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For finite dimensional $V$ and $W$, we know that \begin{equation*} \dim{L(V,W)} = \dim{V}\cdot\dim{W} \end{equation*}

Does this theorem hold for infinite-dimensional vector spaces $V$ and $W$ too?

Ishan Deo
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    How is multiplication on infinity defined? – LegNaiB May 11 '21 at 07:54
  • I presume the multiplication will be of of cardinal numbers – Ishan Deo May 11 '21 at 07:59
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    @lc2r43 What you are saying is not really correct (or at least misleading). You cannot find a basis of $L(V,W)$ as easily as in the finite-dimensional case. This is the same issue as I described in my comment to the current (incorrect) answer. – Captain Lama May 11 '21 at 08:34
  • @CaptainLama Ah, of course (take the algebraic dual, for example). I retract my previous statement. I've been dealing with too much functional analysis lately. – lc2r43 May 11 '21 at 08:43
  • This question https://math.stackexchange.com/questions/58548/why-are-vector-spaces-not-isomorphic-to-their-duals contains the relevant bits of mathematics. – Captain Lama May 11 '21 at 09:01
  • @CaptainLama why can we not find a basis to $L(V,W)$ as easily as in the finite-dimensional case? – Ishan Deo May 11 '21 at 09:58
  • Because as I mentioned in the comment to the first answer, if you do the obvious thing (which would give you a basis of cardinal $\dim(V)\cdot \dim(W)$), you only get a basis of the space of finite-rank linear maps. And that is a very, very small subspace of the actual total space. – Captain Lama May 11 '21 at 10:56
  • I didn't see your 1st answer before you deleted it, so I didn't see your comment there. Also, what are finite-rank linear maps? A Google search isn't giving me any results. – Ishan Deo May 11 '21 at 13:11

1 Answers1

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The answer I linked in my comment shows in particular that if we write $|X|$ for the cardinality of a set $X$, then for a vector space $V$ over $K$: $$ |V| = \max(|K|,\dim(V)).$$

Now if $(e_i)_{i\in I}$ is a basis of $V$, then $L(V,W)$ is in bijection with the functions $I\to W$ (this is what it means to be a basis), so $$ |L(V,W)| = |W|^{\dim(V)}.$$

Since that cardinal is clearly strictly bigger than $|K|$ when $\dim(V)$ is infinite, we have $|L(V,W)| = \dim(L(V,W))$ so $$\dim(L(V,W)) = |W|^{\dim(V)}.$$

So now you have two cases:

  • If $\dim(W)\leqslant |K|$, then $\dim(L(V,W)) = |K|^{\dim(V)}$;
  • If $\dim(W)\geqslant|K|$, then $\dim(L(V,W)) = \dim(W)^{\dim(V)}$.

Note that the behaviour of cardinal exponentiation can be a little tricky: see the problem of the continuum hypothesis. If you are comfortable with assuming the Generalized Continuum Hypothesis (I personally am), then https://en.wikipedia.org/wiki/Continuum_hypothesis#The_generalized_continuum_hypothesis gives you a complete list of formulas to compute exponentiations.

Captain Lama
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  • "Since that cardinal is clearly strictly bigger than $|K|$ when $\dim(V)$ is infinite..." I'm sorry, I don't see how it's clearly strictly bigger. Could you possibly elaborate? – Ishan Deo May 12 '21 at 21:28
  • Also, what would happen if $\dim(V)$ was finite? – Ishan Deo May 12 '21 at 21:39