Show that if the matrix of $T$ with respect to all bases of $V$ is same, then $T = \alpha I$ where $T$ is a linear operator on $V$ Now the only hint I could derive from the question that we might have to use eigenvalues as $Tx = \lambda x$ for $T = \lambda I$. I think eigenvalues are invariant with basis.
Any help would be appreciated

There are several possible proofs. One goes like this - it uses elementary matrices, but implicitly.
Suppose first of all that there is a vector $v$ such that $v, T v$ are linearly independent. Choose a basis that starts with $v, T v$. Then the first column of $T$ will have a $1$ in the second position, and a zero elsewhere. Now choose a basis that starts with $v, v + T v$. This time the first column of $T$ will have two $1$'s in the first two positions, and then all zeroes.
Since this is a contradiction, we will have that $v, T v$ are always dependent, for all vectors $v$. Let $v, w$ be two independent vectors (if there aren't any, the space has dimension $1$, and we are done). Let $T v = a v$, $T w = b v$, and $T (v + w) = c (v + w)$ for some $a, b, c$. Then
$$
 c v + c w = T(v + w) = T v + T w = a v + b w,
$$
so that $a = c = b$, and you are done.
