Proof of injectivity,

Question

Let T be an operator on an NxN matrix that transforms it into a function that maps NxN matrices to NxN matrices.

For example $T(A) = T_A$ where $T_A(B)=A*B$ where A, B are NxN matrices. The proof that T is injective is: $$T_A = T_B$$ $$T_A(I_{nxn})=T_B(I_{nxn})$$ $$A = B$$

I don't understand this proof. The problem I see is that you choose a specific argument of $T_A$ and $T_B$ Hence, this only proves injectivity when the argument is $I_{nxn}$. Why is this enough?

Answer 1

Perhaps that a similar situation will make it clear. Conseider the map $\Psi$ from $\mathbb N$ into the set $\mathcal N$ of all functions from $\mathbb N$ into itself thus defined: if $n\in\mathbb N$, then $\Psi(n)$ is the multiplication by $n$ ($\Psi(n)(m)=n\times m$). Is $\Psi$ injective? Yes, because\begin{align}\Psi(n)=\Psi(m)&\implies\Psi(n)(1)=\Psi(m)(1)\\&\iff n\times1=m\times1\\&\iff m=n.\end{align}

Answer 2

You have $T:\mathbb{R}^{n\times n} \to \{f\mid f\colon \mathbb{R}^{n\times n}\to \mathbb{R}^{n\times n} \}, A\mapsto T_A$. To show that this is injective you assume $T_A=T_B$ for some $A,B\in\mathbb{R}^{n\times n}$, i.e. you have functions $T_A: \mathbb{R}^{n\times n}\to \mathbb{R}^{n\times n}, X\mapsto A\cdot X$, $T_B: \mathbb{R}^{n\times n}\to \mathbb{R}^{n\times n}, X\mapsto B\cdot X$ with $T_B(X)=T_A(X)$ for any matrix $X$.

Now, plugging in the identity matrix $I_n$ you get $$A=A\cdot I_n=T_A(I_n)=T_B(I_n)=B\cdot I_n=B.$$ Therefore, you have the implication $T_A=T(A)=T(B)=T_B\Rightarrow A=B$ and which is the definition of injectivity.