Linear Transformation

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TheWrathOfMath

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Prove that dimKerAB is less than or equal to dimkerA+dimkerB.

I was able to prove that dimKerB is less than or equal to dimkerAB and that dimKerA is less than or equal to dimKerAB.
 
I actually think that the statement "dimKerA is less than or equal to dimKerAB" is false.

I made a correction. Now it's true, I believe.

Though it seem to me like the conclusion that can be drawn is the exact opposite of what I was asked to prove.
 

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You say that v is in ker(AB)
That is, (AB)v=0 or Au=0 where u=Bv. You then conclude that v is in Ker A.
That is pure nonsense. Suppose, for example, that B is the zero matrix (all entries are 0s).
Surely for any vector s (of the appropriate size) we will have Bs=0 and hence A(Bs)=0. But is s guaranteed to be in ker A?
You can only say that u or Bv is in ker A. Do you see that?

If not, consider this. B maps something, say b, to 0. Then A has to map Bb to 0, but A does not have to map b to 0.
If that was true, then kerA C kerB for ANY matrices A and B. What would that imply? Please answer.
 
Steven G said:
You say that v is in ker(AB)
That is, (AB)v=0 or Au=0 where u=Bv. You then conclude that v is in Ker A.
That is pure nonsense. Suppose, for example, that B is the zero matrix (all entries are 0s).
Surely for any vector s (of the appropriate size) we will have Bs=0 and hence A(Bs)=0. But is s guaranteed to be in ker A?
You can only say that u or Bv is in ker A. Do you see that?

If not, consider this. B maps something, say b, to 0. Then A has to map Bb to 0, but A does not have to map b to 0.
If that was true, then kerA C kerB for ANY matrices A and B. What would that imply? Please answer.
Click to expand...
I said that v is in Im(AB), but I suppose that it is still incorrect.
 
Steven G said:
Sorry but your handwriting is hard to read.
Click to expand...
I understand and apologize for it.

But in that case, is my proof correct?
I believe that there must be a mistake, since from what I have done, one can conclude that dimKerAB is *greater than* or equal to dimKerA+dimKerB.
 
Steven G said:
Sorry but your handwriting is hard to read.
Click to expand...
I concluded that dimKerB is less than or equal to dimKer(AB).
Then I proved that dimKerA is less than or equal to dimker(AB) using the fact that Im(AB)C ImA
 
Last edited:
Steven G said:
Sorry but your handwriting is hard to read.
Click to expand...
I can see why the "equal part" is correct, since from the two conclusions I obtained, dimKer(AB) can be equal to dimKerA and can be equal to dimKerB, so if that is the case, we can label dimkerA=dimkerB=k, hence dimKer(AB)=2k, which satisfies dimKer(AB) is less than or equal to dimkerA+dimkerB.

However, I do not see how this can be used to prove that dimKer(AB) can be less than the sum of dimKerA and dimKerB.

I either did something wrong so far or need to further continue this proof.

In fact, even if I use: r(AB) is less than or equal to min{r(A), r(B)} I obtain dimKer(AB) is greater than or less than min{dimKerA, dimKerB}.

So it genuinely seems to me like a case in which dimKerA=1, dimKerB=2 and dimKerAB=4, for instance, is possible.
 
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