# How to replace Hadamard Product and Column wise addition with new Matrix operation?

Let $$A= \begin{bmatrix} a_{1,1} & a_{1,2} & a_{1,3}\\ a_{2,1} & a_{2,2} & a_{2,3}\\ a_{3,1} & a_{3,2} & a_{3,3}\\ \end{bmatrix}$$ and $$B= \begin{bmatrix} b_{1,1} & b_{1,2} & b_{1,3}\\ b_{2,1} & b_{2,2} & b_{2,3}\\ b_{3,1} & b_{3,2} & b_{3,3}\\ \end{bmatrix}$$ and $C=A\circ B$, i.e.,

$$C= \begin{bmatrix} a_{1,1} \times b_{1,1} & a_{1,2}\times b_{1,2} & a_{1,3}\times b_{1,3}\\ a_{2,1}\times b_{2,1} & a_{2,2}\times b_{2,2} & a_{2,3}\times b_{2,3}\\ a_{3,1}\times b_{3,1} & a_{3,2}\times b_{3,2} & a_{3,3}\times b_{3,3}\\ \end{bmatrix}$$

and then I have to column-wise addition of $C$, i.e.,

$C'=[c_{1,1}+c_{2,1}+c_{3,1}, c_{1,2}+c_{2,2}+c_{3,2}, c_{1,3}+c_{2,3}+c_{3,3}]$.

My question is

Is there any other way to represent $C'$ in terms of column-wise addition of $A$ (i.e., $A'$) and column-wise addition of $B$ (i.e., $B'$)? or $A$ and $B'$?

• Isn't $C^\prime$ just the diagonal of $A^TB$? – Dilip Sarwate Aug 19 '14 at 14:15
• @DilipSarwate Exactly! Thanks a lot. That's what I wanted. – Mithun Aug 19 '14 at 15:02
• @DilipSarwate Can it be possible to represent $C′$ in terms of column-wise addition of $A$ (i.e., $A′$) and column-wise addition of $B$ (i.e., $B′$)? or $A$ and $B′$? – Mithun Aug 20 '14 at 15:06