I am willing to prove the following.
Let $V$ be an $n$-dimensional vector space over a field $\mathbb{K}$.
Denote $L(V)$ = the ring of linear maps from $V \rightarrow V$.
$M_n \left( \mathbb{K}\right)$ = the matrices ring of $n \times n$ matrices.
Prove that $L(V) \cong M_n \left( \mathbb{K} \right)$.
I did as the following.
Let $\left \{e_1, e_2, \dots , e_n \right \}$ be an orthonormal basis of $V$.
Assume $f \in L (V)$ then
$f(e_1) = \beta _1$
$f( e_2) = \beta_2$
$\vdots$
$f(e_n) = \beta _n.$
$\forall x \in V, x= a_1 e_1 + \dots a_n e_n$, we have
$f(x) = a_1 \beta_1 + \dots + a_m \beta_n$.
Suppose that
$\beta_1 = b_1^1 e_1 + b_1^2 e_2 + \dots b_1^ne_n$
$\beta_2 = b_2^1 e_1 + b_2^2 e_2 + \dots + b_n^2 e_n$
$\vdots$
$\beta _n = b_n^1 e_1 + b_n ^2 e_2 + \dots b_n ^n e_n$.
Then
$f(x) = a_1 \left( b_1^1 e_1 + b_1 ^2 e_2 + \dots b_1^n e_n \right) + \dots + a_n \left( b_n^1 e_1 + b_n^2 e_2 + \dots + b_n^n e_n \right)$.
I stopped here since I have no idea to construct the corresponding matrix.
Could you please give me a hint?
Thank you in advance.
Best Answer
Let $V$ be a finite-dimensional vector space with basis $\beta=\left\{v_1, \dots, v_n\right\}$.
We construct a map $\phi:M_n(\mathbb{K})\rightarrow L(V)$ by mapping a matrix $A$ to the linear transformation $L_A:V\rightarrow V$ defined by $L_A(v)=AX$ where $X$ is given by $\begin{pmatrix} \lambda_1\\ \vdots \\ \lambda_n \end{pmatrix}$ where $v=\sum_{i=1}^n\lambda_iv_i$.
Conversely, we construct a map $\psi:L(V)\rightarrow M_n(\mathbb{K})$ as follows: Let $f\in L(V)$, then for each $i$ there exist $\mu_{j,i}\in \mathbb{K}$ such that $$f(v_i)=\sum_{i=1}^n\mu_{j,i}v_j.$$ In this way we obtain a matrix $f_{\beta}^{\beta}=(\mu_{j,i})_{1\leq i,j\leq n}^T$. Hence the map $\psi$ is given by $\psi(f)=f^{\beta}_{\beta}$.
It remains to show that these two maps are inverses of each other.
We first show that $(\psi\circ\phi)(A)=A$. Notice that $L_A(v_i)=Ae_i$. Here $e_i=\begin{pmatrix} 0\\ \vdots\\ 1\\ \vdots\\ 0 \end{pmatrix}$, the column vector with zeroes everywhere except at the $i$-th position where there is a $1$. Now $Ae_i=\begin{pmatrix} a_{1i}\\ a_{2i}\\ \vdots\\ a_{ni} \end{pmatrix}$ is simply the $i$-th column of $A$. It follows that $\mu_{j,i}=a_{j,i}$. Hence $f^{\beta}_{\beta}=(\mu_{j,i})_{i,j}=(a_{j,i})_{i,j}^T=(a_{i,j})_{i,j}=A$. This shows that $\psi\circ \phi=Id_{M_n(\mathbb{K})}$.
Can you show $\psi\circ \phi=Id_{L(V)}$ as well?
Understanding the above correspondence is the magical key to understanding everything of basic finite-dimensional linear algebra. All further topics such as eigenvectors, eigenbases, diagonalization and so on become easy once you truly understand this correspondence.