Let $V$ be a vector space and $T:V→V$ a linear transformation such that for any basis $B$ of $V$, the matrix representation of $T$ with respect to $B$ is diagonal. Show that $T$ has only one eigenvalue.
I found this problem on an exam and have been thinking about it for a while, but I don't know how to start. I think the proof may be related to this question. Any help is greatly appreciated.
Suppose it has two different eigenvalues $\lambda_1\neq \lambda_2$. Their associated eigenvectors are linearly independent, call them $v_1$ and $v_2$.
Now choose a basis like this: $u_1=(v_1+v_2)/2$, $u_2=(v_1-v_2)/2$, and choose the rest of the basis arbitrarily.
In this basis, the top $2\times 2$ submatrix of your linear transformation is not diagonal, since $$ Tu_1=T(v_1+v_2)/2=\lambda_1 v_1/2+\lambda_2 v_2/2 $$ which is not a multiple of $u_1$. To see that, observe that $v_1=u_1+u_2$ and $v_2=u_1-u_2$ and therefore $$ Tu_1=\lambda_1 v_1/2+\lambda_2 v_2/2=(\lambda_1 +\lambda_2)u_1/2+(\lambda_1 -\lambda_2)u_1/2, $$ a combination of $u_1$ and $u_2$. The same is true about $u_2$.
Every nonzero vector $v$ is an eigenvector: for we can extend to a basis, the corresponding matrix is diagonal, so $T(v)=\lambda v$ for some scalar $\lambda$.
If $\dim(V)=1$, there is nothing to prove. Assume $\dim(V)\geq 2$.
Let $v_1$ and $v_2$ be two nonzero vectors. Let $T(v_1)=\lambda_1v_1$, $T(v_2)=\lambda_2v_2$. We prove that $\lambda_1=\lambda_2$. If $v_1$ and $v_2$ are scalar multiples of each other, this is immediate. Otherwise, $v_1+v_2$ is nonzero, so there is a scalar $\mu$ such that $$\mu(v_1+v_2) = T(v_1+v_2) = T(v_1)+T(v_2) = \lambda_1 v_1 + \lambda_2v_2.$$ This gives $$(\mu-\lambda_1)v_1 + (\mu-\lambda_2)v_2 =0$$ and because $v_1$ and $v_2$ are linearly independent, we get $\lambda_1=\mu=\lambda_2$, so we are done.
Saying that for any basis $B$ of $V$, the matrix representation of $T$ with respect to $B$ is diagonal, is another way of saying that $Tv = \lambda_v \cdot v$ for any $v \in V$ (where $\lambda_v$ is a scalar. Note the a-priori dependency on the choice of $v$). If $\dim(V) = 1$, there's nothing to prove. Else, I refer you to a previous answer of mine:
