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?

  • 2
    $\begingroup$ The rank is not equal to $2$, it depends on $U,V$. See for example the case $U=0$. To find the other eigenvalues, just write the equality $Ax =\lambda x$ where $A$ is the given matrix and look at what it implies regarding the value of $\lambda$. $\endgroup$ Commented Jul 26, 2022 at 6:11
  • $\begingroup$ That's true. The rank it atmost $2$. $\endgroup$
    – Kroki
    Commented Jul 26, 2022 at 6:12
  • $\begingroup$ @mathcounterexamples.net can you elaborate your comment? $\endgroup$
    – Kroki
    Commented Jul 26, 2022 at 6:18
  • $\begingroup$ What do you don’t understand ? I’m just suggesting you to use the definition of an eigenvalue / eigenvector. $\endgroup$ Commented Jul 26, 2022 at 6:31
  • $\begingroup$ Ah yes! I did that indeed but I was not yet able to extract $\lambda$ from the equation. $\endgroup$
    – Kroki
    Commented Jul 26, 2022 at 6:32

3 Answers 3


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 $\mathbb{C}$.

  • $\begingroup$ Using $A^2$ seems over killing for me. $\endgroup$ Commented Jul 26, 2022 at 8:30
  • 1
    $\begingroup$ With block multiplication, you have $A^2=\begin{bmatrix} UV^T & \mathbf{0} \\ \mathbf{0}^T & U^TV \end{bmatrix}$ which is much simpler to manage. $\endgroup$
    – egreg
    Commented Jul 26, 2022 at 8:42
  • $\begingroup$ @egreg, thanks for the tip. I incorporated it into the answer. $\endgroup$ Commented Jul 26, 2022 at 8:53

Denote by $A$ the given matrix and suppose that the base field is $\mathbb R$.

Case 1: $U=V=0$

In that case, $A=0$ is diagonalizable.

Case 2: only one of $U,V$ is equal to zero

Here, $A$ is a non-zero triangular matrix with the main diagonal equal to zero. Hence $A$ is not diagonalizable.

Case 3: non of $U,V$ is zero

In that case, the rank of $A$ is equal to $2$. Its null space is of dimension $n-2$ and defined by the equations:

$$\begin{cases} x_{n+1} &= 0\\ \sum_{i=1}^n V_i x_i &= 0 \end{cases}$$

Let's have a look at the other eigenspaces. If $x \neq 0$ is an eigenvector associated to the eigenvalue $\lambda \neq 0$, we have the equations $$\begin{cases} U_i x_{n+1} &= \lambda x_i \text{ for } 1 \le i \le n\\ \sum_{i=1}^n V_i x_i &= \lambda x_{n+1} \end{cases}$$

$x_{n+1} \neq 0$ as otherwise $x =0$. Replacing $x_i= \frac{U_i x_{n+1}}{\lambda}$ in the last equation, we get

$$\lambda^2 = \sum_{i=1}^n U_i V_i = U^TV$$

If $U^TV \lt 0$, $0$ is the only eigenvalue of $A$ which is non-zero, and $A$ is not diagonalizable.

If $U^TV = 0$, we get $\lambda = 0$. $0$ is the only eigenvalue and the associated eigenspace is of dimension $n-2$. Hence $A$ is not diagonalizable.

Finally if $U^TV \gt 0$, we get two additional non-zero eigenvalues, namely $\pm \sqrt{U^TV}$ with associated distinct eigenspaces of dimension $1$. Therefore $A$ is diagonalizable.


First if $U=0$ or $V=0$, then the matrix is nilpotent, as $A^2=O$, and hence cannot be diagonalized, unless $A=O$ itself. Now we assume $U\not=0$ and $V\not=0$, which implies $A$ has rank $2$, and therefore $A$ can be diagonalized iff it has two linearly independent eigenvectors with nonzero eigenvalues, and in fact we can explicitly compute the nonzero eigenvalues.

$$A-\lambda I = \begin{pmatrix} -\lambda & \dots & U_1\\ \vdots & \ddots & \vdots\\ V_1 & \dots & -\lambda \end{pmatrix} =\frac{1}{\lambda} \begin{pmatrix} -1 & \dots & U_1/\lambda\\ \vdots & \ddots & \vdots\\ V_1/\lambda & \dots & -1 \end{pmatrix}$$

Now we can use the $-1$ on the diagonal in each row to cancel $U_i/\lambda$ at the end of each row, by adding the $i$-th column multiplied by $U_i/\lambda$ to the last column, which creates the following lower triangular matrix $$\begin{pmatrix} -1 & & \\ \vdots & \ddots & \\ V_1/\lambda & \dots & -1+\sum_{i=1}^n \frac{U_iV_i}{\lambda^2} \end{pmatrix}$$ whose determinant is up to a sign $$-1+\sum_{i=1}^n \frac{U_iV_i}{\lambda^2}=0$$ Hence $$\lambda^2=\sum_{i=1}^n U_iV_i=U^tV$$

If $U^tV=0$, then it has no nonzero solution, therefore the matrix cannot be diagonalized.

In any general field with char $\not=2$, it either doesn't have any solution or two distinct solutions. In the latter case, the two eigenvectors corresponding to distinct eigenvalues must be linearly independent. Therefore, a necessary and sufficient condition for the matrix to be diagonalizable over $F$ (with $\text{char }F\not=2$) is $U^tV$ is a nonzero square in $F$.

In particular, when $F=\mathbb R$, the above is equivalent to $U^tV>0$.

In fact, we can explicitly write down the eigenvectors corresponding to eigenvalues $\alpha=\pm \sqrt{U^tV}$: $$\begin{pmatrix} V \\ \alpha \end{pmatrix}$$

Update: Once the above is done, I realized we can directly work with the eigen-equation $$A\begin{pmatrix} W \\ w \end{pmatrix} = \begin{pmatrix} wV \\ U^tW \end{pmatrix} = \lambda \begin{pmatrix} W \\ w \end{pmatrix}$$

From $wW=\lambda V$, by a proper scaling, we may assume the $W=V$, hence $\lambda = w$ is the eigenvalue with the property $\lambda^2=U^tW=U^tV$. In particular, in characteristic $2$, as there can be at most one such $\lambda$, it has at most one linearly independent eigenvector with nonzero eigenvalue, hence the matrix is never diagonalizable (unless it's the zero matrix).


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