Exercise.
Suppose $V$ is finite-dimensional with $\text{dim } V > 1$. Show that if $\varphi : \mathcal{L}(V) \to \mathbb{F}$ is a linear map such that $\varphi(ST) = \varphi(S)\varphi(T)$ for all $S, T \in \mathcal{L}(V)$, then $\varphi = 0$.
Source.
Linear Algebra Done Right, 4th edition, by Sheldon Axler.
Chapter 3, section 3B, exercise 32.
Notation used.
- $\mathcal{L}(V)$ is the set of all linear transformations from a vector space $V$ to $V$.
- $\mathbb{F}$ represents either the set of real numbers, $\mathbb{R}$, or the set of complex numbers, $\mathbb{C}$.
What I've tried.
Axler suggests to use the definition of a two-sided ideal which goes as follows:
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 $T \in \mathcal{L}(V)$.
I have shown before that $\mathcal{L}(V)$ is a two-sided ideal of $\mathcal{L}(V)$. This implies that for any $S,T \in \mathcal{L}(V)$, we have $ST \in \mathcal{L}(V)$ and $TS \in \mathcal{L}(V)$.
This implies that $\varphi(TS)$ is well-defined. Since $\varphi(ST) = \varphi(S)\varphi(T)$, we must have then $\varphi(TS) = \varphi(T)\varphi(S)$. It is easy to show that
$$\varphi(S)\varphi(T) = \varphi(T)\varphi(S)$$
This implies that $\varphi(S)\varphi(T) – \varphi(T)\varphi(S) = 0$ which, by the distributive property of linear maps, grants us
$$\varphi(S)[\varphi(T) – \varphi(T)] = 0$$
Question.
Obviously, this will not lead me to the conclusion $\varphi = 0$. Is this the direction I am meant to go in? Could someone give me a hint?
The only place where I have used the fact that $V$ is finite-dimensional is when showing $\varphi(S)\varphi(T) = \varphi(T)\varphi(S)$ (I omitted this here, but have shown it on paper). Am I supposed to use it elsewhere to help me?
Best Answer
Let's first show that $\text{ker } \varphi$ is a two-sided ideal of $\mathcal{L}(V)$.
By definition, $$\text{ker } \varphi = \{E \in \mathcal{L}(V) : \varphi(E) = 0\}$$
Consider any $S \in \text{ker } \varphi$ and $T \in \mathcal{L}(V)$. Since $(S\in \text{ker } \varphi) \implies (S \in \mathcal{L}(V))$ we have, by definition of $\varphi$, that $\varphi(ST) = \varphi(S)\varphi(T)$. But $\varphi(S) = 0$ which implies
$$ \varphi(ST) = 0\varphi(T) = 0 $$
This means $ST \in \text{ker } \varphi$. I have shown in my previous response that $\varphi(ST) = \varphi(TS)$. This implies that $TS \in \text{ker } \varphi$ as well. Thus, $\text{ker } \varphi$ is a two-sided ideal of $\mathcal{L}(V)$.
The only two-sided ideals of $\mathcal{L}(V)$ are $\{0\}$ and $\mathcal{L}(V)$. Let's now show $\text{ker } \varphi \neq \{0\}$ by showing there exists a non-zero mapping $S \in \text{ker } \varphi$.
Let $v_1,\ldots, v_n$ be a basis of $V$. Define a linear mapping $S: V \to V$ on the basis of $V$ as follows:
If $n$ is even: $$ Sv_j = \begin{cases} v_{j+1}, & j=1, 3, \ldots, n-1 \\ 0, & j=2, 4, \ldots, n \end{cases} $$
If $n$ is odd: $$ Sv_j = \begin{cases} v_{j+1}, & j=1, 3, \ldots, n-2 \\ 0, & j=2, 4, \ldots, n-3, n-1, n \end{cases} $$
In either case, we'll have that $\text{range }S \subseteq \text{ker }S$. This implies that $S^2 = 0$, which further implies $S^2 \in \text{ker }\varphi$. But by definition of $\varphi$, we have that $\varphi(S^2) = \varphi(S)\varphi(S) = 0$. This implies $\varphi(S) = 0$ and so, $S \in \text{ker }\varphi$. Since $S$ is non-zero, we have that $\text{ker }\varphi \neq \{0\}$. This must mean that $\text{ker }\varphi = \mathcal{L}(V)$, which means that for every linear map $T \in \mathcal{L}(V)$, we have that $\varphi(T) = 0$, as was to be proven.