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
 A: $B^{-1}AB=A$ is the same as $AB=BA$ for every invertible $B$. Now use elementary non-diagonal matrices $B$ to prove the result.
A: To complement the other answers, here is a straightforward method, that just uses the definition of the matrix of $T$.
Fix a basis $v_1,\ldots,v_n$. We have
\begin{align}
Tv_1&=T_{11}v_1+T_{21}v_2+\ldots+T_{n1}v_n\\
Tv_2&=T_{12}v_1+T_{22}v_2+\ldots+T_{n2}v_n\\
\end{align}
As the matrix of $T$ is the same with respect to the basis $v_2,v_1,v_3,\ldots,v_n$,
\begin{align}
Tv_2&=T_{11}v_2+T_{21}v_1+\ldots+T_{n1}v_n\\
Tv_1&=T_{12}v_2+T_{22}v_1+\ldots+T_{n2}v_n\\
\end{align}
Comparing the expressions, and using the linear independence of the basis, we get $T_{11}=T_{22}$ and $T_{21}=T_{12}$. As we can do this with any pair of elements in the basis we get
$$
T_{11}=T_{22}=\cdots=T_{nn}
$$
and $T_{kj}=T_{jk}$ for any $k\ne j$. Now consider the basis
$$
v_1-v_2,v_1+v_2,v_3,\ldots,v_n. 
$$
Using this basis,
\begin{align}
Tv_1-Tv_2&=T_{11}(v_1-v_2)+T_{21}(v_1+v_2)+T_{31}v_3+\cdots+T_{n1}v_n\\
Tv_1+Tv_2&=T_{12}(v_1-v_2)+T_{22}(v_1+v_2)+T_{32}v_3+\cdots+T_{n2}v_n\\
\end{align}
Adding,
$$
2Tv_1=(T_{11}+T_{12}+T_{21}+T_{22})v_1+(T_{21}+T_{22}-T_{11}-T_{12})\,v_2+(T_{31}+T_{32})v_3+\cdots
$$
The previous relations we found allow us to reduce this to
$$
Tv_1=(T_{11}+T_{12})v_1+0\,v_2+T_{32}v_3+\cdots
$$
Comparing with the first $Tv_1$ we obtain $T_{11}+T_{12}=T_{11}$, so $T_{12}=0$. As we can do this for any pair $k\ne j$ of indices, we obtain $T_{kj}=0$. Then $T=\alpha\,I$, with $\alpha=T_{11}$.
A: It seems that this is related to the isotropic matrices, i.e. matrices proportional to the identity. When you use a change of basis you transform $T$ into something else (in general components of the transformed matrix are different, but this does not occur in here). Recall the transformation of $T$ into  $Q$: $Q=V^{-1}T \: V$. You can replace $T$ by $\alpha I$ to check the result for any $V$ (hint: pull out the scalar and use properties of the identity).
A: 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.
