Find the Eigenvalues and eigenvectors of $T(x,y)=(kx,ky)$ where $k$ is some scalar.
Some googling tells me that the eigenvalue should be $k$ and the Eigenspace is the entire $\mathbb{R}^2$.
Some patient explanation would be appreciated as I am new to this.
An eigenvalue is by definition a number $\lambda$ such that at least one non-zero vector is expanded by $T$ by a factor of $\lambda$: $Tx=\lambda x$ for some $x\neq \vec 0$.
The eigenspace is then the set of all such vectors, including $\vec 0$ this time.
In your case there are definitely such non-zero vectors: any non-zero vector fits.
Note that you can rewrite $Tx=\lambda x$ as $(T-\lambda\text{Id})x=\vec 0$, and since $x\neq \vec 0$ this means that the kernel of $T-\lambda\text{Id}$ is non-trivial, (equivalently this transformation is not injective), hence $\det(T-\lambda\text{Id})=0$. This is why to find the eigenvalues we solve the (polynomial!) equation $\det(T-\lambda\text{Id})=0$.