I was trying to solve this problem: find all vectors $U, V\in \mathbb R^{n}$ such that $$\begin{bmatrix}\mathbf{O} & U \\
V^T & 0\end{bmatrix}$$ is a diagonalisable matrix.
I was able to see that the above matrix is of rank at most $2$ so I tried to find the remainning two eigenvalues of the matrix, but I did not find them in terms of $U$ and $V$. Is there any way to do it?
Best Answer
I assume we are considering diagonalizability over $\mathbb{R}$. Furthermore, assume that neither $U$ nor $V$ are zero.
A necessary and sufficient condition for diagonalizability is that the minimal polynomial is a product of distinct linear factors.
The nonzero eigenvalues are $\pm \sqrt{U^T V}$. To arrive at this, consider $A^2,$ the eigenvalues of which are the eigenvalues of $A$ squared.
$A^2 = \begin{pmatrix} U V^T & \mathbf{0} \\ \mathbf{0}^T & U^T V \end{pmatrix}.$
From this, it is clear that the vector of all ones is an eigenvector. Furthermore, the vector with only 1 in the last entry and zeros elsewhere is an eigenvector. In both cases, the corresponding eigenvalue of $A^2$ is $U^T V$. Taking the positive and negative square root yields the eigenvalues of $A$.
Interestingly, for any $k \in \mathbb{N}$, if $(i, j)$ is a nonzero entry of $A^k$, then the $(i,j)$ entry of $A^{k+1}$ is zero and vice versa. To see this in practice, computing $A^3$ yields
$A^3 = \begin{pmatrix} \mathbf{0}\mathbf{0}^T & U^TV U \\ U^TV V^T & 0 \end{pmatrix}.$
In other words, odd powers of $A$ have the "opposite" nonzero pattern than the even powers of $A$.
From this, assuming neither $U$ nor $V$ equals $0$, we can conclude that the minimal polynomial of $A$ will not include even terms (the only way to cancel even terms would be even terms of higher order, which would be unnecessary).
It is clear from these calculations that $x^3 - U^T V x$ is an annihilating polynomial for A. Because $x$ is not an annihilating polynomial (neither $U$ nor $V$ are zero), this polynomial must be the minimal polynomial.
Thus, diagonalizability of $A$ is determined by the factorization of $x^3 - U^TV x$ into linear factors. We have
$x^3 - U^TV x = x(x^2 - U^TV).$
For $U^TV$ positive, $x^2 - U^TV$ factors into $(x + \sqrt{U^TV})(x - \sqrt{U^TV})$. For $U^TV$ negative, no such factorization is possible over $\mathbb{R}$. For $U^TV = 0$, the minimal polynomial is $x^3,$ and thus not a product of distinct linear factors.
Therefore, the matrix is diagonalizable over $\mathbb{R}$ if and only if $U^TV$ is positive.
For the case where both $U$ and $V$ are zero, the matrix is trivially diagonalizable. For the case where one of $U$ or $V$ are zero, the matrix is nilpotent and the minimal polynomial is $x^2,$ and thus the matrix is not diagonalizable. For $U^TV$ negative, the matrix is diagonalizable over $\mathbb{C}$.