf xinmathsf{Ker}BA$ or equivalently $BAmathbf x=mathbf0$ implies that $Amathbf xinmathsf{Ker}B$.



Also we have $Amathbf xinmathsf{Ran}A$, so $Amathbf xinmathsf{Ran}Acapmathsf{Ker}B={mathbf0} $.



So actually we have $Amathbf x=mathbf0$ or equivalently $mathbf xinmathsf{Ker}A$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 2 '18 at 13:06









drhabdrhab

99.1k544130




99.1k544130












  • $begingroup$
    Could you maybe explain this a bit more in detail?
    $endgroup$
    – Mathmeeeeen
    Dec 2 '18 at 13:09










  • $begingroup$
    Do you understand and agree with the first line?
    $endgroup$
    – drhab
    Dec 2 '18 at 13:10












  • $begingroup$
    Yes I do understand it.
    $endgroup$
    – Mathmeeeeen
    Dec 2 '18 at 13:15










  • $begingroup$
    What I actually prove is that for every vector $x$ in $mathsf{Ker}BA$ the vector $Ax$ is an element of $mathsf{Ker}B$ and an element of $mathsf{Ran}A$. Then we can draw the conclusion that $Ax=0$. This comes to the same as $xinmathsf{Ker}A$. Proved is then that every vector that is an element of $mathsf{Ker}BA$ is also an element of $mathsf{Ker}A$, q.e.d..
    $endgroup$
    – drhab
    Dec 2 '18 at 13:16












  • $begingroup$
    Thank you very much that helped me a lot!
    $endgroup$
    – Mathmeeeeen
    Dec 2 '18 at 13:26


















  • $begingroup$
    Could you maybe explain this a bit more in detail?
    $endgroup$
    – Mathmeeeeen
    Dec 2 '18 at 13:09










  • $begingroup$
    Do you understand and agree with the first line?
    $endgroup$
    – drhab
    Dec 2 '18 at 13:10












  • $begingroup$
    Yes I do understand it.
    $endgroup$
    – Mathmeeeeen
    Dec 2 '18 at 13:15










  • $begingroup$
    What I actually prove is that for every vector $x$ in $mathsf{Ker}BA$ the vector $Ax$ is an element of $mathsf{Ker}B$ and an element of $mathsf{Ran}A$. Then we can draw the conclusion that $Ax=0$. This comes to the same as $xinmathsf{Ker}A$. Proved is then that every vector that is an element of $mathsf{Ker}BA$ is also an element of $mathsf{Ker}A$, q.e.d..
    $endgroup$
    – drhab
    Dec 2 '18 at 13:16












  • $begingroup$
    Thank you very much that helped me a lot!
    $endgroup$
    – Mathmeeeeen
    Dec 2 '18 at 13:26
















$begingroup$
Could you maybe explain this a bit more in detail?
$endgroup$
– Mathmeeeeen
Dec 2 '18 at 13:09




$begingroup$
Could you maybe explain this a bit more in detail?
$endgroup$
– Mathmeeeeen
Dec 2 '18 at 13:09












$begingroup$
Do you understand and agree with the first line?
$endgroup$
– drhab
Dec 2 '18 at 13:10






$begingroup$
Do you understand and agree with the first line?
$endgroup$
– drhab
Dec 2 '18 at 13:10














$begingroup$
Yes I do understand it.
$endgroup$
– Mathmeeeeen
Dec 2 '18 at 13:15




$begingroup$
Yes I do understand it.
$endgroup$
– Mathmeeeeen
Dec 2 '18 at 13:15












$begingroup$
What I actually prove is that for every vector $x$ in $mathsf{Ker}BA$ the vector $Ax$ is an element of $mathsf{Ker}B$ and an element of $mathsf{Ran}A$. Then we can draw the conclusion that $Ax=0$. This comes to the same as $xinmathsf{Ker}A$. Proved is then that every vector that is an element of $mathsf{Ker}BA$ is also an element of $mathsf{Ker}A$, q.e.d..
$endgroup$
– drhab
Dec 2 '18 at 13:16






$begingroup$
What I actually prove is that for every vector $x$ in $mathsf{Ker}BA$ the vector $Ax$ is an element of $mathsf{Ker}B$ and an element of $mathsf{Ran}A$. Then we can draw the conclusion that $Ax=0$. This comes to the same as $xinmathsf{Ker}A$. Proved is then that every vector that is an element of $mathsf{Ker}BA$ is also an element of $mathsf{Ker}A$, q.e.d..
$endgroup$
– drhab
Dec 2 '18 at 13:16














$begingroup$
Thank you very much that helped me a lot!
$endgroup$
– Mathmeeeeen
Dec 2 '18 at 13:26




$begingroup$
Thank you very much that helped me a lot!
$endgroup$
– Mathmeeeeen
Dec 2 '18 at 13:26


















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