Prove tr(AA^T)=Sum of all elements in A^2

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Randyyy

Junior Member
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113
Hey, so my task is to prove that [MATH]tr(AA^T)=\sum_{i=1}^{n}\sum_{k=1}^{n}(a_{ik})^2 [/MATH]. In essence (in case my equality is incorrectly written, I need to prove that [MATH]tr(AA^T)[/MATH] is equal to the sum of all elements in A squared. This is what I have attempted.

proof.jpg

But then I get stuck, not sure how to continue or even if I am on the right track.
 
Sorry but the statement of the theorem is not correct. You clearly stated that I need to prove that tr(AAT) is equal to the sum of all elements in A squared.

After you state what you meant to state correctly I will help you.
 
My apologies for being unclear Jomo, I'll try again to explain exactly what it is I am trying to prove.
I want to prove that given a [MATH]m \times n[/MATH] matrix A, that the [MATH]tr(AA^T)[/MATH] will be equal to the sum of all elements individually squared in A. Example, suppose A is of the form [MATH]2 \times 2[/MATH] and has got the following elements,
\begin{equation*}
A =
\begin{pmatrix}
1 & 2 \\
4 & 5 \\
\end{pmatrix}
\end{equation*}
If we called A with each elements squared for B we get the following matrix:
\begin{equation*}
B=
\begin{pmatrix}
1 & 4\\
16& 20\\
\end{pmatrix}
\end{equation*}
Finally, if we take the sum of B we get [MATH]1+4+16+25=46[/MATH].
[MATH]Tr(AA^T)=46[/MATH]Of course saying it is [MATH]A^2[/MATH] like my post title is incorrect. Because that would imply [MATH]Tr(AA^T)=A \cdot A[/MATH] which is not true!
 
Actually, after taking a break and looking back, I think I have already proved it. Suppose we have a matrix A that is a [MATH]n\times m[/MATH] matrix, and we want to sum the elements squared, we can write that using sigma notation as follows: [MATH]\sum^n_{i=i}\sum^n_{k=i}(a_{ik})^2[/MATH] But notice that it is exactly what I have concluded [MATH]Tr(AA^T)[/MATH] can be rewritten as. So unless I am overlooking something (which is VERY possible) It looks like I proved the theorem? (Doing math late at night seems to be a bad idea but it is hard to resist sometimes due to how fun it can be..).
 
Actually you stated it correctly. I never realized that it was sum of the squares of all the entries in A. I knew that your proofs was correct for the trace(AA^T) but now I learned that is also the sum of the squares of all the entries in A.
 
Jomo said:
Actually you stated it correctly. I never realized that it was sum of the squares of all the entries in A. I knew that your proofs was correct for the trace(AA^T) but now I learned that is also the sum of the squares of all the entries in A.
Click to expand...

I didn't even realize myself I had proved it until looking at it again. I actually think this proof was really interesting, it isn't intuitive to me to think that Tr(AA^T) would end up being the sum of all entries in A squared, very interesting indeed.

Thank you Jomo, I appreciate you taking the time to respond and helping me once again! :)
 
Check out this video I made up, especially part e. I did not realize the consequence of my own video which is that the trace(AA^T) is the sum of the squares of all the entries in A.
 
Very good video indeed Jomo! I actually sat all night yesterday proving those properties, your video is a nice way to confirm that my proofs are correct and also give a clearer view on why it is the way it is.
 
Randyyy said:
Very good video indeed Jomo! I actually sat all night yesterday proving those properties, your video is a nice way to confirm that my proofs are correct and also give a clearer view on why it is the way it is.
Click to expand...
Did you subscribe to my channel? Thanks!
 
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