Matrices -- Small Problem in Proof

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TheWrathOfMath

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I was asked to prove that if A is a real matrix, and given that A*A^T=0, A=0.
I proved it for i=j by using the sigma notation.
But how do I prove the case for i=/=j?
 
You last line says [imath]a_{ik} = 0 \forall i,k[/imath] -- is not that the proof you are looking for?
 
TheWrathOfMath said:
I was asked to prove that if A is a real matrix, and given that A*A^T=0, A=0.
I proved it for i=j by using the sigma notation.
But how do I prove the case for i=/=j?
Click to expand...
Have you considered that perhaps you don't need to use the case [imath]i\ne j[/imath]?

You've shown that if all the elements on the diagonal of the product matrix are zero, then A has to be zero.

But if the product matrix is the zero matrix, then all the elements on the diagonal are zero! So ...

By the way, your notation is bad:

1651351112558.png

This sum doesn't give the whole matrix, but only the i,j element of it, right? That left me quite confused at first. Writing it clearly might have left you less confused, too.
 
You wrote I was asked to prove that if A is a real matrix, and given that A*A^T=0, A=0.
I think it should be I was asked to prove that if A is a real matrix, and given that A*A^T=0, then A=0 or I was asked to prove that if A is a real matrix, and given that A*A^T=0, then show A=0.
 
Steven G said:
You wrote I was asked to prove that if A is a real matrix, and given that A*A^T=0, A=0.
I think it should be I was asked to prove that if A is a real matrix, and given that A*A^T=0, then A=0 or I was asked to prove that if A is a real matrix, and given that A*A^T=0, then show A=0.
Click to expand...
I agree that I should have phrased the question more clearly and accurately.
 
Dr.Peterson said:
Have you considered that perhaps you don't need to use the case [imath]i\ne j[/imath]?

You've shown that if all the elements on the diagonal of the product matrix are zero, then A has to be zero.

But if the product matrix is the zero matrix, then all the elements on the diagonal are zero! So ...

By the way, your notation is bad:

View attachment 32496

This sum doesn't give the whole matrix, but only the i,j element of it, right? That left me quite confused at first. Writing it clearly might have left you less confused, too.
Click to expand...
Why is my notation bad, if I may please ask?
 
Look at the definition of the product of two matrices. f you are still unclear then please post your definition.
 
Dr.Peterson said:
I thought I made that clear:

The left-hand side should say [imath]c_{ij}[/imath], not just C, which means the entire matrix. You need to specify what element you are calculating.

See the definition as stated here:
en.wikipedia.org

Matrix multiplication - Wikipedia

en.wikipedia.org en.wikipedia.org
Click to expand...
I fixed the error.
Thank you for pointing it out : )

Other than this, is the proof free of errors?
Because I shown that aik is equal to zero, which is not matrix A.
Rather it is an entry in row i.
 

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Last edited:
TheWrathOfMath said:
I fixed the error.
Thank you for pointing it out : )

Other than this, is the proof free of errors?
Click to expand...
I would want to see some words explaining how you conclude that [imath]a_{ik}=0\;\forall i,k[/imath]. It's "obvious", but not without explanation.

Also, I'm still unhappy with your notation in the last line:

1651359025259.png​

Is the matrix A actually equal to [imath]a_{ik}[/imath]? Or might it be better to refer to A separately on the next line? What do you think? What do other proofs you've seen look like at the end?
 
Dr.Peterson said:
Is the matrix A actually equal to [imath]a_{ik}[/imath]? Or might it be better to refer to A separately on the next line? What do you think? What do other proofs you've seen look like at the end?
Click to expand...
You are correct. I noticed it before you pointed it out and deleted it. Please note the the new attachment below (which I posted above as well, by the way):
 

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I had a different reason to say that your notation is not clear.
I still don't like your proof as something is really bothering me.

Let C=A*B.
Now \(\displaystyle c_{ij}=\sum_{k=1}^n a_{ik}b_{kj}\ not\ c_{ij}=\sum_{k=1}^n a_{ik}b_{jk}\).
You need to explain how you got around that.
 
