If matrix is diagonalizable, eigenvalue? 
Let $A$ be an $n \times n$ matrix and suppose $A$ is diagonalizable and the only eigenvalue is $\lambda = k$, what can you say about matrix $D$ where $A = P^{-1} D P$, for invertible matrix $P$.

So if the only eigenvalue of $A$ is $\lambda = k$, what can I say about $D$? 
I know that $D$ is a diagonal matrix, but is it necessarily true that $D = \text{diag } (k, k, ... , k)$ ?
 A: Yes, it is. One way to see this is that eigenvalues are invariant under conjugation. This is a fancy way to say that if 
$$
A=PDP^{-1}
$$
then $A$ and $D$ have the same eigenvalues. 
Now, what are the eigenvalues of a diagonal (or upper triangular) matrix? 
Edit: Proof of fact mentioned in comments:
suppose 
$$
\lambda I=PAP^{-1}
$$
for some change of basis matrix $P$. Then 
$$
A=P^{-1}\lambda IP=\lambda P^{-1}P=\lambda I
$$
A: The first question is: What can you say about $A$?
You know it's diagonalizable, so write it out in a diagonal form.
You know the only possible eigenvalue, so you know the diagonal entries.
What does it look like?
Notice that $A$ was not assumed to be diagonal, just diagonalizable.
What happens to it after you change the basis?
Once you have found $A$, you can find $D$ easily.
A: An $n\times n$ diagonal matrix $D$ has $n$ pairwise orthogonal eigenvectors (namely, the canonical basis), since $De_j=D_{jj}e_j$. In the case from the question, we have 
$$
APe_j=PDe_j=D_{jj}\,Pe_j
$$
so for each $j$, $D_{jj}$ is an eigenvalue of $A$ with eigenvector $Pe_j$. If the only eigenvector of $A$ is $k$, then $D_{jj}=k$ for all $j$, i.e., $D=kI$. Now 
$$
A=PDP^{-1}=kPP^{-1}=kI.
$$
