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Matrix multiplication is defined as row by column multiplication. It represents linear transformation. Why isn't matrix addition defined in a similar way: row by column addition. Given to matrices A and B (of the same size). What transformation, if any, does (element by element) addition A+B represent? What would row by column addition represent? How about the Hadamard product (element by element matrix multiplication) represent?

Hass
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  • Row by column addition is a part of tropical matrix multiplication. – Chris Culter Jun 06 '17 at 22:27
  • If $A$ is a the matrix representation of the linear transformation $f$ (wrt two bases) and $B$ is the matrix representation of the linear map $g$ (wrt the same bases), then yes, $A+B$ is the matrix representation of the linear map $f+g$. It's not so hard to prove (try it). –  Jun 06 '17 at 23:36

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I would be careful here. A matrix itself represents a linear transformation. When you multiply a vector $n\times 1$ vector by a $n\times n$ matrix you are doing a linear transformation of the vector. However, the notion of doing a linear transformation on a $n\times n$ matrix $B$ by another $A$ usually means that you are writing $B$ in the basis induced by the transformation $A$. In that case the linear transformation of $B$ is given by $ABA^{-1}$, not $AB$.

A natural way to look at multiplication of two matrices is as a composition of linear transformations. So if $A$ and $B$ are linear transformations then $AB$ the transformation where you first apply the transformation $B$ then $A$.

I’m unaware of anything simple within the paradigm of linear transformations represented by the element-by-element product (which is not to say there isn’t anything). The only thing that has this flavor is the Kronecker product that is somewhat more complicated.

$A+B$ is a linear transformation. Again, careful: addition itself isn’t the linear transformation, two linear transformations represented by $A$ and $B$ can be added to produce a third linear transformation $A+B$.

What $A+B$ does as a transformation when applied to an object is just to take the object transformed by $A$ and the object transformed by $B$ and add them together. This preserves the linearity of $A$ and $B$.

Row-by-column addition (however you define it) isn’t anything useful I’m aware of. I say “however you define it” because remember that matrix multiplication doesn’t just include multiplication, but addition when you take the inner product of the rows.

EDIT I see as I wrote this a commenter gave an example of something like row-by-column addition that is useful.

Rócherz
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spaceisdarkgreen
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  • This answer gets to the heart of the reason. Matrix addition is defined the way it is because that's the most natural and most useful definition when you think about what matrices are (frequently but not always) used for. – Ethan Bolker Jun 06 '17 at 22:54
  • "To do a linear transformation on a matrix $B$ by the matrix $A$, you don't multiply $AB$, you do $ABA^{−1}$" -- this is confusing: both operations are linear and it's not clear what exactly is meant by "linear transformation on a matrix $B$ by the matrix $A$". – Trevor Gunn Jun 06 '17 at 23:54
  • @T.Gunn True. I wrote the next sentence after that to clarify that I meant the usual notion of a transformation of an $n\times n$ matrix (as interpreted as a linear operator on $\mathbb R^n$) to a new basis but perhaps it wasn't clear. I agree that $B\to AB$ is a valid interpretation of the ambiguous "linear transformation of $B$ by $A$." – spaceisdarkgreen Jun 07 '17 at 00:56
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Matrix addition as it is currently defined allows for matrix multiplication to distribute over matrix multiplication and allows for the set of $n\times n$ matrices to be a ring.

It also allows for the set of $m\times n$ matrices to be a vector space.

I would have to see if the definition you propose for matrix addition, and see if these properties still hold.

One place where matrix addition as currently defined come with eigenvectors.

Suppose for some matrix $A$ there exists a $v$ such that $Av = \lambda v$ then we can say

$Av = \lambda I v\\ Av - \lambda I v = \mathbf 0\\ (A - \lambda I) v = \mathbf 0$

and knowing that $(A - \lambda I)$ helps us to find first possibilities for $\lambda$ and then one we know the eigenvalues $v$

Doug M
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When considering two linear maps $f, g$ between two vector spaces, and if the bases are chosen, there are two matrices $A, B$ represent these two maps. Now we want to know what the matrix represents $f + g$ under these bases looks like. By a direct computation the matrix represents $f + g$ is exactly $A + B$ in the sense of addition element by element.

Zhiping Xu
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Let $f$ be a linear function that maps column vectors to column vectors. By linear we mean that $f(v_1+rv_2)=f(v_1)+rf(v_2)$ for any vectors $v_1,v_2$ and any number $r.$ Then there is a unique matrix $M$ such that $f(v)=Mv$ for all $v,$ where the product $Mv$ is defined in the usual way.

Conversely if $M$ is a matrix then $f_M(v)=Mv$ is a linear function.

Suppose $f_M$ and $f_N$ are linear with $f_M(v)=Mv$ and $f_N(v)=Nv$ for all v.. Then with the usual def'n of $MN$ and of $M+N,$ we have $f_M(f_N(v))=M(Nv)=(MN)v$ and $f_M(v)+f_N(v)=(Mv)+(Nv)=(M+N)v.$

This gives us a nice algebraic structure on the set of linear functions, correspondingly to the algebra of the matrices, that is useful for finding out a great deal about them.

DanielWainfleet
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