[Math] Matrix multiplication and generalized inner product

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first time asking in this community, so please bear with me if the question is off-topic and/or meaningless.

My understanding of linear and abstract algebra is quite rudimentary, but as far as I understand, one can define an inner product over a vector space as any function taking a pair of elements in that space and producing a scalar value from its associated field (as long as it satisfies the relevant axioms, etc.).

In $R^n$, the inner product corresponds to the dot product between two vectors.

On the other hand, every discussion of matrix multiplication that I have been able to find seems to discuss this in the context of $R^n$, by reference to the dot product. E.g. the wikipedia article for "matrix multiplication" says: "if $A$ is an $n × m$ matrix and $B$ is an $m × p$ matrix, their matrix product $AB$ is an $n × p$ matrix, in which the $m$ entries across a row of $A$ are multiplied with the $m$ entries down a column of $B$ and summed to produce an entry of AB", which I understand to basically say that the entries in $AB$ are the dot-product between the row vectors in $A$ and the column vectors in $B$.

But if the dot-product is merely the inner product defined over $R^n$, why can't matrix multiplication be defined with some other function $f(v_i,v_j) \rightarrow s$? where $v_i$ and $v_j$ are elements of some space defined over a field $F$ and $s$ is a scalar from $F$?

Does matrix multiplication lose all meaning when applied to fields other than R? Is there a name for this "generalized matrix product" that I'm not aware of?

I get the feeling that I'm lacking the proper nomenclature, but after searching for discussion of such an operation for weeks I gave up and decided to come here and ask what I'm missing.

Thanks!

Best Answer

The most common case, where your $f(v_i, v_j)$ is a bilinear form, would simply become $$ ACB, $$ where $C$ is a square matrix expressing the bilinear form for the bases that gave $A,B.$

In this case, when $C$ is symmetric and $A = B^T,$ we say that $C$ represents $B^T CB$ as quadratic forms.

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