Determinant of infinity

Question

I'm currently studying Linear Algebra. My question is this: If I multiply the basis vectors of a vector space of infinite dimensions, by infinity, will the determinant be a larger infinity than the other two or is it considered the same seeing as we are still operating in the real number space? To my mind, I can't see how it would be of the same size infinity as the cardinality of real number set, as the basis vectors and the dimensions would be.

Let me know if this question makes sense, i'll amend accordingly.

Answer 1

Multiplication in a vector space is only defined for scalars of the ground field. You can use the symbol infinity to denote an element of your field but there is nothing like multiplication with "an arbitrary large number", since there is not field containing infinity and not just an element denoted by the same symbol.

You can easily check if you want to "define" a field with an element $\infty$ that has an interpretation as "arbitrary large number". For example you would interpret $\infty\cdot \infty=\infty$, $\infty-\infty=0$, $\infty-1=\infty$. But then, $$\infty(\infty-1)=\infty\cdot \infty=\infty$$ when evaluating the parenthesis first and $$\infty(\infty-1)=\infty\infty-\infty=\infty-\infty=0$$ by using distributive law which holds in any field. You now get $\infty =0$.

By the way, the convention $\infty-x=\infty$ for all $x\neq\infty$ gives, by substracting $\infty$ on both sided that every element of the field had to be zero, another contradiction since a field contains at least two elements.

Answer 2

Sorry, your question doesn't make sense.

There's no determinant in infinite-dimensions. There are many ways to define/characterise determinants in finite dimensions, but none work in infinite dimensions. There's no nice way to extend the Lebesgue measure (which is area in $\Bbb{R}^2$ and volume in $\Bbb{R}^3$) into infinite dimensions, further there's no such thing as orientation either.

Even if there were some kind of determinant idea, it would not be appropriate to talk about larger and smaller infinities. The different sizes of infinities is a cardinality concept; it's talking about counting the number of things. It makes sense to distinguish between different sizes of infinity, as there are some infinite collections of objects that cannot be matched up with other infinite collections of objects (in the same way that we can always match up any two collections of $5$ objects, say).

Determinants, on the other hand, are not counting anything, they're measuring. They assign a number to give you an idea of how much the linear operator grows or shrinks a subset of the space. It's a real or complex number. The infinities attached to $\Bbb{R}$ and $\Bbb{C}$ are more an artefact of the ordering on these fields (a partial order in the case of $\Bbb{C}$, defined by $z \le w$ if and only if $|z| \le |w|$). The $\infty$s are just abstract constructs to make something that is greater (or lesser, in the case of $-\infty$) than every number in the field. It makes no sense to talk about a larger $\infty$, because essentially the structure of the field stops before $\infty$ (which is not true for sets; they can be made larger and larger).

Also, to back up Crostul and James, this $\infty$ is an order-related construct, but not an element of the field itself. Scalar multiplication has no definition for $\infty$, or indeed any other object that lies outside the scalar field.