Compact and linear operators

Question

How could I demonstrate the following property? I'm a little bit stuck here.

If $X$ and $Y$ are normed spaces of finite dimension $\Rightarrow$ $\sigma(X,Y) = L(X,Y)$

where $\sigma(X,Y)$ is the space of compact operators.

Thanks

Answer 1

Would this be correct? I'll add the definitions justo to see if I understood it correctly.

The definition of a compact operator: If X,Y are Banach spaces. A linear operator C: X $\rightarrow$ Y is said to be compact if for each bounded sequence $\left \{x_{n}\right \} \subset X$, there is a subsequence of $\left \{Cx_{n} \right \}$ that is convergent.

Now, let X, Y be of finite dimension and T: X $\rightarrow$ Y a linear bounded operator. Let $\left \{x_{n}\right \}$ be a bounded sequence in X. Using the inequality $||Tx_{n}|| \leq ||T|| ||x_{n}||$ we can see that $\left \{Tx_{n}\right \}$ is bounded in Im(T).

Now, using the theorem of Bolzano Weierstrass (Each bounded sequence in $\mathbb{R^{n}}$ has a convergent subsequence), it is clear, that $\left \{Tx_{n}\right \}$ has a convergent subsequence.

$\Rightarrow$ T is compact.

$\Rightarrow$ $\sigma(X,Y)=L(X,Y)$