My issue with this is the only way I know how to prove it is to set $\dim V=n$, but then that wouldn't make sense because the second part is finding the $\dim$.
What I was thinking is using the rank-nullity theorem $$\rm rank(A)+N(A)=n \ \text{or} \ \dim(\rm Im(A))+\dim(\ker(A))=\dim V$$ Any suggestions on how to prove that if $(V,K)$ is a finite-dimensional vector space, then the space of all linear transformations on $V$ is finite dimensional and to find its dimension?
[Math] If $(V,k)$ is a finite-dimensional vector space, then the space of all linear transformations on $V$ is finite dim and find its dim
functionslinear algebranumerical linear algebra
Best Answer
It sounds like you're trying to show that, given a finite dimensional vector space $V$ over a field $F$, the space $S$ of linear transformations $T: V \to V$ is finite dimensional.
Here's a suggestion for how to proceed: First, show that $S$ is a vector space (if you haven't already). You can do this straight from the definition. (Show that if $T, U \in S$ and $\alpha, \beta \in F$ then $\alpha T + \beta U \in S$, which is to say that $(\alpha T + \beta U)(v) = \alpha T(v) + \beta U(v)$ is a linear transformation.)
Let $n = \dim V \in \mathbb{N}$. Fixing a basis allows us to identify $V$ with $F^n$ and $S$ with the set of all $n\times n$ matrices with entries in $F$. If you think about it, there should be a clear choice for a basis of this space $S$ over $F$ now.
If you're not seeing it, imagine taking the rows of an $n\times n$ matrix and laying them out in a line. Note you can do this for any $n\times n$ matrix, and so this allows you to think of $S$ as being essentially the same vector space as $F^{n^2}$. You now have an element of $F^{n^2}$. What's a basis for $F^{n^2}$? Relate this back to a basis for $S$. What's the order of the basis you found? It should be a function of $n$.
Ah. I just saw your edit. Note you're trying to find the dimension of $\dim S$, and so it's perfectly fine to set $\dim V = n$. In fact, you should expect that $\dim S$ is going to be a function of $\dim V$, and so this is the natural thing to do.