Proof Verification: Suppose $V$ is finite-dimensional. Show that the only two-sided ideals of $\mathcal{L}(V)$ are $\{0\}$ and $\mathcal{L}(V)$.

Question

Could someone please verify if my proof of the below exercise is valid? It comes from the 4th edition of Linear Algebra Done Right by Sheldon Axler - it recently became available and is open access if you are interested.  

Exercise. Suppose $V$ is finite-dimensional. Show that the only two-sided ideals of $\mathcal{L}(V)$ are $\{0\}$ and $\mathcal{L}(V)$.

A subspace $\mathcal{E}$ of $\mathcal{L}(V)$ is called a two-sided ideal of $\mathcal{L}(V)$ if $TE \in \mathcal{E}$ and $ET \in \mathcal{E}$ for all $E \in \mathcal{E}$ and all $T \in \mathcal{L}(V)$.

$\mathcal{L}(V)$ is the set of linear mappings from a vector space $V$ to $V$.

Proof.

Step 1

Let's first show that $\{0\}$ and $\mathcal{L}(V)$ are two-sided ideals of $\mathcal{L}(V)$.

Consider first $\{0\}$. It is clearly a subspace of $\mathcal{L}(V)$. Furthermore, the only element in $\{0\}$ is $0$, that is, the zero linear map. Let $T \in \mathcal{L}(V)$ be any mapping. Then $\forall v \in V$,

$$ 0T(v)= 0(Tv) = 0 \in \{0\} \implies 0T \in \{0\} $$

and

$$ T0(v) = T(0v) = T(0) = 0 \in \{0\} \implies T0 \in \{0\} $$

Since $0T \in \{0\}$ and $T0 \in \{0\}$, it implies $\{0\}$ is a two-sided ideal of $\mathcal{L}(V)$.

Now let us consider $\mathcal{L}(V)$. Clearly, $\mathcal{L}(V)$ is a subspace of $\mathcal{L}(V)$.

Let $S, T \in \mathcal{L}(V)$ be any two linear mappings. By definition of linear map products (composition), $ST \in \mathcal{L}(V)$ and $TS \in \mathcal{L}(V)$. Thus, $\mathcal{L}(V)$ is a two-sided ideal of $\mathcal{L}(V)$.

Step 2.

Now we must show they are the only two-sided ideals of $\mathcal{L}(V)$.

Suppose $\mathcal{E}$ is a subspace of $\mathcal{L}(V)$ such that $\mathcal{E} \neq \{0\}$ and $\mathcal{E} \neq \mathcal{L}(V)$. This, along with the assumption that $V$ is finite-dimensional, implies there exists a nonzero linear map $T \in \mathcal{L}(V)$ such that $T \notin \mathcal{E}$ which we are free to define on a basis of $V$, say, the list of vectors $v_1, \ldots, v_n$ such that

$$ T(a_1v_1 + \cdots + a_nv_n) = v_1 $$

Also, let $E \in \mathcal{E}$ such that

$$ E(a_1v_1 + \cdots + a_nv_n) = v_n $$

I claim that $TE \notin \mathcal{E}$, since if otherwise, that would imply $\forall v \in V$,

\begin{align} TE(v) = TE(a_1v_1 + \cdots + a_nv_n) &= T(E(a_1v_1 + \cdots + a_nv_n)) \\ &= T(v_n) \\ &= v_1 \end{align}

showing that both $T(a_1v_1 + \cdots + a_nv_n) = v_1$ and $TE(a_1v_1 + \cdots + a_nv_n) = v_1$, implying that $TE = T$, where $TE \in \mathcal{E}$ implies $T \in \mathcal{E}$, contradicting the assumption that $T \notin \mathcal{E}$. Thus, we must have $TE \notin \mathcal{E}$ which is enough to show that $\mathcal{E}$ cannot be a two-sided ideal of $\mathcal{L}(V)$.

Thus, $\{0\}$ and $\mathcal{L}(V)$ are the only two-sided ideals of $\mathcal{L}(V)$.