# $\begin{bmatrix}\mathbf O & U\\ V^T & 0\end{bmatrix}$ is diagonalizable?

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?

• 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 $\lambda$. Jul 26, 2022 at 6:11
• That's true. The rank it atmost $2$. Jul 26, 2022 at 6:12
• @mathcounterexamples.net can you elaborate your comment? Jul 26, 2022 at 6:18
• What do you don’t understand ? I’m just suggesting you to use the definition of an eigenvalue / eigenvector. Jul 26, 2022 at 6:31
• Ah yes! I did that indeed but I was not yet able to extract $\lambda$ from the equation. Jul 26, 2022 at 6:32

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 $$\mathbb{C}$$.

• Using $A^2$ seems over killing for me. Jul 26, 2022 at 8:30
• 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. Jul 26, 2022 at 8:42
• @egreg, thanks for the tip. I incorporated it into the answer. 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).