describe all normal nxn matrices hat have only one eigenvalue.
This is over the complex.
If it has only one eigenvalue, $\displaystyle \lambda \in \mathbb{C}$, then in a basis that diagonalises the original matrix, we would have :
$\displaystyle \begin{pmatrix}
\lambda & 0 & \dots & 0\\
0 & \lambda & \dots & 0\\
\vdots &\vdots&\ddots& 0\\
0 & 0&\dots & \lambda
\end{pmatrix}=\lambda \begin{pmatrix}
1 & 0 & \dots & 0\\
0 & 1 & \dots & 0\\
\vdots &\vdots&\ddots& 0\\
0 & 0&\dots & 1
\end{pmatrix}=\lambda I_{n}$
This explains half the answer of CaptainBlack (so you should understand it at half at least ), but I don't know what a "normal matrix hat" is ~
A normal matrix is one such that AA*=A*A where A* denotes the conjugate transpose.
When you diagonalize a normal matrix A, the diagonal matrix contains the eigenvalues for the map.
Since these matrices are restricted to having only one eigenvalue then:
All normal nxn matrices over the Complex space are similar to a diagonal matrix that has the single eigenvalue for every entry in the diagonal.
Is this correct? Is there more that can be said? Captain. THANK YOU
And a matrix $\displaystyle A$ is normal if and only if there exists a unitary matrux $\displaystyle U$ and a diagonal matrix $\displaystyle D$ such that:
$\displaystyle A=UDU^H$
(You can treat your definition of normality as the definition then this is a theorem, or this as the definition then yours is a theorem)
RonL