Commutativity of Invertible Matrices Show that the only matrices that commute with all other matrices in $GL_{n}(\mathbb{R})$ will be multiples of the identity matrix. Or:
$Z(GL_{n}(\mathbb{R}))=\{\lambda I_{n}:\lambda \in \mathbb{R}\}$

One direction is clearly obvious. However, to show that every element $g$ such that $gh = hg$ for $h \in GL_{n}(\mathbb{R})$ has to be a multiple of the identity is causing me trouble as I don't know where to begin with. What would be a way to handle this problem? I was thinking of using contradiction, but that doesn't take me anywhere.
Thanks for the help.

The interesting (characterizing) property of scalar matrices (operators) is that every one dimensional subspace is invariant under them (i.e. eigenspace). This is easy to prove.
On the contrary, suppose that $A$ is an invertible matrix with some non-invariant one dimensional subspace, say $A(v_1)=v_2$ with $v_2\notin \langle v_1\rangle$.
Now the subspace $\langle v_1\rangle$ and $\langle v_2\rangle$ are different, and $A$ carries former to later. In particular, $\{v_1,v_2\}$ are linearly independent, hence $\{v_1+v_2, v_2\}$ are also linearly independent.
Let $B$ be an invertible matrix (transformation) such that 
$B(v_1)=v_1+v_2$ and $B(v_2)=v_2$. Then 
$$AB(v_1)=A(v_1+v_2)=v_2+A(v_2)$$
and 
$$BA(v_1)=B(v_2)=v_2.$$
So $AB=BA$ would imply that they agree at $v_1$, and hence $A(v_2)=0$, a contradiction (why?)
Hence, scalar matrices are in center, and non-scalar matrices are not in center of $GL_n$.
