Endomorphisms of a finite dimensional vector space

From Humphreys' Introduction to Lie Algebras and Representation Theory:

If $V$ is a finite dimensional vector space over $F$, denote by $\text{End }V$ the set of linear transformations $V\rightarrow V$. As a vector space over $F$, $\text{End }V$ has dimension $n^2$ ($n=\text{dim }V$), and $\text{End }V$ is a ring relative to the usual product operation.

1) Why does $\text{End }V$ have dimension $n^2$? I think of an example like $V=\mathbb{R}^n$, with the basis $e_1=(1,0,\ldots,0),e_2=(0,1,\ldots,0),\ldots,e_n=(0,0,\ldots,1)$. Then a linear transformation $V\rightarrow V$ can take $e_1$ to $e_j$ for any $j=1,2,\ldots,n$, take $e_2$ to $e_j$ for any $j=1,2,\ldots,n$, and so on. This should give rise to $n^n$ independent linear transformations.

2) "$\text{End }V$ is a ring relative to the usual product operation." What is the product operation referred to here?

When you choose a basis for $V$, you identify $\operatorname{End}{V}$ with the set of $n\times n$ matrices over $F$, which has dimension $n^2$. Your linear transformations are not all linearly independent.

The product operation is composition of maps, i.e. if $f,g\in\operatorname{End}{V}$, then the product $fg$ is the map defined by $(fg)(v)=g(f(v))$.

1. Hint

Let $f_{p,q}:V\to V$ defined by $$f_{p,q}(e_i)=\delta_{pi}e_q$$

Prove that $(f_{p,q})_{1\leq p,q\leq n}$ is a basis for $\mathrm{End}(V)$. Conclude.

2.The product operation is the composition of the endomorphisms

• You're welcome. – user63181 Aug 6 '13 at 15:12

1) For each $e_i$, there are $n$ components of ${e_i}'$, the image of $e_i$, that must be specified. There are $n$ basis vectors $e_i$, therefore you get $n \times n = n^2$.

2) The product operation is composition. When written as matrices, composition is computed through matrix multiplication.