# Why is the space of linear operators on $n$-dimensional complex vector space $V$ finite dimensional

I'm reading Axler Sheldon's paper Down with Determinants! and on page 7 he states the following:

4. The Minimal Polynomial

Because the space of linear operator on $V$ is finite dimensional, there is a smallest positive integer $k$ such that

$$I, T, T^2, ..., T^k$$

are not linearly independent.

In above $T$ is a linear operator on $V$ and $V$ is a $n$-dimensional complex vector space. The author doesn't seem to explain why the space of linear operators on $V$ is finite. He simply seems to states this.

Question: Why is it finite dimensional? Is this a consequence from something?

Subquestion: The author also uses on some places of the paper the notation $T|_{V_2}$ where $V_2$ is a subspace of V. What does this notation mean? Is it the restriction of $T$ to $V_2$?

• subanswer: yes, it is the restriction – Bananach Aug 11 '16 at 10:26
• Thank you appreciate it! =) – jjepsuomi Aug 11 '16 at 10:28

Choose a basis of $V$, say $e_1, \dots, e_n$. Any linear operator $T$ on $V$ is determined uniquely by its action on the basis, i.e. by the values $$T(e_i) = a_{i1} e_1 + \dots + a_{in} e_n, \quad 1 \le i \le n,$$ for some $a_{ij} \in \mathbb{C}$. In other words, $T$ is uniquely determined by the $n^2$ complex numbers $a_{ij}$. This implies that the space of operators has dimension $n^2$, and so is finite-dimensional. Using your basis to view the operators as matrices is another easy way to see the space is of dimension $n \times n = n^2$.

• Understood :) Thank you! – jjepsuomi Aug 11 '16 at 10:28

That is because, over any field $K$, in the finite dimensional vector space $V$ choose a basis $\mathcal B=(e_1,\dots,e_n)$ of $V$. Then, for any vector space $W$, we have a (non-canonical) isomorphism: $$\operatorname{Hom}_K(V,W)\cong W^{\lvert\mathcal B\rvert}=W^n.$$ In particular $$\operatorname{End_K}(V)=\operatorname{Hom_K}(V,V)\cong V^n.$$

Actually, this theorem is more generally true for any commutative ring $R$ and any finitely generated free $R$-module.
• that explains why the space of operators from $V$ to $\mathbb{R}$ is finite-dimensional. The question is about operators from $V$ to $V$. – Bananach Aug 11 '16 at 10:38