What is the difference between an inner product space and an Algebra over a field?

Question

From what I’ve gathered an algebra over a field is a vector space equipped with a bilinear form, and an inner product space is the same but that bilinear form satisfies certain extra axioms (symmetric, nondegenerate, etc.) . Is every inner product space an algebra? Are most typical algebras inner product spaces?

Answer 1

An algebra (over a field) and an inner-product space are quite different beasts. Let me quote the first line from the wikip.

An Algebra (over a field) is a vector space equipped with a bilinear product.

You are basically confusing the bilinear product with bilinear form. Let's drop the word bilinear from the above quote and say

An algebra is a vector space together with a multiplication.

This is more or less correct. First thing to note is that by a multiplication we mean an operation which takes two elements from you algebra (to be!) and yields another element from your algebra. More technically, it is a map $$m:V\times V\to V.$$ But, then one can define all sort of weird multiplications, say $m(x,y)=e$ for all $x,y\in V,$ where $e$ is some vector in $V.$ The whole point is that $V$ is at the least a vector space and we are putting an extra structure on $V.$ It would only be good to have this extra structure respect some the vector space structure of $V.$ Here, we hit upon the condition that $m:V\times V\to V$ be bi-linear. For example, take the set of all real-valued continuous functions on $\mathbb{R},$ say $C(\mathbb{R}).$ It is a vector space over $\mathbb{R}.$ But, more than that you can multiply two continuous functions in an obvious way and get another continuous function on $\mathbb{R}.$ This is your additional 'product' structure. Check that is is bilinear. The key thing here is that after you multiply two vectors (here two continuous functions) you get a new vector (here a new continuous function).

Coming to an inner product space, it is also a vector space but equipped with an inner product. Inner product is a bilinear form with some additional conditions. But the key thing to note here is that when you take inner product of two vectors $u,v\in V,$ you do not get a new vector, but you get some scalar. I will not type the definition of inner product here, but what you must remember is that an inner product $I$ on $V$ is a map $$I: V\times V\to F,$$ where $F$ is your underlying field.

Towards the end, $\mathbb{R}^3$ is an inner-product space with the dot-product but is an algebra with cross-product. If you know what dot-product and cross-product are, it might be easier to remember the difference between an algebra and an inner-product space.

Answer 2

An algebra $L$ over a field $k$ is a $k$-vector space that is also a ring. So we have also multiplication \begin{align} \cdot \colon L\times L & \to L\\ (v,w) &\mapsto v\cdot w \end{align} This map being bilinear means that we have distributive property; being symmetric means that the algebra is commutative; and so on.

Three examples I believe you should have in mind are

1. The $\mathbb{R}$-algebra $(\mathbb{R}^3, +, \times)$ where $+$ is the usual sum of vectors and $\times$ stands for the vector product.
2. The $k$-algebra $(M_2(k),+,\cdot)$ of $2\times2$ matrices with entries in $k$ where $+$ is the usual sum and $\cdot$ is the product of matrices.
3. The $k$-algebra $k[X_1, \dots, X_n]$ of polynomials with coefficients in $k$.

On the other hand, an inner product in $L$ is a $k$-bilinear symmetric map $L\times L \to k$.