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I have two matrices $\mathbf{A}$ and $\mathbf{B}$. They are full and not square, though their dimensions implies that their product is : $$ \dim (\mathbf{A}) = (n,p) \quad \text{and} \quad \dim (\mathbf{B}) = (p,n) \implies \dim (\mathbf{A}\mathbf{B})=(n,n) $$

Now, I am interested in calculating the diagonal terms of this product, but without having to calculate all the other terms in the process (this will be in a code where $n$ and $p$ are very large). I wrote it on a simple example and ended up with : $$ diag(\mathbf{A}\mathbf{B}) = \sum Col\left(\mathbf{A} \cdot \mathbf{B}^\mathsf{T} \right) $$

with $\cdot$ the inner product between the two matrices, and $\sum Col$ the sum on the columns of the matrix. Sorry if my formalism is bad, I wrote it in a coding style, so please feel free to correct me.

I tested it on my code and it matches, but :

  • How could I demonstrate that properly?
  • Is there a theorem or a property for that? I failed to find it on the internet...

Thanks for your time!

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The $i$th term on the diagonal is the scalar product of the $i$th row of $A$ with the transpose of the $i$th column of $B$. Is that what you need?

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  • $\begingroup$ Thanks for your answer! Could be, would you have a reference for that? I could dig in to tell you if this is what I seek. I did not even find a wikipedia page for the "diag" operator ^^" $\endgroup$ – jeannej Sep 11 at 22:23
  • $\begingroup$ This is basically just the definition of the matrix product en.wikipedia.org/wiki/Matrix_multiplication $\endgroup$ – S. Dolan Sep 11 at 22:26
  • $\begingroup$ Ok I already stated that $diag(\mathbf{A}\mathbf{B})=\left[\sum_{k=1}^p a_{ik} b_{ki} \right]_{i=1..n}$ (hope the formalism is ok here). I also have $\mathbf{A}\cdot\mathbf{B} = \left[ a_{ik} b_{ik} \right]_{i=1..n \, , \, k=1..p}$, but I fail to see how to properly link the twos (like said I had to write it down... well I didn't say I was good at maths ^^) $\endgroup$ – jeannej Sep 11 at 22:35
  • $\begingroup$ Your diag(AB) result is fine. However A.B. does **not ** give you this. $\endgroup$ – S. Dolan Sep 11 at 22:43
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    $\begingroup$ Yes but you need to add elements like $a_{12}b_{21}$, not $a_{21}b_{21}$ $\endgroup$ – S. Dolan Sep 11 at 22:56
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With the help of @S. Dolan, I finally got the demonstration I was searching for. So in case others need it, here it is step by step.

Let $\mathbf{A}$ and $\mathbf{B}$ two matrices such as:

$$\dim (\mathbf{A}) = (n,p) \quad \text{and} \quad \dim (\mathbf{B}) = (p,n)$$

As their dimensions match, their multiplication gives:

$$\mathbf{A} \mathbf{B}= \sum_{k=1}^p a_{ik}b_{kj} \quad \forall i\in[1,n] \text{ and } \forall j\in[1,n]$$

were $a$ and $b$ are the elements of $\mathbf{A}$ and $\mathbf{B}$. Taking only the diagonal terms of this product, hence the terms for which $i=j$, yields:

$$diag\left(\mathbf{A} \mathbf{B}\right)= \sum_{k=1}^p a_{ik}b_{ki} \quad \forall i\in[1,n]$$

Besides, their dot/inner product $\mathbf{A} \cdot \mathbf{B}^\mathsf{T}$ is the term-by-term product: $$\mathbf{A} \cdot \mathbf{B}^\mathsf{T} = a_{ik} b_{ki} \quad \forall i\in[1,n] \text{ and } \forall k\in [1,p] $$

So that the sum on its columns (with index $k$) is: $$\sum Col \left(\mathbf{A} \cdot \mathbf{B}^\mathsf{T}\right) = \sum_{k=1}^p a_{ik} b_{ki} \quad \forall i\in[1,n] $$

It can then be concluded that: $$diag\left(\mathbf{A} \mathbf{B}\right)= \sum Col \left(\mathbf{A} \cdot \mathbf{B}^\mathsf{T}\right)$$

QED

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