The problems asks that one prove if the eigenvalues of a Hermitian or skew-Hermitian matrix $A$ are all equal to $c$, then $A=cI$.

I came up with a proof but I'm not sure it's logically valid, and even if it is, probably not the best way of tackling the problem. I was wondering if someone could check the validity of this proof and explain if there are errors in reasoning, and provide a better proof if possible.

**Thm A**: Assume $\dim V = n$ and let $T: V\rightarrow V$ be Hermitian or skew-Hermitian. Then there exist n eigenvectors that form an orthonormal basis for $V$. The matrix of $T$ relative to this basis is the diagonal matrix of the eigenvalues $\lambda\_k$, where $\lambda\_k$ is the eigenvalue associated with the kth eigenvector in the basis.

Let $T$ be the linear transformation represented by the matrix $A$ for the basis $E= (e\_1,e\_2,...,e\_k)$. We know that there is a diagonal matrix $cI$ for $T$ with an orthonormal basis formed from the eigenvectors of $c$; let this basis be $U = (u\_1,u\_2,...,u\_k)$ from Thm A.

Then we have $e\_j = \sum\_{i=1}^k a\_iu\_i$ for $j=1,2,...,k$. Then $T(e\_j) = \sum\_{i=1}^k a\_iT(u\_i) = \sum\_{i=1}^k a\_icu\_i$, from the definition of $T$. Then $T(e\_j) = c\sum\_{i=1}^k a\_iu\_i = ce\_j$. But then $e\_j$ is an eigenvector of the transformation $T$ so by Thm A, $A=cI$.