Steven G said:
I had a different reason to say that your notation is not clear.
I still don't like your proof as something is really bothering me.

Let C=A*B.
Now \(\displaystyle c_{ij}=\sum_{k=1}^n a_{ik}b_{kj}\ not\ c_{ij}=\sum_{k=1}^n a_{ik}b_{jk}\).
You need to explain how you got around that.
Click to expand...
I will take note of that.
I appreciate you pointing it out.

The main issue is that I have is that I obtained a proof for an entry aik (an entry in row i in matrix A) being equal to zero rather than one for all entries of matrix A.
 
TheWrathOfMath said:
I will take note of that.
I appreciate you pointing it out.

The main issue is that I have is that I obtained a proof for an entry aik (an entry in row i in matrix A) being equal to zero rather than one for all entries of matrix A.
Click to expand...
I'll echo Dr.Peterson. You have more than you know by calculating the diagonal. For example:
[imath]\left ( \begin{matrix} a & b \\ c & d \end{matrix} \right )^T \left ( \begin{matrix} a & b \\ c & d \end{matrix} \right ) = \left ( \begin{matrix} a^2 + c^2 & ab + cd \\ ab + cd & b^2 + d^2 \end{matrix} \right )[/imath]

As you argued (correctly), if this is 0 then [imath]a^2 + c^2 = 0[/imath] and [imath]b^2 + d^2 = 0[/imath]. What does this mean about a, b, c, and d? And thus about the whole matrix A?

-Dan
 
TheWrathOfMath said:
The main issue is that I have is that I obtained a proof for an entry aik (an entry in row i in matrix A) being equal to zero rather than one for all entries of matrix A.
Click to expand...
I'm not sure what you mean here. A proof that [imath]a_{ik}[/imath] is zero for any i and k is a proof that all entries are zero, and therefore that A is zero. That's what I wanted you to add as a last step; dropping mention of A at the end doesn't fix it. And apparently it doesn't convince you, either!

This means that, at the start, in addition to correcting the notation for [imath]c_{ij}[/imath], you needed to make it clear what was given. Rather than just say "if [imath]i\ne j[/imath]", where it is not clear what they are, you might have started by saying, Let i be any index from 1 to m. Then you are evaluating entry [imath]c_{ii}[/imath], and using that to conclude that [imath]a_{ik}=0\;\forall i,k[/imath]. That is, you picked any i, and have shown that all entries in that row are zero. Then you have to make a conclusion about A from that.

Steven G said:
I had a different reason to say that your notation is not clear.
I still don't like your proof as something is really bothering me.

Let C=A*B.
Now \(\displaystyle c_{ij}=\sum_{k=1}^n a_{ik}b_{kj}\ not\ c_{ij}=\sum_{k=1}^n a_{ik}b_{jk}\).
You need to explain how you got around that.
Click to expand...
I believe the point here is that you didn't explain where you got your summation (namely, you ought to say something explicitly about the transpose).
 
Dr.Peterson said:
I'm not sure what you mean here. A proof that [imath]a_{ik}[/imath] is zero for any i and k is a proof that all entries are zero, and therefore that A is zero. That's what I wanted you to add as a last step; dropping mention of A at the end doesn't fix it. And apparently it doesn't convince you, either!

This means that, at the start, in addition to correcting the notation for [imath]c_{ij}[/imath], you needed to make it clear what was given. Rather than just say "if [imath]i\ne j[/imath]", where it is not clear what they are, you might have started by saying, Let i be any index from 1 to m. Then you are evaluating entry [imath]c_{ii}[/imath], and using that to conclude that [imath]a_{ik}=0\;\forall i,k[/imath]. That is, you picked any i, and have shown that all entries in that row are zero. Then you have to make a conclusion about A from that.


I believe the point here is that you didn't explain where you got your summation (namely, you ought to say something explicitly about the transpose).
Click to expand...
I made some correction.
Please see the attached file below:
 

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Looks good. You should add a statement as to why each of the [imath]a_{ik}[/imath] are zero.

-Dan
 
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