How to prove the space of all linear transformations from $V$ to itself isomorphic to $M(n \times n, F)$?

Question

Dimension of $V$ is $n$ here. I know that the two spaces have the same dimension. Does this suffice to say they're isomorphic. How so?

Answer 1

For each linear transformations $T:V\rightarrow V$, consider the transformation matrix $M_{n\times n}(F)$ for each $T$. $M_{n\times n}(F)$ is unique for certain base. Moreover for any $\alpha T_1+\beta T_2$, the transformation matrix is $\alpha M_{T_1}+\beta M_{T_2}$.

So space of all linear transformations from $V$ to itself is isomorphic to $M_{n \times n}(F)$.

Answer 2

first suppose you have $T\in L(V)$, define matrix $A$ that column $j$ is equal $T(e_j)$ where $e_j$ is $j$ th elemet of basis $V$. conversely for given matrix let $T(e_j)$ equal to jth column $A$.(you know for knowing an operator, it is enough to know how that operate on element of basis)

And, yes your reason is also right, Theorem: Two finite-dimensional vector spaces are isomorphic if and only if they have the same dimension.