Matrix which has every vector in the space as an eigenvector is square, diagonal. [duplicate]

I've been trying to solve this linear algebra problem for some time and have gotten stuck. I've been asked to either prove or disprove the following statement: For $V$ an $\mathbb{ R }$ vector space, if $A$ is a matrix with the property that $Ax \in \langle x \rangle = \{ax \mid a \in \mathbb{ R } \}$ for every vector $x \in V$, then $A$ is square and diagonal. I suspect the statement to be true due to lack of counterexamples. I can pretty easily show that $A$ must be square, otherwise the dimensionality doesn't work out right. I haven't really got much of an idea where to go on the diagonal part though. My great apologies if this is a duplicate question-I searched somewhat thoroughly and didn't find it anywhere, however. Thanks in advance!
From the linear transformation view point, for every $x \in V$ there is an $a \in \mathbb{ R }$ such that $T(x) = ax$. We have the expression $$x = c_1 e_1 + \ldots +c_n e_n$$ where $e_i$ is the ith standard basis vector for $V$. Therefore, $$ax = T (c_1 e_1 + \ldots +c_n e_n)$$ whereby $$a\left( \sum_{i=1} ^n c_i e_i \right) = \sum_{i=1} ^n c_i T (e_i)$$ Therefore, $$\begin{pmatrix} a c_1 \\ \vdots\\ 0 \end{pmatrix} + \ldots + \begin{pmatrix} 0 \\ \vdots \\ a c_n \end{pmatrix} = c_1 \begin{pmatrix} a_{1,1} \\ \vdots \\ a_{1,n} \end{pmatrix} + \ldots + c_n \begin{pmatrix} a_{n,1} \\ \vdots \\ a_{n,n} \end{pmatrix}$$ where the matrix representation of $T$ is the matrix $A = \left[ a_{i,j} \right]$, $1 \le i, j \le n$. From here I'm having some difficulty once more. Does this in fact give that $a_{i,j} = 0$ for $i \ne j$? I'm having difficulty seeing how.

• Hint: the result might be stronger, consider whether it is possible for $u+v$ to be an eigenvector if $u,v$ are eigenvectors with distinct eigenvalues. – JHance Jul 30 '14 at 22:10

Hint: Remember that the $i$-th column of your matrix corresponds to image of the $i$-th basis vector under the transformation represented by the matrix.
• @aherring, for the purposes of determining what the matrix "looks" like, it might be easier if you consider things one basis vector at a time. Applying your matrix to the vector $e_1=[1,0,0,...,0]^{T}$ gives you the first row, applying it to $e_2=[0,1,0,0,..,0]^{T}$ gives you the second row, etc. (Note that I take the transpose so that my vectors are columns, not rows). – vociferous_rutabaga Jul 30 '14 at 23:48
• @aherring, I have noticed a typo in my previous comment too late.. I meant, of course, that applying the matrix to $e_1$ gives the first column, applying the matrix to $e_2$ gives the second column, etc. – vociferous_rutabaga Jul 30 '14 at 23:55