I'm having trouble finding a matrix for a transformation $L:\mathbb{R}^3\rightarrow\mathbb{R}^3$ I'm given the following facts:
$$L\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}5\\0\\0\end{bmatrix}, L\begin{bmatrix}1\\1\\0\end{bmatrix}=\begin{bmatrix}6\\6\\0\end{bmatrix},L\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}7\\7\\7\end{bmatrix}$$
I found L to be the following
$$L=\begin{bmatrix}5&1&1\\0&6&1\\0&0&7\end{bmatrix}$$
Does this seem right? There is another method that utilizes a change in basis using gauss-jordan elimination. If anyone can provide insight into how that method works to find the transformation given any nonstandard bases, then that would be awesome.
This is correct. We can gather a few things about this mapping.
First, look at the images - they're multiples of the inputs! If you haven't learned about eigenstuff yet, here's a quick and dirty show. The vector $x$ is an eigenvector of the matrix $A$ with eigenvalue $\lambda$ if $Ax = \lambda x$. Basically, the linear mapping reduces to a scaling.
We do see that the given vectors $(1, 0, 0), (1, 1, 0),$ and $(1, 1, 1)$ do indeed form a basis (check!). Given a basis of $V$, we can construct a matrix for a linear mapping on $V$. We do this by "gluing" the images of the basis elements into a matrix.
We can also use Gauss-Jordan elimination to map into the standard basis and find a corresponding mapping, but I don't quite suggest it.