Existence of symmetric matrix for linear operator

Question

Given any linear operator $ T: ℝ^n →ℝ^n$ I'm asked to show that there exists an n x n matrix A such that T(x)= A x for all x in $ℝ^n.$
I don't even know where to start this proof, yet it seems terribly straightforward. Could anyone please give me a hint/ guide me through the proof? Unfortunately I'm behind on my lectures, so any help is appreciated!

Answer 1

$\textbf{Hint}$:Consider a basis $\{v_1,...,v_n\}$ of $\mathbb{R}^n$ and use that $Tv_j=\sum_{j=1}^{n}\alpha_{ij}v_j$ where $\alpha_{ij}$,$i=1,...,n$ are unique for $j=1,...,n$.

Answer 2

The matrix of your transformation is the matrix $M$ whose columns are vectors found by the images of the standard basis $$ { T(1,0,0,...,0), T(0,1,0,...,0),...,T(0,0,0,...,1) }.$$

The first column is $ T( 1,0,0,...,0)$

The second column is $ T( 0,1,0,...,0)$

And so forth.